AP-Level Chemistry

Three numbers that refuse to line up

Stripping one electron from a mole of hydrogen atoms costs 1312 kJ. Doing the same job on helium costs 2372 kJ. Do it on lithium, which has three protons to helium's two, and the price collapses to 520 kJ. More protons, less than a quarter of the energy. If the nucleus is what holds an electron in place, that number is backwards.

It is not backwards. It is the first evidence that an electron in an atom does not feel the full nuclear charge, and that how far out the electron sits matters as much as how much charge is pulling on it. This lesson builds the tool that makes all three numbers obvious, and that tool then explains most of the periodic table for the rest of the course.

You met ionization energy in CHEM 100 as a trend to memorise: it rises across a period and falls down a group. Here you will stop memorising it and start calculating it.

Coulomb's law, in the only form you need

Two charges attract or repel with a force that grows with the size of the charges and falls off with the square of the distance between them. The energy of that interaction is what chemists actually use, and it goes as

E proportional to q₁q₂ divided by r

where q₁ and q₂ are the two charges and r is the distance between them. When the charges have opposite signs the product is negative, which means the energy is lowered by bringing them together: that is attraction. Two things follow immediately, and you will use both of them in every lesson from here to electrochemistry.

• Bigger charges bind harder. Double one charge and you double the interaction energy.
• Closer binds harder, and it is not a gentle effect. Halve the distance and you double the energy, because r sits in the denominator.

Key idea: Every question in this course about which electron is held more tightly, which bond is stronger, or which ionic solid melts higher, reduces to the same two variables: how much charge, and how far apart.

Why lithium is cheap to ionize

Return to the three numbers. Hydrogen has one proton and one electron, and that electron sits in the n = 1 shell. Helium has two protons and two electrons, both in n = 1. Doubling the nuclear charge while keeping the distance roughly the same should roughly quadruple the binding energy of a hydrogenic system, and 1312 times four is 5248 kJ per mole. The measured value is 2372. So helium's electrons do bind harder than hydrogen's, but nothing like four times harder.

The reason is that helium's two electrons repel each other. Each one partly cancels, from the other's point of view, some of the nucleus's pull. Chemists call this shielding, and the charge that is left over after the cancellation is the effective nuclear charge, written Z_eff.

Now lithium. Its third electron cannot join the first two in n = 1, so it goes into n = 2, considerably farther out. It is also shielded by two electrons that sit between it and the nucleus almost all of the time. Large r, small Z_eff, and by Coulomb's law the energy holding it is small. 520 kJ per mole small.

Putting a number on Z_eff

The crude version is Z_eff = Z minus the number of core electrons, which gives every element in a group the same answer and is right about the shape of the trend and wrong about its size. The useful version is a set of counting rules published by John C. Slater in 1930. For an electron in an s or p subshell of shell n, add up a shielding constant S like this:

• Other electrons in the same shell n contribute 0.35 each.
• Electrons in shell n minus 1 contribute 0.85 each.
• Electrons in shells n minus 2 and lower contribute 1.00 each.
• The electron you are calculating for contributes nothing to its own shielding.

Then Z_eff = Z minus S. Work lithium, whose configuration is 1s²2s¹. The 2s electron has no companions in shell 2, so the 0.35 term is zero. It has two electrons in shell 1, so S = 2 times 0.85 = 1.70. Therefore Z_eff = 3 minus 1.70 = 1.30. Three protons in the nucleus, and the outer electron feels the pull of about 1.3 of them.

Now beryllium, 1s²2s². The 2s electron now has one companion in shell 2 and two electrons in shell 1: S = (1 times 0.35) + (2 times 0.85) = 0.35 + 1.70 = 2.05, so Z_eff = 4 minus 2.05 = 1.95. Add one proton, and the outer electron gains 0.65 units of effective charge, because the new electron that came with the proton only shields 0.35 of it.

That 0.65 per step is the whole trend. Here it is across period 2, with the measured first ionization energies alongside.

Element	Z	S (Slater)	Zeff	First IE (kJ/mol)
Li	3	1.70	1.30	520
Be	4	2.05	1.95	900
B	5	2.40	2.60	801
C	6	2.75	3.25	1086
N	7	3.10	3.90	1402
O	8	3.45	4.55	1314
F	9	3.80	5.20	1681
Ne	10	4.15	5.85	2081

Read the last two columns together. Z_eff climbs steadily and so, mostly, does the ionization energy. The word doing the work is mostly: boron is lower than beryllium, and oxygen is lower than nitrogen. Slater's rules cannot see either dip, because they count electrons without asking which subshell those electrons occupy or how their spins are arranged. Lesson 3 takes both exceptions apart. For now, notice that a rule which explains six of eight steps and visibly fails at the other two is more useful than a rule you cannot check, because the failures tell you exactly what the model left out.

The point: Slater's rules are an approximation, and a rough one. More accurate shielding constants come from solving the atom numerically, as Clementi and Raimondi did in 1963. Use Slater's numbers to reason about direction and rough size, never to predict an ionization energy to three figures.

Down a group: distance wins

Run the rule on sodium, 1s²2s²2p⁶3s¹. The 3s electron has no companions in shell 3. Shell 2 holds eight electrons at 0.85 each, and shell 1 holds two at 1.00 each. S = (8 times 0.85) + (2 times 1.00) = 6.80 + 2.00 = 8.80, so Z_eff = 11 minus 8.80 = 2.20.

Sodium's outer electron therefore feels a larger effective charge than lithium's 1.30, yet sodium's first ionization energy is 496 kJ per mole, lower than lithium's 520. Coulomb's law has two variables and only one of them moved in the direction you expected. The 3s electron sits in a shell farther from the nucleus, and that increase in r more than cancels the increase in Z_eff. Going down a group, distance beats charge every time, which is why the alkali metals get easier to ionize the further down you go: Li 520, Na 496, K 419, Rb 403, Cs 376 kJ per mole.

Successive ionizations: the shells announce themselves

Everything above treats shells as real. Here is the measurement that makes them undeniable. Keep pulling electrons off the same atom and record what each one costs.

Element	IE1	IE2	IE3	IE4
Na	496	4562	6910	9543
Mg	738	1451	7733	10540
Al	578	1817	2745	11577

All values in kJ per mole. Every ionization costs more than the one before it, because you are pulling a negative charge away from an ion that has just become more positive. That part is ordinary. What is not ordinary is where the jumps are enormous. Sodium's second ionization costs nine times its first. Magnesium's first two are comparable, and then the third jumps by a factor of five. Aluminium's first three rise gently, and the fourth quadruples.

Count valence electrons: sodium has one, magnesium two, aluminium three. The cliff falls immediately after the valence shell empties, because the next electron has to come out of a complete inner shell that sits much closer to the nucleus and is barely shielded at all. Small r, large Z_eff, enormous energy.

Why this matters: You can identify an unknown element's group from its ionization energies alone, without knowing anything else about it. Find the big jump; the number of ionizations before it is the number of valence electrons.

Worked example: identify the element

Problem. An element in period 3 has successive ionization energies, in kJ per mole, of 1012, 1907, 2914, 4964, 6274, 21267, 25431. Which element is it?

Step 1. Look for the discontinuity. The steps from 1012 to 6274 rise by factors of roughly 1.9, 1.5, 1.7 and 1.3. The step from 6274 to 21267 is a factor of 3.4. That is the cliff.

Step 2. Count the ionizations before it. Five values sit below the cliff, so the atom has five valence electrons.

Step 3. Place it. Period 3, five valence electrons, group 15. The element is phosphorus, and 1012 kJ per mole is indeed the measured first ionization energy of phosphorus.

Notice what you did not need: the atomic mass, the density, the colour, or any chemistry at all. Energies alone located the element on the table.

Worked example: rank without a table

Problem. Put these in order of increasing first ionization energy: Ca, K, Mg, Sr. Then justify each comparison with Coulomb's law.

Step 1. Group comparisons first, because distance dominates. Mg is above Ca, which is above Sr, so among the three group 2 elements the order is Sr less than Ca less than Mg.

Step 2. Period comparison. K and Ca are both period 4. Ca has one more proton and its outer electrons are in the same shell, so Z_eff is larger for Ca: K is lower than Ca.

Step 3. Position K against Sr. K is a group 1 element with a single 4s electron and low Z_eff. Sr is group 2 in period 5, farther out but with a higher effective charge. Measured values settle it: K 419, Sr 549 kJ per mole.

Answer. K (419) less than Sr (549) less than Ca (590) less than Mg (738). The two rules, charge across and distance down, got you three of the four comparisons; only the diagonal one needed data.

Common misconceptions

• "An electron feels the full nuclear charge." Only in a one-electron system. In every other atom, inner electrons cancel part of the nuclear pull, and the leftover is Z_eff. Lithium has three protons; its outer electron feels about 1.3.
• "Adding a proton always makes the atom harder to ionize." Across a period, yes, because the added electron shields only about 0.35 of the added proton. Down a group, no: sodium has eight more protons than lithium and is easier to ionize, because the outer electron moved to a larger shell.
• "Electrons in the same shell shield each other well." They shield each other poorly, roughly 0.35 on Slater's counting, because they are at similar distances from the nucleus rather than between the nucleus and each other. This is exactly why Z_eff rises across a period.
• "Shielding and penetration are the same thing." Shielding is what an outer electron experiences. Penetration is what an inner-looking part of an outer orbital does: a 2s orbital has electron density close to the nucleus, so it penetrates the 1s shell and is shielded less than 2p. That difference sets subshell energy order.
• "Slater's rules give the true effective nuclear charge." They give an estimate built from counting, published in 1930 before any atom had been solved numerically. Treat the number as a direction and a rough magnitude, not a measurement.

Putting it together

• Coulomb's law is the engine: interaction energy goes as the product of the charges over the distance, so more charge or less distance means tighter binding.
• An electron in a many-electron atom feels an effective nuclear charge Z_eff, not the full Z, because the other electrons shield it.
• Slater's rules estimate the shielding by counting: 0.35 per same-shell electron, 0.85 per electron one shell in, 1.00 per electron further in.
• Across period 2, Z_eff rises about 0.65 per element, which is why ionization energy rises across a period.
• Down a group, r grows faster than Z_eff, so ionization energy falls: Li 520, Na 496, K 419, Rb 403, Cs 376 kJ per mole.
• Successive ionization energies show a large jump immediately after the valence shell empties, which lets you read an element's group straight off the data.

Sources

1. OpenStax. (2019). Electronic structure of atoms (electron configurations) (Section 6.4). In Chemistry 2e. Rice University. openstax.org
2. OpenStax. (2019). Periodic variations in element properties (Section 6.5). In Chemistry 2e. Rice University. openstax.org
3. National Institute of Standards and Technology. (n.d.). Atomic spectra database: ionization energies. U.S. Department of Commerce. physics.nist.gov
4. Slater, J. C. (1930). Atomic shielding constants. Physical Review, 36(1), 57-64.
5. Clementi, E., and Raimondi, D. L. (1963). Atomic screening constants from SCF functions. The Journal of Chemical Physics, 38(11), 2686-2689.

Five numbers that spell out a configuration

The photoelectron spectrum of neon has three peaks. They sit at 84.0, 4.68 and 2.08 megajoules per mole, and the heights are in the ratio 2 to 2 to 6. From those five numbers, and nothing else, you can write neon's electron configuration: 1s²2s²2p⁶. Three peaks means three subshells are occupied. The heights count the electrons in each. The positions say how tightly each set is held.

That is the whole technique, and it is worth pausing on how unusual it is. Lesson 1 inferred shell structure indirectly, from the cliff in successive ionization energies. Photoelectron spectroscopy does not infer. It measures every occupied subshell of a neutral atom in one experiment.

The measurement

Take a sample of gaseous atoms and illuminate it with photons that all carry the same energy. If a photon carries more energy than an electron's binding energy, the electron is ejected, and it leaves carrying the difference away as kinetic energy. Energy is conserved, so

photon energy = binding energy + kinetic energy of the ejected electron

The instrument measures the kinetic energy. You know the photon energy because you chose the light source. Subtract, and you have the binding energy. Do it for millions of ejected electrons and plot how many arrive at each binding energy, and you have a spectrum.

Why does one sample give several peaks rather than one? Because a neon atom has electrons in three different subshells, held at three different energies. An electron pulled out of the 1s orbital leaves with much less kinetic energy than one pulled from 2p, because far more of the photon's energy went into freeing it. Each subshell therefore produces its own peak.

Remember: A photoelectron spectrum is a census of one atom's electrons sorted by how hard they are to remove. Position on the axis is energy. Area under the peak is population.

Reading the axis, which runs backwards

Chemistry spectra of this kind are conventionally drawn with binding energy increasing to the left, so the tightly held core electrons appear on the left and the loosely held valence electrons on the right. The unit is usually megajoules per mole, which keeps the numbers between about 0.4 and a few hundred for the elements you will meet. One megajoule per mole is 1000 kJ per mole, so neon's rightmost peak at 2.08 MJ per mole is 2080 kJ per mole.

Compare that with the first ionization energy of neon in Lesson 1: 2081 kJ per mole. They are the same measurement. The rightmost peak of any photoelectron spectrum is the first ionization energy, because the most loosely held electron is the one that comes off first when you ionize the atom by any other method. That is a free consistency check you can run on every spectrum in this course.

The period 2 table, and what it shows

Binding energies in megajoules per mole, rounded to three significant figures, with peak heights in electrons.

Element	1s	2s	2p	3s	Total electrons
H	1.31				1
He	2.37				2
Li	6.26	0.52			3
Be	11.5	0.90			4
B	19.3	1.36	0.80		5
C	28.6	1.72	1.09		6
N	39.6	2.45	1.40		7
O	52.6	3.12	1.31		8
F	67.4	3.88	1.68		9
Ne	84.0	4.68	2.08		10
Na	104	6.84	3.67	0.50	11

Run the check first. Lithium's outermost value, 0.52 MJ per mole, is 520 kJ per mole, its first ionization energy. Boron's 0.80 is 801. Oxygen's 1.31 is 1314. Sodium's 0.50 is 496. Every one matches the NIST table you used in Lesson 1, which tells you the whole table is real data rather than a teaching fiction.

Now read three things off it.

• 2s and 2p are genuinely different energies. In boron, 1.36 against 0.80. Nothing about a shell model alone predicts that gap. It exists because a 2s orbital penetrates the 1s region and is therefore shielded less. Photoelectron spectroscopy makes penetration a measurement rather than an argument.
• The oxygen dip appears here too. Nitrogen's 2p peak is at 1.40 and oxygen's is at 1.31, so oxygen's outermost electron is easier to remove despite the extra proton. Lesson 3 explains why; for now, notice that the spectrum shows it as plainly as the ionization energies did.
• Core electrons are barely shielded by valence electrons. This is the most interesting pattern in the table, and the next section extracts it.

Working an effective charge out of a core peak

For a one-electron atom, binding energy scales as the square of the nuclear charge, and the hydrogen 1s value of 1.31 MJ per mole is the reference. So if you divide a measured 1s binding energy by 1.31 and take the square root, you get an apparent effective nuclear charge for that 1s electron.

Worked, for neon. 84.0 divided by 1.31 is 64.1. The square root of 64.1 is 8.01. Neon has ten protons, so its 1s electrons feel about 8.0 of them.

Do the whole period and the pattern is striking.

Element	Z	1s binding energy (MJ/mol)	Apparent Zeff	Change
Li	3	6.26	2.19
Be	4	11.5	2.96	0.77
B	5	19.3	3.84	0.88
C	6	28.6	4.67	0.83
N	7	39.6	5.50	0.83
O	8	52.6	6.34	0.84
F	9	67.4	7.17	0.83
Ne	10	84.0	8.01	0.84

Each added proton raises the 1s effective charge by about 0.83. Since each proton arrives with an electron that lands in shell 2, the arithmetic says an n = 2 electron shields an n = 1 electron by only about 0.17 units of charge. Compare Lesson 1, where a same-shell electron shielded by 0.35 and an inner-shell electron by 0.85. The ordering is now complete and it is entirely Coulombic: electrons shield those outside them well, those beside them poorly, and those inside them hardly at all.

The upshot: Shielding is not a property an electron has. It is a relationship between two positions, and the only question that matters is which one is closer to the nucleus.

Worked example: name the element

Problem. A photoelectron spectrum shows four peaks. Their binding energies are 126, 9.07, 5.31 and 0.74 MJ per mole, and their relative heights are 2, 2, 6 and 2. Identify the element and write its configuration.

Step 1. Count the electrons. 2 + 2 + 6 + 2 = 12. A neutral atom with twelve electrons has twelve protons, so Z = 12.

Step 2. Assign the subshells by height and order. Peaks holding at most two electrons are s subshells; a peak of six is a filled p subshell. Working from the highest binding energy down: 126 is 1s², 9.07 is 2s², 5.31 is 2p⁶, and 0.74 is 3s².

Step 3. Write it out. 1s²2s²2p⁶3s², which is magnesium.

Step 4. Check. The rightmost peak, 0.74 MJ per mole, is 740 kJ per mole. Magnesium's measured first ionization energy is 738 kJ per mole. The identification holds.

Worked example: the spectrum of an ion

Problem. How would the spectrum of Mg²⁺ differ from that of Mg?

Reasoning. Removing both 3s electrons removes the rightmost peak entirely, so the spectrum drops from four peaks to three, with heights 2, 2 and 6. But that is not the only change. The remaining ten electrons now share the same twelve protons with two fewer companions to help repel them, so every surviving peak shifts to a higher binding energy. A spectrum of an ion is not the neutral spectrum with a peak deleted; the whole thing moves left.

This is a reliable way to catch a careless answer. Any process that removes electrons without changing the nucleus raises the binding energy of everything that remains.

What the idealisation leaves out

The peak heights used in teaching spectra are proportional to the number of electrons in the subshell, and that is close enough to true to identify elements. In a real instrument, the probability that a photon of a given energy ejects a given electron, called the ionization cross section, varies from subshell to subshell and with photon energy. A laboratory X-ray photoelectron spectrum, of the kind catalogued in the NIST database, therefore needs correction factors before areas can be turned into atom counts. It also shows features that a simple model does not predict, including small shifts in a core peak that depend on the chemical environment of the atom, which is precisely what makes the technique useful for analysing surfaces.

Worth holding on to: The teaching version, heights equal electron counts, is an idealisation that is safe for identifying an element and unsafe for quantitative surface analysis.

Common misconceptions

• "The tallest peak is the most tightly held." Height counts electrons, position measures binding energy. Neon's tallest peak, the 2p at 2.08, is its most loosely held set.
• "A brighter light source ejects electrons with more energy." Brighter means more photons, so more electrons come off and every peak grows. Kinetic energy per electron depends only on the photon energy, which is set by the wavelength, not the intensity.
• "Each peak is one shell." Each peak is one subshell. Sodium's spectrum has four peaks for three shells, because shell 2 contributes both a 2s and a 2p peak.
• "Two isotopes give different spectra." Photoelectron spectroscopy probes electrons. Carbon 12 and carbon 14 have identical electron configurations and their spectra are indistinguishable at this resolution.
• "The binding energy of an electron is the same as the energy of the orbital it left." Close, but the remaining electrons relax after the ejection, so the measured binding energy is slightly smaller than the orbital energy of the intact atom. The approximation that ignores this is called Koopmans' theorem.

What to carry forward

• Photon energy equals binding energy plus the kinetic energy carried off by the ejected electron; the instrument measures the kinetic energy and you subtract.
• Peak position gives the binding energy of a subshell; peak height gives the number of electrons in it; the number of peaks gives the number of occupied subshells.
• Binding energy is conventionally plotted increasing to the left, so core electrons sit on the left and valence electrons on the right.
• The rightmost peak always equals the element's first ionization energy, which is a check you can run on any spectrum.
• Dividing a 1s binding energy by 1.31 MJ per mole and taking the square root gives an apparent Z_eff; across period 2 it rises about 0.83 per proton, showing that outer electrons shield inner ones by only about 0.17.
• Removing electrons without changing the nucleus shifts every remaining peak to higher binding energy.

Sources

1. Khan Academy. (n.d.). Photoelectron spectroscopy [Unit topic]. In AP College Chemistry. khanacademy.org
2. OpenStax. (2019). Electronic structure of atoms (electron configurations) (Section 6.4). In Chemistry 2e. Rice University. openstax.org
3. National Institute of Standards and Technology. (n.d.). Atomic spectra database: ionization energies. U.S. Department of Commerce. physics.nist.gov
4. National Institute of Standards and Technology. (n.d.). NIST X-ray photoelectron spectroscopy database, SRD 20. U.S. Department of Commerce. srdata.nist.gov
5. Atkins, P., and de Paula, J. (2014). Atkins' Physical Chemistry (10th ed.). Oxford University Press.

Two places the trend breaks, and they are not mistakes

First ionization energies across period 2 run 520, 900, 801, 1086, 1402, 1314, 1681, 2081 kJ per mole. Six of the seven steps rise. Two fall: beryllium to boron drops 99 kJ, and nitrogen to oxygen drops 88. Both falls are reproducible, both appear again in period 3 at aluminium and at sulfur, and both have specific causes that no amount of counting protons will produce.

This lesson takes the four classic trends, states what drives each one, and then spends most of its time on the exceptions, because the exceptions are where the underlying model is doing real work rather than curve fitting.

Radius, and the awkward fact that there is no such thing

An atom has no edge. The electron density falls off smoothly and never reaches zero, so any radius you quote is a definition, not a measurement. Three definitions are in common use and they do not agree.

Definition	How it is obtained	Example: chlorine
Covalent radius	Half the distance between two bonded like atoms	102 pm
Van der Waals radius	Half the closest approach of two non-bonded atoms	175 pm
Ionic radius	Apportioned from crystal spacings in ionic solids	181 pm for Cl-

Numbers differ by nearly a factor of two depending on which question you asked. Compare like with like, always. The covalent radii below come from a single consistent set, Cordero and colleagues in 2008, so the comparisons within the table are sound.

Period 2 (pm)	Li 128	Be 96	B 84	C 76	N 71	O 66	F 57
Period 3 (pm)	Na 166	Mg 141	Al 121	Si 111	P 107	S 105	Cl 102
Group 1 (pm)	Li 128	Na 166	K 203	Rb 220	Cs 244

Across a period the radius shrinks, because Z_eff rises while the electrons stay in the same shell: more pull, same address, tighter orbit. Lithium to fluorine is a 55 percent reduction. Down a group the radius grows, because a new shell is added and that dominates: lithium to caesium nearly doubles.

What matters here: The same two variables from Lesson 1 explain both directions. Across, charge changes and distance does not. Down, distance changes and wins.

The two ionization energy anomalies

Beryllium to boron. Beryllium's outermost electron is a 2s electron; boron's is a 2p electron. Lesson 2 measured the gap directly: in boron, the 2s peak is at 1.36 MJ per mole and the 2p peak at 0.80. The 2p orbital penetrates the core less, so it is shielded more, so its electron is held less tightly. Boron's extra proton raises Z_eff, but the change of subshell more than cancels it. The same thing happens at magnesium to aluminium, where 738 falls to 578.

Nitrogen to oxygen. Nitrogen's three 2p electrons occupy the three 2p orbitals singly, one each, with parallel spins. Oxygen's fourth 2p electron has nowhere new to go, so it must pair with one that is already there. Two electrons in the same small orbital repel each other hard, and that repulsion partly cancels the extra nuclear attraction. Removing the paired electron relieves the repulsion, so it comes off unexpectedly cheaply. The same pattern recurs at phosphorus to sulfur, where 1012 falls to 1000.

Notice the structure of both explanations. Neither says the trend is wrong. Both identify a second effect that opposes the first and, for one step only, happens to be larger.

Electron affinity, where the trend is weakest

Electron affinity is the energy change when a gaseous atom gains an electron. Reported as a positive number when energy is released, the halogens dominate: F 328, Cl 349, Br 325, I 295 kJ per mole.

Chlorine beats fluorine, which is not what the general rule predicts. The reason is fluorine's own smallness. Its 2p subshell is so compact that adding a ninth valence-region electron forces it into a crowded space where electron-electron repulsion is severe. Chlorine's 3p subshell is roomier, so the incoming electron pays a smaller repulsion penalty even though chlorine's nucleus pulls a little less. Compactness, which makes fluorine the most electronegative element, works against it here.

Nitrogen makes the opposite point. Its electron affinity is essentially zero and slightly unfavourable: N^- is not a bound species under ordinary conditions, because the incoming electron would have to pair up in a half-filled 2p set, the same repulsion that produced the nitrogen-to-oxygen dip. Beryllium and the noble gases are similar; adding an electron would mean starting a new subshell or a new shell entirely, and nothing about that is favourable.

In short: Electron affinity trends across a period in the same direction as ionization energy, but with far more exceptions, because the added electron changes the repulsion balance rather than just feeling the nucleus.

Electronegativity, and what it is not

Electronegativity measures an atom's pull on the shared electrons of a bond it is already in. It is a derived scale, not a measured energy; Pauling built it from bond energies. On the Pauling scale, F 3.98, O 3.44, N 3.04, Cl 3.16, C 2.55, H 2.20, Al 1.61, Mg 1.31, Na 0.93, Cs 0.79.

It rises across a period and falls down a group, for the same reasons as ionization energy. Do not confuse the three quantities, which describe three different situations:

• Ionization energy is about an isolated atom losing an electron completely.
• Electron affinity is about an isolated atom gaining one completely.
• Electronegativity is about an atom in a bond pulling shared electrons toward itself.

Fluorine tops the electronegativity scale but not the electron affinity table, and the distinction above is exactly why.

Ionic radii and isoelectronic series

Cations are smaller than their parent atoms, often dramatically. Sodium's covalent radius is 166 pm; Na⁺ has an ionic radius of 102 pm. Losing the 3s electron does not shave a bit off the outside, it removes the entire outer shell, so the ion is a neon-like core held by eleven protons. Anions are larger than their parents: chlorine 102 pm becomes Cl^- at 181 pm, because the added electron increases repulsion without adding any nuclear charge.

The clean test case is an isoelectronic series: ions with identical electron counts and different nuclear charges. Take the ten-electron series and its six-coordinate radii.

Ion	Protons	Electrons	Radius (pm)
O2-	8	10	140
F-	9	10	133
Na+	11	10	102
Mg2+	12	10	72
Al3+	13	10	53.5

Ten electrons in every row, so shielding is identical and only the nuclear charge changes. Radius falls monotonically as protons are added, and it falls hard: aluminium has five more protons than oxygen and its ion is barely a third the size. This is Coulomb's law with one variable held fixed, which is as close to a controlled experiment as the periodic table offers.

Worked example: rank four species

Problem. Rank K⁺, Ca²⁺, Cl^- and S^2- by increasing radius, and explain the ordering.

Step 1. Count electrons. K⁺ has 19 minus 1 = 18. Ca²⁺ has 20 minus 2 = 18. Cl^- has 17 plus 1 = 18. S^2- has 16 plus 2 = 18. All four are isoelectronic with argon.

Step 2. Order by nuclear charge, largest first. Ca 20, K 19, Cl 17, S 16.

Step 3. Invert, because more protons means a smaller ion. Ca²⁺ is smallest, then K⁺, then Cl^-, then S^2- is largest.

Answer. Ca²⁺ less than K⁺ less than Cl^- less than S^2-. Nothing here required a radius table; the electron count and the proton count settled it.

Worked example: which comparison needs data?

Problem. Predict, with reasoning, whether each comparison can be settled by the trends alone. (a) IE of P against S. (b) Radius of Se against Br. (c) IE of Ca against Br.

(a) P and S are adjacent in period 3 and sit at the half-filled-p boundary, so this is the pairing anomaly. The trend alone predicts S higher; the anomaly predicts P higher. Measured: P 1012, S 1000, so phosphorus is higher. This comparison needs the anomaly, not just the trend.

(b) Se and Br are adjacent in the same period with no subshell change and no pairing change at that step, so the plain trend applies: Se is larger. Covalent radii 120 pm against 120 pm are close enough that the safer statement is that they are nearly equal, which itself tells you the trend flattens on the right of a long period.

(c) Ca is in period 4 group 2 and Br is in period 4 group 17. Same period, so Z_eff decides and there are no anomalies between them: Br has the higher ionization energy, 1140 against Ca's 590 kJ per mole.

Bottom line: Use the trend first, then ask whether the step crosses a subshell boundary or a pairing boundary. If it does, expect the trend to lose.

Common misconceptions

• "Atomic radius is a measured property of an atom." It is a convention. Chlorine is 102, 175 or 181 pm depending on whether you asked about bonded atoms, non-bonded contact, or the chloride ion in a crystal.
• "Fluorine has the highest electron affinity because it is the most electronegative." Chlorine's electron affinity, 349 kJ per mole, exceeds fluorine's 328. Electronegativity describes an atom inside a bond; electron affinity describes an isolated atom taking on a whole extra electron into a very small space.
• "Anomalies mean the trend is unreliable." Both period 2 anomalies are predictable from configuration. Cross a subshell boundary and the trend dips; cross a pairing boundary and it dips again. The rule is not broken, it has a second term.
• "Cations are smaller because electrons are removed from the outside of the same shell." For a main-group metal the whole outer shell goes. Na⁺ is not a slightly smaller sodium atom, it is a neon-sized core with an extra proton.
• "Effective nuclear charge explains everything." It explains the shape of every trend and none of the exceptions, because it is built by counting electrons rather than by asking which orbital they are in and how their spins are arranged.

The short version

• Atomic radius has no unique definition; compare covalent with covalent and ionic with ionic.
• Across a period, radius shrinks and ionization energy and electronegativity rise, because Z_eff grows while n stays fixed. Down a group, all three reverse, because n grows.
• Each period has two ionization energy dips: at the s to p boundary (Be to B, Mg to Al) and at the start of p pairing (N to O, P to S).
• Electron affinity follows the same broad direction with far more exceptions, and chlorine beats fluorine because fluorine's 2p subshell is too cramped to accept an extra electron comfortably.
• Cations are much smaller than their parent atoms and anions much larger.
• In an isoelectronic series, radius falls as nuclear charge rises: O^2- 140, F^- 133, Na⁺ 102, Mg²⁺ 72, Al³⁺ 53.5 pm.

Sources

1. OpenStax. (2019). Periodic variations in element properties (Section 6.5). In Chemistry 2e. Rice University. openstax.org
2. OpenStax. (2019). Covalent bonding (Section 7.2). In Chemistry 2e. Rice University. openstax.org
3. National Center for Biotechnology Information. (n.d.). PubChem periodic table of elements. National Library of Medicine. pubchem.ncbi.nlm.nih.gov
4. Cordero, B., Gomez, V., Platero-Prats, A. E., Reves, M., Echeverria, J., Cremades, E., Barragan, F., and Alvarez, S. (2008). Covalent radii revisited. Dalton Transactions, (21), 2832-2838.
5. Shannon, R. D. (1976). Revised effective ionic radii and systematic studies of interatomic distances in halides and chalcogenides. Acta Crystallographica Section A, 32(5), 751-767.

Two elements, three substances, three orders of magnitude

Sodium metal melts at 98 degrees Celsius. Chlorine boils at minus 34. Put them together and sodium chloride melts at 801. The same two elements, arranged three ways, and the energy needed to pull the particles apart spans a factor of many hundreds. Nothing about the atoms changed. What changed is where the bonding electrons went.

CHEM 100 taught you to label these bonds: metal plus non-metal gives ionic, two non-metals give covalent, metal plus metal gives metallic. Those labels are useful and they are also a simplification of a continuum. This lesson replaces the labels with the electrons.

Three destinations for a valence electron

• Transferred. One atom takes the electron outright. You are left with a cation and an anion, and the attraction between them is straight electrostatics. This is ionic bonding.
• Shared between two nuclei. The electron density piles up in the region between two atoms, and both nuclei are attracted to it. This is covalent bonding.
• Delocalised over many nuclei. The electrons leave their parent atoms but belong to no particular partner, moving through a lattice of cations. This is metallic bonding.

The core of it: All three are the same Coulombic attraction between positive nuclei and negative electrons. They differ only in how localised the electrons are, and every property difference follows from that.

The continuum, and the number that is not a law

Electronegativity difference tells you how unevenly a bonding pair is shared. On the Pauling scale, sodium is 0.93 and chlorine 3.16, a difference of 2.23: the pair is so lopsided that calling the electron transferred is a good description. In HCl the difference is 3.16 minus 2.20, or 0.96: shared, but pulled toward chlorine, giving a polar covalent bond with a partial negative charge on Cl. In Cl₂ the difference is zero and the sharing is even.

You will see a rule that a difference above 1.7 means ionic. Treat it as a rough dividing line someone drew, not a physical threshold. Hydrogen fluoride has a difference of 1.78 and is a molecular gas at room temperature that dissolves in water as a weak acid, which is not ionic behaviour. Aluminium chloride, with a difference of 1.55, sublimes at 180 degrees Celsius as Al₂Cl₆ molecules despite pairing a metal with a non-metal. The continuum is real; the cut point is bookkeeping.

Lattice energy: what holds an ionic solid together

The lattice energy of an ionic compound is the energy released when gaseous ions come together into one mole of the solid, or equivalently the energy required to blow the solid apart into gaseous ions. It is the single number that predicts an ionic compound's melting point, hardness and solubility better than anything else, and it is Coulomb's law applied to a crystal:

lattice energy scales as the product of the ionic charges divided by the distance between the ion centres.

Compound	Charges	Ion radii (pm)	Separation (pm)	Lattice energy (kJ/mol)
KCl	1 and 1	138 and 181	319	715
NaCl	1 and 1	102 and 181	283	787
NaF	1 and 1	102 and 133	235	923
LiF	1 and 1	76 and 133	209	1030
MgO	2 and 2	72 and 140	212	3795

Read the top four rows first. The charges are identical throughout, so only the distance changes, and the lattice energy tracks it inversely: as separation falls from 319 to 209 pm, lattice energy climbs from 715 to 1030 kJ per mole. Now compare NaCl and MgO, whose separations are similar, 283 against 212 pm, but whose charge products differ by a factor of four. The lattice energy jumps almost five-fold, and MgO melts at 2852 degrees Celsius against sodium chloride's 801.

Push the arithmetic and it does not quite work, which is worth knowing. Scaling NaCl's 787 by the charge factor of 4 and the distance factor of 283 divided by 212, or 1.34, predicts about 4200 kJ per mole for MgO. The measured value is 3795, roughly ten percent low. Simple point-charge Coulombics overestimates here because the Mg-O bond has appreciable covalent character; the ions are not hard spheres of pure charge.

Key idea: Charge beats distance. Doubling both charges multiplies the lattice energy by four, while realistic changes in ionic radius move it by tens of percent.

Measuring what cannot be measured: the Born-Haber cycle

You cannot put gaseous sodium ions and gaseous chloride ions in a calorimeter. Lattice energy is obtained indirectly, by building a closed loop of steps whose energies you can measure and letting the unknown be the one that closes it. Because enthalpy is a state function, the total around the loop must equal the direct route, which is Hess's law used before you formally meet it in Module 5.

For sodium chloride, the direct route is the formation reaction, Na(s) plus half Cl₂(g) giving NaCl(s), with a standard enthalpy of formation of minus 411 kJ per mole. The indirect route has five steps.

1. Sublime the sodium. Na(s) to Na(g), plus 109 kJ per mole.
2. Ionize it. Na(g) to Na⁺(g) plus an electron, plus 496 kJ per mole. That is the first ionization energy from Lesson 1.
3. Break half a mole of chlorine bonds. Half of Cl₂(g) to Cl(g), plus 122 kJ per mole, half the 243 kJ per mole bond dissociation energy.
4. Add the electron to chlorine. Cl(g) plus an electron to Cl^-(g), minus 368 kJ per mole, the electron affinity from Lesson 3 with the sign convention flipped.
5. Assemble the lattice. Na⁺(g) plus Cl^-(g) to NaCl(s), minus L, where L is the lattice energy we want.

Solve. 109 + 496 + 122 + (minus 368) + (minus L) = minus 411. The first four terms sum to plus 359, so 359 minus L = minus 411, giving L = 770 kJ per mole.

Tabulated values run from about 769 to 787 kJ per mole depending on which sublimation and electron affinity data a source uses. That spread is not sloppiness; it is what happens when a quantity is assembled from five measurements, each with its own uncertainty. Quote the source with the number.

Metallic bonding, and the properties nobody else can explain

In sodium metal, each atom contributes its 3s electron to a shared pool. The result is a regular lattice of Na⁺ cores immersed in electrons that belong to the whole crystal. That single picture accounts for four properties at once, and it is worth seeing each one fall out.

• Electrical conductivity. Apply a potential difference and delocalised electrons drift. Nothing has to break. Contrast an ionic solid, which conducts only when melted or dissolved so that the ions themselves can move, and a molecular solid, which does not conduct at all.
• Thermal conductivity. The same mobile electrons carry kinetic energy through the metal far faster than lattice vibrations alone could.
• Malleability. Hammer a metal and one plane of cations slides over another. The electron sea simply flows with them and the bonding survives. Hammer an ionic crystal and a slip of one layer brings like charges face to face; the repulsion splits the crystal, which is why salt is brittle and copper is not.
• Lustre. Delocalised electrons absorb and re-emit light across a wide range of frequencies, giving a reflective surface rather than a colour.

Metallic bond strength varies enormously, and the same two variables govern it. Sodium contributes one electron per atom from a large cation and melts at 98 degrees Celsius. Magnesium contributes two from a smaller cation and melts at 650. Aluminium contributes three and melts at 660. Tungsten, with a small core and many bonding electrons, melts at 3422. Mercury, whose 6s electrons are held unusually tightly, is a liquid at room temperature at minus 39.

So what?: Alloys work because the electron sea does not care much which cations sit in the lattice. Substituting zinc for some copper gives brass; wedging small carbon atoms into the gaps of an iron lattice gives steel, whose interstitial carbon blocks the planes from sliding and makes it far harder than pure iron.

Worked example: rank four solids

Problem. Rank CaO, KBr, NaCl and KCl by increasing melting point, using bonding arguments only.

Step 1. Separate by charge product. CaO pairs a 2 plus with a 2 minus, a charge product of 4. The other three are all 1 and 1, product 1. CaO is therefore in a different class and will melt highest.

Step 2. Order the singly charged three by ion size. Larger ions mean a larger separation and a weaker lattice. K⁺ (138 pm) is larger than Na⁺ (102 pm), and Br^- (196 pm) is larger than Cl^- (181 pm). So KBr has the largest separation, then KCl, then NaCl the smallest.

Step 3. Assemble. KBr, then KCl, then NaCl, then CaO.

Check against data. KBr melts at 734, KCl at 771, NaCl at 801, CaO at 2613 degrees Celsius. The prediction holds, and the size of the gap between the first three and the fourth is the charge effect made visible.

Common misconceptions

• "Ionic and covalent are two separate categories." They are the ends of one continuum of electron sharing. Every real bond between different elements sits somewhere between, and where you draw the boundary is a convention.
• "An electronegativity difference above 1.7 means the compound is ionic." HF has 1.78 and is a molecular gas; AlCl₃ has 1.55 and sublimes as covalent dimers. Use the number as a guide, then check the actual behaviour.
• "Lattice energy depends mostly on ion size." Charge dominates. Going from 1 and 1 to 2 and 2 multiplies the Coulombic term by four, an effect no realistic change of radius can match.
• "Metals conduct because electrons are passed from atom to atom." The electrons are not handed along; they are already delocalised over the whole crystal and simply drift when a field is applied.
• "Ionic solids conduct electricity." Solid sodium chloride is an insulator, because its ions are locked in place. Melt it or dissolve it and it conducts well, because now the charge carriers can move.
• "Lattice energy is a directly measured quantity." It is calculated from a Born-Haber cycle out of five separately measured quantities, which is why published values for the same compound differ by twenty kilojoules or more.

Where this leaves us

• Ionic, covalent and metallic bonding are one electrostatic phenomenon distinguished by how localised the bonding electrons are.
• Electronegativity difference places a bond on the continuum; the 1.7 cut point is a convention with counterexamples on both sides.
• Lattice energy scales as the charge product over the internuclear separation, and charge dominates: MgO at 3795 kJ per mole against NaCl at 787.
• A Born-Haber cycle obtains lattice energy indirectly; for NaCl the five measured steps give about 770 kJ per mole.
• Metallic bonding explains conductivity, malleability and lustre together, because the electron sea moves without breaking when the lattice deforms.
• Ionic solids are brittle for the mirror-image reason: sliding a plane brings like charges together and the crystal splits.

Sources

1. OpenStax. (2019). Ionic bonding (Section 7.1). In Chemistry 2e. Rice University. openstax.org
2. OpenStax. (2019). Strengths of ionic and covalent bonds (Section 7.5). In Chemistry 2e. Rice University. openstax.org
3. OpenStax. (2019). Lattice structures in crystalline solids (Section 10.6). In Chemistry 2e. Rice University. openstax.org
4. National Center for Biotechnology Information. (n.d.). Sodium chloride, PubChem compound summary CID 5234. National Library of Medicine. pubchem.ncbi.nlm.nih.gov
5. Shannon, R. D. (1976). Revised effective ionic radii and systematic studies of interatomic distances in halides and chalcogenides. Acta Crystallographica Section A, 32(5), 751-767.

A paper from 1916 that you are still using

Gilbert N. Lewis published a paper in the Journal of the American Chemical Society in 1916 that drew atoms as cubes with electrons at the corners. The cubes did not survive. The proposal in the same paper, that a chemical bond is a pair of electrons shared between two atoms, became the most useful bookkeeping device in chemistry, and the dot diagrams you are about to draw still carry his name.

What follows is a procedure, not an art. Done in the same order every time it will produce a defensible structure for almost anything you meet, and formal charge will tell you when you have the best one.

The procedure, in five steps

1. Count the total valence electrons. Sum the group valence electrons of every atom. Add one electron for each unit of negative charge; subtract one for each unit of positive charge.
2. Choose the central atom. It is the least electronegative atom that is not hydrogen. Hydrogen is never central, because it can form only one bond.
3. Connect everything with single bonds and subtract two electrons per bond from your total.
4. Complete the octets of the outer atoms with lone pairs, then place any electrons still left on the central atom.
5. If the central atom is short of an octet, convert lone pairs on outer atoms into additional bonds until it is satisfied.

Worked: carbon dioxide. Step 1: carbon has 4 valence electrons, each oxygen has 6, total 16. Step 2: carbon is central, being less electronegative. Step 3: two C-O single bonds use 4 electrons, leaving 12. Step 4: three lone pairs on each oxygen uses all 12, leaving none for carbon, which now has only 4 electrons around it. Step 5: move one lone pair from each oxygen into the bonding region, making two double bonds. Carbon now has 8, each oxygen has 8, and the structure is O=C=O.

Worked: the sulfate ion. Step 1: sulfur 6, four oxygens 24, plus 2 for the charge, total 32. Step 2: sulfur is central. Step 3: four S-O single bonds use 8, leaving 24. Step 4: three lone pairs on each oxygen uses exactly 24. Sulfur has an octet from its four bonds and every oxygen has one. The structure closes with all single bonds and the charge of 2 minus spread over the ion.

Formal charge: the referee

When two structures both obey the rules, formal charge decides which is a better description of the electron distribution. For each atom,

formal charge = valence electrons in the free atom, minus lone pair electrons, minus half the bonding electrons

Three rules follow. The best structure has formal charges as close to zero as possible. Where charges are unavoidable, negative formal charge should sit on the more electronegative atom. And structures with like charges on adjacent atoms are poor.

Worked: which structure for CO₂? Compare O=C=O with the alternative O-C(triple bond)O, which also uses 16 electrons and gives everyone an octet.

Structure	Left O	C	Right O
O=C=O	6 minus 4 minus 2 = 0	4 minus 0 minus 4 = 0	0
O-C(triple)O	6 minus 6 minus 1 = minus 1	4 minus 0 minus 4 = 0	6 minus 2 minus 3 = plus 1

All zeroes beats a separated pair of charges, so O=C=O is the preferred structure, which matches the experimental fact that both C-O bonds in carbon dioxide are the same length.

Why this matters: Formal charge is not a real charge. It is the answer to a bookkeeping question, namely what charge each atom would carry if every bonding pair were split exactly down the middle. It is useful precisely because it is a fixed convention, so two structures can be compared on the same terms.

The sulfate argument, which is not settled

Return to sulfate. The all-single-bond structure gives sulfur a formal charge of 6 minus 0 minus 4 = plus 2, and each oxygen 6 minus 6 minus 1 = minus 1, summing to the 2 minus of the ion. Many textbooks reject that and redraw sulfate with two S=O double bonds, which brings sulfur's formal charge to zero and gives sulfur ten electrons, an expanded octet.

The traditional justification was that sulfur's empty 3d orbitals accept the extra pair. Computational work since the 1990s has found that 3d participation in second-row main-group bonding is very small, and that the bonding is better described as highly polar single bonds. Several general chemistry texts have since moved back toward the all-single-bond structure with the plus 2 formal charge on sulfur. Others keep the double-bonded picture because it reproduces the observed short S-O bond lengths more intuitively.

You should be able to draw and defend both, and you should know that a chemistry course teaching only one of them is presenting a live disagreement as a settled fact.

Resonance, and what ozone actually looks like

Some species cannot be described by any single Lewis structure. Ozone, O₃, has 18 valence electrons and a bent shape. Draw it with a double bond on the left and a single bond on the right, and you can equally draw the mirror image. Neither is right on its own, because measurement says the two bonds in ozone are identical, both 128 pm.

Put that number between two references. In O₂, a genuine double bond, the length is 121 pm. In hydrogen peroxide, a genuine O-O single bond, it is 148 pm. Ozone's 128 pm sits between them, which is exactly what you would expect if each bond were something like one and a half bonds.

The resonance structures are not two forms the molecule flips between. Ozone does not spend Monday as one and Tuesday as the other. It is one molecule whose electrons are delocalised over three atoms, and our notation, which insists on drawing pairs in fixed places, cannot represent that in a single picture. Resonance is a limitation of the drawing system, not a behaviour of the molecule.

Remember: Resonance means the true structure is a single average of the contributors, weighted toward the ones with the best formal charges. It never means the molecule oscillates.

The nitrate ion works the same way. Three equivalent structures place the double bond on each of the three oxygens in turn, and all three N-O bonds are measured to be identical, which no single Lewis structure can express.

Three families of exception

Exception	Example	Electron count at the central atom	Why it happens
Incomplete octet	BF3	6	Boron has only three valence electrons; adding a double bond would put positive formal charge on fluorine, the most electronegative element.
Odd electron count	NO2	7	17 valence electrons cannot be paired. One electron is left unpaired, making it a radical.
Expanded valence	SF6	12	Sulfur is in period 3 and physically large enough to bond six fluorines; the modern description uses polar bonds rather than d-orbital involvement.

Note the asymmetry. Period 2 elements never expand beyond eight, because n = 2 offers only one s and three p orbitals, four orbitals and therefore at most eight electrons. Nitrogen cannot form NF₅ no matter how you draw it. Period 3 and beyond can exceed eight, and whether you attribute that to d orbitals or to the atom's size and the polarity of its bonds is the argument above.

Worked example: the thiocyanate ion

Problem. Draw the resonance structures of SCN^- and use formal charge to say which contributes most.

Step 1. Count. S 6, C 4, N 5, plus 1 for the charge: 16 valence electrons.

Step 2. Skeleton. Carbon is the least electronegative of the three, so the arrangement is S-C-N.

Step 3. Three candidates that all use 16 electrons and give every atom an octet.

• A: S with a triple bond to C, C single bonded to N. Formal charges: S = 6 minus 2 minus 3 = plus 1, C = 0, N = 5 minus 6 minus 1 = minus 2.
• B: S=C=N with double bonds on both sides. Formal charges: S = 6 minus 4 minus 2 = 0, C = 0, N = 5 minus 4 minus 2 = minus 1.
• C: S single bonded to C, C triple bonded to N. Formal charges: S = 6 minus 6 minus 1 = minus 1, C = 0, N = 5 minus 2 minus 3 = 0.

Step 4. Judge. A is poor: a plus 1 and a minus 2 is a large charge separation. B and C both carry a single minus 1. Nitrogen is more electronegative than sulfur, so B, which puts the negative charge on nitrogen, is generally taken as the largest contributor, with C significant as well. The real ion is a blend dominated by B and C.

The point: When several structures survive the octet test, formal charge ranks them, and electronegativity breaks the tie between structures with equal charge magnitudes.

Common misconceptions

• "Resonance means the molecule flips between structures." It does not. The molecule has one structure, which our notation cannot draw. Ozone's two bonds are both 128 pm at all times.
• "Formal charge is the actual charge on the atom." Formal charge assumes bonding pairs are split evenly, which is false for every polar bond. Actual partial charges follow electronegativity and are usually smaller.
• "The octet rule is a law." It is a pattern with three whole families of exception, and it never applies to hydrogen, which is complete at two electrons, or to boron and beryllium, which are routinely content with six and four.
• "Any atom can expand its octet if you need it to." Period 2 elements cannot, because n = 2 supplies only four orbitals. There is no valid Lewis structure for NF₅ or for a nitrogen with five bonds.
• "The structure with more double bonds is always better." BF₃ is the counterexample: adding a B=F double bond would give fluorine a positive formal charge, which is worse than leaving boron short of an octet.

Pulling it together

• Count valence electrons first, adjusting for charge, then build outward: skeleton, outer octets, leftovers on the centre, then multiple bonds if needed.
• Formal charge equals valence electrons minus lone pair electrons minus half the bonding electrons, and the best structure minimises it.
• When charges cannot be avoided, put the negative one on the more electronegative atom.
• Resonance is required when no single structure fits the measurements; ozone's two identical 128 pm bonds sit between the 121 pm of O₂ and the 148 pm of hydrogen peroxide.
• Three exception families: incomplete octets like BF₃, odd electron species like NO₂, and expanded valence shells like SF₆.
• Whether expanded octets involve d orbitals is genuinely disputed; modern calculations favour polar single bonds, and sulfate is the case where the disagreement is most visible.

Sources

1. OpenStax. (2019). Lewis symbols and structures (Section 7.3). In Chemistry 2e. Rice University. openstax.org
2. OpenStax. (2019). Formal charges and resonance (Section 7.4). In Chemistry 2e. Rice University. openstax.org
3. National Center for Biotechnology Information. (n.d.). Ozone, PubChem compound summary CID 24823. National Library of Medicine. pubchem.ncbi.nlm.nih.gov
4. Lewis, G. N. (1916). The atom and the molecule. Journal of the American Chemical Society, 38(4), 762-785.
5. Housecroft, C. E., and Sharpe, A. G. (2018). Inorganic Chemistry (5th ed.). Pearson. See the discussion of hypervalence and d-orbital participation.

Two oxygens, two answers

Carbon dioxide is a straight line, 180 degrees, with no dipole moment at all. Sulfur dioxide is bent at 119 degrees and has a dipole moment of 1.63 debye. Both are a central atom flanked by two oxygens. The difference is a single lone pair on sulfur that carbon does not have, and that one pair changes the shape, the polarity, the boiling point and the chemistry.

Lesson 5 gave you the electron bookkeeping. This lesson turns that bookkeeping into three dimensions.

The one premise of VSEPR

Valence shell electron pair repulsion rests on a single claim: regions of electron density around a central atom repel each other, so they arrange themselves as far apart as possible. Count the regions, called electron domains, and the arrangement follows.

A domain is any of the following: a lone pair, a single bond, a double bond, a triple bond, or a lone unpaired electron. Notice that a double bond counts once. The two pairs in a double bond are held between the same two nuclei, so they point the same direction and act as one region.

Domains	Electron geometry	Ideal angles	Lone pairs	Molecular shape	Example
2	Linear	180	0	Linear	CO2
3	Trigonal planar	120	0	Trigonal planar	BF3
3	Trigonal planar	120	1	Bent	SO2
4	Tetrahedral	109.5	0	Tetrahedral	CH4
4	Tetrahedral	109.5	1	Trigonal pyramidal	NH3
4	Tetrahedral	109.5	2	Bent	H2O
5	Trigonal bipyramidal	90 and 120	0	Trigonal bipyramidal	PCl5
5	Trigonal bipyramidal	90 and 120	1	Seesaw	SF4
5	Trigonal bipyramidal	90 and 120	2	T-shaped	ClF3
5	Trigonal bipyramidal	90 and 120	3	Linear	XeF2
6	Octahedral	90	0	Octahedral	SF6
6	Octahedral	90	1	Square pyramidal	BrF5
6	Octahedral	90	2	Square planar	XeF4

Key idea: Electron geometry counts all domains. Molecular shape describes only where the atoms are. Water has tetrahedral electron geometry and a bent molecular shape, and confusing the two is the single most common error in this topic.

Lone pairs push harder

The ideal tetrahedral angle is 109.5 degrees, and methane hits it exactly. Ammonia, with one lone pair, closes to 107. Water, with two, closes to 104.5. The pattern is consistent: each lone pair squeezes the bond angles by roughly two and a half degrees.

The explanation is that a lone pair is held by only one nucleus, so its electron cloud spreads wider near the central atom than a bonding pair does, which is pinched between two nuclei. A wider cloud pushes harder on its neighbours. The repulsion ordering that follows is

lone pair to lone pair, stronger than lone pair to bonding pair, stronger than bonding pair to bonding pair.

That ordering explains why lone pairs go equatorial in a trigonal bipyramid. An equatorial position has two neighbours at 90 degrees; an axial position has three. Put the lone pair where it faces the fewest close neighbours, and SF₄ comes out as a seesaw rather than as a trigonal pyramid with an axial gap.

Hybridisation: making the orbitals match the shape

There is a problem VSEPR does not address. Carbon's ground state configuration is 1s²2s²2p², which offers two unpaired electrons in p orbitals at 90 degrees to each other. Methane has four identical bonds at 109.5 degrees. The atomic orbitals as they come do not have the right count or the right angles.

Valence bond theory answers by mixing them. One 2s and three 2p orbitals combine into four equivalent sp³ hybrid orbitals pointing at the corners of a tetrahedron. The count is preserved: four orbitals in, four orbitals out. Read hybridisation straight off the domain count.

Domains	Hybridisation	Geometry of the hybrids	Example
2	sp	Linear	BeCl2, CO2
3	sp2	Trigonal planar	BF3, ethene carbons
4	sp3	Tetrahedral	CH4, NH3, H2O
5	sp3d	Trigonal bipyramidal	PCl5
6	sp3d2	Octahedral	SF6

The last two rows carry the caveat from Lesson 5. Labelling PCl₅ as sp³d assumes real d-orbital participation, and modern calculations find that contribution small. The labels remain standard in general chemistry and on examinations; treat them as descriptions of geometry rather than as claims about which orbitals are genuinely mixed.

Sigma, pi, and why a double bond will not rotate

Hybrid orbitals overlap end to end along the internuclear axis, producing a sigma bond. Unhybridised p orbitals overlap side to side, above and below that axis, producing a pi bond. The accounting is fixed:

• A single bond is one sigma bond.
• A double bond is one sigma and one pi.
• A triple bond is one sigma and two pi.

In ethene, each carbon is sp² hybridised with one leftover p orbital perpendicular to the molecular plane. Those two p orbitals overlap sideways to form the pi bond, and that sideways overlap is destroyed if one end of the molecule twists. This is why there is no free rotation about a carbon-carbon double bond, and therefore why 1,2-dichloroethene exists as two separable compounds, cis and trans, with different boiling points. A single bond, being pure sigma, rotates freely at room temperature.

In short: Sigma bonds set the shape and permit rotation. Pi bonds shorten and stiffen the bond and forbid it.

Polarity: adding vectors, not counting bonds

A polar bond has a dipole, a vector pointing from the partial positive atom toward the partial negative one, with magnitude set by the electronegativity difference. A molecule's overall dipole is the vector sum of its bond dipoles. Symmetry can cancel every one of them.

Molecule	Shape	Bonds polar?	Dipole moment (D)	Why
CO2	Linear	Yes	0	The two C=O dipoles point opposite ways and cancel exactly.
SO2	Bent	Yes	1.63	Bent geometry leaves a resultant pointing away from the lone pair.
CCl4	Tetrahedral	Yes	0	Four identical dipoles arranged symmetrically sum to zero.
CHCl3	Tetrahedral	Yes	1.04	Replacing one Cl with H breaks the symmetry.
H2O	Bent	Yes	1.85	Two O-H dipoles plus two lone pairs, all pointing broadly the same way.
NH3	Trigonal pyramidal	Yes	1.47	Bond dipoles point toward N and the lone pair adds to them.
NF3	Trigonal pyramidal	Yes	0.24	Bond dipoles point away from N, opposing the lone pair, so they nearly cancel.
XeF4	Square planar	Yes	0	Four dipoles in a plane at 90 degrees cancel in pairs.

The NH₃ and NF₃ comparison is the one to remember. Same shape, same number of polar bonds, and dipole moments differing by a factor of six. In ammonia the nitrogen is the electronegative end of every bond, so the three bond dipoles point inward toward nitrogen, in the same direction as the lone pair's contribution. In nitrogen trifluoride, fluorine is more electronegative than nitrogen, so the bond dipoles point outward, against the lone pair, and most of the effect cancels.

Worked example: three species, three questions each

SF₄. Valence electrons: 6 plus 28 equals 34. Four S-F bonds use 8; 24 complete the fluorine octets; 2 remain as a lone pair on sulfur. Five domains, one lone pair, so the electron geometry is trigonal bipyramidal and the shape is seesaw. Hybridisation sp³d. Four sigma bonds, no pi bonds. The lone pair sits equatorially, so the dipoles do not cancel and the molecule is polar.

XeF₄. Valence electrons: 8 plus 28 equals 36. Four bonds use 8; 24 complete the fluorines; 4 remain as two lone pairs on xenon. Six domains, two lone pairs, so the electron geometry is octahedral and the shape is square planar, with the two lone pairs opposite each other. Hybridisation sp³d². The four Xe-F dipoles lie at 90 degrees in one plane and cancel in opposite pairs, so the molecule is non-polar despite four polar bonds.

HCN. Valence electrons: 1 plus 4 plus 5 equals 10. The structure is H-C(triple)N. Carbon has two domains, so the shape is linear and the hybridisation is sp. Bond count: two sigma bonds, H to C and C to N, plus two pi bonds inside the triple bond. Both bonds are polar and they point the same way along one axis, so the molecule is strongly polar.

Common misconceptions

• "Electron geometry and molecular shape are the same." Water is tetrahedral in its electron geometry and bent in its shape. Ammonia is tetrahedral and trigonal pyramidal. Only when there are no lone pairs do the two names agree.
• "A double bond counts as two domains." It counts as one. Both pairs point along the same internuclear axis, so they behave as one region of repulsion. This is why CO₂ is linear.
• "Polar bonds make a polar molecule." Carbon tetrachloride has four strongly polar bonds and a dipole moment of zero, because symmetry cancels them. Geometry decides, not bond count.
• "Hybridisation is something an atom does before bonding." It is a mathematical recombination of atomic orbitals used to describe a molecule we already observe. No atom sits in isolation and hybridises.
• "Lone pairs and bonding pairs repel equally." Lone pairs are held by one nucleus and spread wider, so they repel more strongly. That is why the tetrahedral angle falls from 109.5 in CH₄ to 107 in NH₃ to 104.5 in H₂O.
• "NF₃ should be more polar than NH₃ because fluorine is more electronegative." The measured values are 0.24 and 1.47 debye. In NF₃ the bond dipoles oppose the lone pair's contribution instead of adding to it.

What you now know

• Count electron domains, where a multiple bond counts once, and the electron geometry follows directly.
• Subtract the lone pairs to get the molecular shape, and remember that lone pairs compress bond angles by about two and a half degrees each.
• In a trigonal bipyramid, lone pairs take equatorial positions, giving the seesaw, T-shaped and linear series.
• Hybridisation follows the domain count: 2 gives sp, 3 gives sp², 4 gives sp³, 5 and 6 give the d-labelled sets whose d participation is disputed.
• A single bond is one sigma; a double is one sigma plus one pi; a triple is one sigma plus two pi. Pi bonding blocks rotation and creates cis and trans isomers.
• Molecular polarity is the vector sum of bond dipoles, so symmetric molecules such as CO₂, CCl₄ and XeF₄ are non-polar despite polar bonds.

Sources

1. OpenStax. (2019). Molecular structure and polarity (Section 7.6). In Chemistry 2e. Rice University. openstax.org
2. OpenStax. (2019). Valence bond theory (Section 8.1). In Chemistry 2e. Rice University. openstax.org
3. OpenStax. (2019). Hybrid atomic orbitals (Section 8.2). In Chemistry 2e. Rice University. openstax.org
4. OpenStax. (2019). Multiple bonds (Section 8.3). In Chemistry 2e. Rice University. openstax.org
5. Gillespie, R. J., and Hargittai, I. (1991). The VSEPR Model of Molecular Geometry. Allyn and Bacon.

A boiling point that is 160 degrees out of place

Water boils at 100 degrees Celsius. Hydrogen sulfide, which sits directly below it in group 16 and weighs nearly twice as much, boils at minus 60. Follow the group further and the numbers behave: H₂Se boils at minus 41, H₂Te at minus 2. Heavier means higher, steadily, and then water breaks the pattern by 160 degrees.

The same thing happens in group 15, where ammonia at minus 33 towers over phosphine at minus 88, and in group 17, where hydrogen fluoride at plus 20 towers over hydrogen chloride at minus 85. Group 14 provides the control: methane minus 162, silane minus 112, germane minus 89, a clean upward march with no anomaly at all.

Something is present in H₂O, NH₃ and HF that is absent in CH₄. Finding it is the point of this lesson, and once you have it you can predict boiling points, viscosities, vapour pressures and solubilities from a structural formula.

What boiling actually costs

First, a distinction that decides half the questions in this topic. Boiling water does not break O-H bonds. The steam coming off a kettle is still H₂O. What boiling breaks are the attractions between molecules, and those are far weaker than the bonds inside them.

Put numbers on it. Breaking one mole of O-H bonds costs roughly 460 kJ. Vaporising one mole of water costs 40.7 kJ, about eleven times less. A single hydrogen bond is worth roughly 20 kJ per mole, against a covalent bond's several hundred.

Why this matters: Melting and boiling points, vapour pressure, viscosity, surface tension and solubility are all controlled by intermolecular forces. Chemical reactivity is controlled by the bonds inside molecules. Confusing the two produces answers that are wrong by an order of magnitude.

The three forces, weakest first

London dispersion forces exist in every substance without exception, including helium and including molecules with no dipole at all. At any instant, the electrons in a molecule are not perfectly evenly distributed, which gives a fleeting dipole; that dipole induces a matching one in a neighbour, and the two attract. The effect is tiny for one pair and enormous when summed over a large, floppy electron cloud.

Two structural features control its strength.

• Polarisability, which rises with the number of electrons and with how loosely they are held. The noble gases show it cleanly: He boils at minus 269, Ne at minus 246, Ar at minus 186, Kr at minus 153, Xe at minus 108. The halogens show it more dramatically still: F₂ and Cl₂ are gases, Br₂ is a liquid boiling at 59, and I₂ is a solid melting at 114 and boiling at 184 degrees Celsius. Only dispersion forces are at work in all of these.
• Shape, which sets how much surface two molecules can bring into contact. Pentane, 2-methylbutane and 2,2-dimethylpropane all have the formula C₅H₁₂ and identical molar masses, and they boil at 36, 28 and 9.5 degrees Celsius respectively. The straight chain lies alongside its neighbours over its whole length; the compact, ball-shaped isomer touches only at a point.

Dipole-dipole forces act between molecules that carry a permanent dipole, aligning positive ends to negative ends. They are stronger than dispersion forces between molecules of similar size, which is why propanone (acetone), with a mass of 58 and a strong C=O dipole, boils at 56 degrees Celsius while butane, mass 58 and non-polar, boils at minus 0.5.

Hydrogen bonding is the special case that solves the opening puzzle. It occurs when a hydrogen atom bonded to nitrogen, oxygen or fluorine is attracted to a lone pair on another nitrogen, oxygen or fluorine. Only those three elements qualify, for two reasons that work together: they are electronegative enough to strip the hydrogen almost bare, and they are small enough that the resulting concentrated positive charge can get very close to a lone pair. Chlorine is nearly as electronegative as nitrogen but much larger, and HCl shows no hydrogen bonding worth the name.

Ion-dipole forces complete the set. They act between an ion and a polar molecule, and they are what makes sodium chloride dissolve: each Na⁺ is surrounded by water molecules turning their oxygen ends inward, each Cl^- by water molecules turning their hydrogen ends inward.

Ranking, in the right order of operations

Given two substances, work through this sequence.

1. Are they the same phase of matter and the same kind of substance? An ionic solid or a covalent network will beat any molecular substance outright. NaCl at 801 degrees Celsius and SiO₂ at about 1710 are not competing in the same league as molecules.
2. Does one hydrogen bond and the other not? If so, that usually decides it, unless the masses are wildly different. Ethanol (mass 46) boils at 78; dimethyl ether, an isomer with the same mass 46 and no O-H bond, boils at minus 24.
3. If both hydrogen bond or neither does, compare polarity. The polar one wins among molecules of similar size.
4. If the forces are of the same type, compare electron count and then shape. More electrons and more extended shape mean stronger dispersion.

The upshot: Molar mass is a proxy for electron count, and it works only when the force type is held constant. The moment hydrogen bonding enters, mass stops predicting anything.

Water, and four properties that follow from one force

Water hydrogen bonds twice over: each molecule has two O-H bonds to donate and two lone pairs to accept, so a water molecule can participate in up to four hydrogen bonds at once. Almost every unusual property of water traces back to that number.

Property	Value	What hydrogen bonding does
Boiling point	100 degrees Celsius	160 degrees above the group trend, because a network must be dismantled.
Specific heat capacity	4.184 J per gram per degree	Added energy goes into stretching hydrogen bonds before it raises the temperature, which is why coastal climates are mild.
Enthalpy of vaporisation	40.7 kJ per mole	Escaping the liquid means breaking several hydrogen bonds, so sweating cools you efficiently.
Density of ice	0.917 g per cubic centimetre	The solid holds each molecule at exactly four hydrogen bonds in an open hexagonal lattice, which is less dense than the liquid.

The last row is the strange one and it is worth being precise about. In liquid water near freezing, molecules are close and hydrogen bonds are constantly breaking and reforming, so molecules can crowd in. On freezing, each molecule locks into an arrangement with four hydrogen bonds at fixed angles, and that geometry has holes in it. The ice is about 9 percent less dense than the liquid, so it floats, and liquid water reaches its maximum density at about 4 degrees Celsius rather than at its freezing point. Lakes therefore freeze from the top down, which is why anything lives in them through a winter.

Worked example: order five liquids

Problem. Rank by increasing boiling point: butane C₄H₁₀, propanone C₃H₆O, propan-1-ol C₃H₈O, ethane C₂H₆, and ethanoic acid C₂H₄O₂.

Step 1. Sort by force type. Ethane and butane are non-polar hydrocarbons: dispersion only. Propanone is polar with no O-H: dispersion plus dipole-dipole. Propan-1-ol and ethanoic acid both have an O-H group: hydrogen bonding.

Step 2. Within the dispersion-only pair, compare size. Butane has more electrons and a longer chain than ethane, so butane is higher.

Step 3. Place propanone. Dipole-dipole beats dispersion at comparable size, so propanone sits above butane.

Step 4. Within the hydrogen-bonding pair, count donors and acceptors. Ethanoic acid has both an O-H and a C=O, so two molecules can form a pair joined by two hydrogen bonds at once, a dimer. That doubled interaction puts it above propan-1-ol.

Answer. Ethane, then butane, then propanone, then propan-1-ol, then ethanoic acid. Measured boiling points in degrees Celsius: minus 89, minus 0.5, 56, 97, 118. The reasoning got the order right without a single data point.

Common misconceptions

• "Boiling water breaks the bonds in H₂O." Steam is still water. Vaporisation costs 40.7 kJ per mole; breaking O-H bonds costs about 460 kJ per mole each. Boiling separates molecules from each other.
• "A hydrogen bond is a type of covalent bond." It is an intermolecular attraction, roughly twenty times weaker than the O-H covalent bond it involves. The name is unfortunate and has confused students since it was coined.
• "Any molecule containing hydrogen can hydrogen bond." The hydrogen must be bonded to nitrogen, oxygen or fluorine. Methane contains four hydrogens and hydrogen bonds not at all, which is why it boils at minus 162.
• "Heavier molecules always boil higher." True only when the force type is unchanged. H₂S is heavier than H₂O and boils 160 degrees lower.
• "Dispersion forces only matter in non-polar substances." They act in everything, and in large molecules they can dominate. Iodine, with no dipole at all, is a solid at room temperature while polar HCl is a gas.
• "Ice floats because it traps air." Pure ice with no bubbles still floats. The open hexagonal lattice of four hydrogen bonds per molecule is intrinsically less dense than the liquid, by about nine percent.

Summing up

• Intermolecular forces control physical properties; covalent bonds control chemical reactivity, and the energies differ by roughly a factor of ten to twenty.
• Dispersion forces act in everything and grow with electron count and with how much surface two molecules can bring into contact.
• Dipole-dipole forces add to dispersion in polar molecules; ion-dipole forces are what dissolve salts in water.
• Hydrogen bonding needs H attached to N, O or F, and only those, because the partner must be both very electronegative and small.
• Rank by force type first, then by size within a type; molar mass predicts nothing once hydrogen bonding is in play.
• Water's boiling point, specific heat, enthalpy of vaporisation and floating ice all follow from each molecule's ability to make four hydrogen bonds.

Sources

1. OpenStax. (2019). Intermolecular forces (Section 10.1). In Chemistry 2e. Rice University. openstax.org
2. OpenStax. (2019). Properties of liquids (Section 10.2). In Chemistry 2e. Rice University. openstax.org
3. OpenStax. (2019). Phase transitions (Section 10.3). In Chemistry 2e. Rice University. openstax.org
4. National Center for Biotechnology Information. (n.d.). Water, PubChem compound summary CID 962. National Library of Medicine. pubchem.ncbi.nlm.nih.gov
5. National Institute of Standards and Technology. (n.d.). NIST Chemistry WebBook, SRD 69. U.S. Department of Commerce. webbook.nist.gov

A label that does not give you what you need

The bottle of concentrated hydrochloric acid on a stockroom shelf is labelled two ways: 37 percent HCl by mass, and density 1.19 grams per millilitre. Neither number is a molarity, and you cannot dilute anything until you have one. Turning that label into 12.1 mol per litre is a four-line calculation, and doing it carefully is a good model for every concentration problem in this course.

Step 1. Choose a convenient amount. Take exactly 1.000 L of the solution.

Step 2. Find its mass. 1000 mL times 1.19 g per mL = 1190 g of solution.

Step 3. Find the mass of solute in it. 1190 g times 0.37 = 440 g of HCl.

Step 4. Convert to moles and divide by the volume. The molar mass of HCl is 36.46 g per mole, so 440 g divided by 36.46 g per mole = 12.1 mol, and 12.1 mol in 1.000 L is 12.1 M.

Notice that the density was doing essential work. Percent by mass says nothing about volume, and molarity is defined per litre of solution, so any conversion between them must pass through the density.

Five ways to say how much

Unit	Definition	Depends on temperature?	Used for
Molarity (M)	moles of solute per litre of solution	Yes, because volume expands	Almost all bench work and titrations
Molality (m)	moles of solute per kilogram of solvent	No, mass does not expand	Freezing and boiling point calculations
Mole fraction	moles of one component over total moles	No	Gas mixtures, vapour pressure
Mass percent	mass of solute over mass of solution, times 100	No	Commercial reagent labels
Parts per million	milligrams of solute per kilogram of solution	No	Trace contaminants in water and air

Bottom line: Molarity is per litre of solution, not per litre of solvent. Dissolving 1 mole of solute in 1 litre of water gives something slightly more than 1 litre and therefore slightly less than 1 M. Volumetric flasks exist to make this distinction operational: you add solute, then top up to the mark.

Preparing and diluting, with the moles kept in view

Worked: make 250.0 mL of 0.150 M NaCl. Moles needed = 0.2500 L times 0.150 mol per L = 0.0375 mol. Mass = 0.0375 mol times 58.44 g per mol = 2.19 g. Weigh 2.19 g of sodium chloride into a 250 mL volumetric flask, dissolve in less water than you need, then fill to the mark.

Worked: make 500.0 mL of 0.100 M HCl from the 12.1 M stock. Dilution adds solvent and changes nothing about the amount of solute, so the moles before equal the moles after:

M₁V₁ = M₂V₂

12.1 times V₁ = 0.100 times 500.0 mL, so V₁ = 50.0 divided by 12.1 = 4.13 mL of stock, diluted to 500.0 mL. Both volumes are in millilitres on each side, so the units cancel and you never need to convert. Add the acid to the water, never the reverse.

Worked: mixing two solutions of the same solute. Combine 50.0 mL of 0.200 M KNO₃ with 150.0 mL of 0.100 M KNO₃. Moles from the first: 0.0500 L times 0.200 = 0.0100 mol. From the second: 0.1500 L times 0.100 = 0.0150 mol. Total 0.0250 mol in 0.2000 L, so the mixture is 0.125 M. Add moles, add volumes, then divide. Never average the molarities.

Why anything dissolves, and why some of it gets cold

Dissolving an ionic solid takes two energy steps in opposite directions. Pulling the lattice apart into gaseous ions costs the lattice energy, which is always endothermic and often large. Surrounding those ions with solvent molecules, called hydration when the solvent is water, releases energy. The enthalpy of solution is the sum.

The sum can land either way, and both cases are sold commercially. Ammonium nitrate has an enthalpy of solution of about plus 25.7 kJ per mole, so it absorbs heat and an instant cold pack chills your skin. Anhydrous calcium chloride is about minus 81.3 kJ per mole, so it releases heat, which is why it is used in self-heating cans and to melt ice on roads.

This raises a question worth flagging now. If dissolving ammonium nitrate absorbs energy, why does it happen at all? Nothing about enthalpy explains it. The answer is entropy: ions dispersed randomly through a solvent are a far more disordered arrangement than ions locked in a crystal, and that gain outweighs the enthalpy cost. Lesson 13 makes this quantitative with free energy, but hold the observation now: an endothermic process that happens spontaneously is a signal that entropy is driving it.

Temperature, pressure and gases

Most ionic solids dissolve better in hot water. Gases do the opposite: warming a solution drives dissolved gas out, which is why a fizzy drink goes flat faster when warm and why heated river water carries less dissolved oxygen for fish.

Pressure matters only for gases, and Henry's law states the relationship directly: the concentration of a dissolved gas is proportional to its partial pressure above the liquid. A sealed carbonated drink is bottled under carbon dioxide at above atmospheric pressure; open it and the partial pressure of CO₂ above the liquid drops to near zero, so the solubility collapses and the gas leaves as bubbles.

Measuring concentration with light

A coloured solution absorbs light, and it absorbs more when there is more of it in the path. The Beer-Lambert law makes that quantitative:

A = (epsilon) b c

where A is absorbance, a unitless number; b is the path length through the sample in centimetres, usually 1.00 cm for a standard cuvette; c is the molar concentration; and epsilon is the molar absorptivity, a constant for a given substance at a given wavelength.

Worked: find an unknown. A standard solution of copper(II) sulfate at 2.00 times 10^-4 M gives an absorbance of 0.480 at 810 nm in a 1.00 cm cuvette. An unknown of the same substance, measured the same way, gives 0.300. What is its concentration?

Step 1. Get epsilon from the standard. epsilon = A divided by (b times c) = 0.480 divided by (1.00 cm times 2.00 times 10^-4 mol per L) = 2400 L per mol per cm.

Step 2. Apply it to the unknown. c = A divided by (epsilon times b) = 0.300 divided by (2400 times 1.00) = 1.25 times 10^-4 M.

Step 3. Sanity check. The unknown absorbs 0.300 divided by 0.480, or 62.5 percent as much as the standard, and 1.25 times 10^-4 is 62.5 percent of 2.00 times 10^-4. Because absorbance is directly proportional to concentration, the ratio method works and is faster than computing epsilon at all.

What matters here: The relationship is linear only over a limited range, typically up to an absorbance of about 1. Above that, so little light reaches the detector that the measurement becomes unreliable, and a concentrated sample must be diluted by a known factor before it is read.

Colligative properties: counting particles, not identifying them

Some solution properties depend only on how many solute particles are present, not on what they are. Freezing point depression is the clearest.

Delta T_f = i K_f m, where K_f for water is 1.86 degrees Celsius per molal, m is the molality, and i is the van't Hoff factor, the number of particles each formula unit produces on dissolving.

Worked. A 0.100 m solution of sodium chloride. NaCl dissociates into Na⁺ and Cl^-, so i is 2 in the ideal case. Delta T_f = 2 times 1.86 times 0.100 = 0.372 degrees Celsius, and the solution freezes at minus 0.372.

Compare 0.100 m glucose, which does not dissociate: i is 1, and the depression is only 0.186 degrees. Same molality, half the effect, because glucose supplies half as many particles.

Measured van't Hoff factors for sodium chloride come out a little below 2, near 1.9, because at real concentrations some Na⁺ and Cl^- spend part of their time associated as ion pairs rather than moving independently. The deviation grows with concentration and with ionic charge, and it is a good early example of a model that is exact only in the limit of infinite dilution.

Common misconceptions

• "Molarity and molality are interchangeable." Molarity is per litre of solution and changes with temperature; molality is per kilogram of solvent and does not. Freezing point work uses molality precisely because the sample is being cooled.
• "Diluting a solution reduces the number of moles of solute." It reduces the concentration by adding solvent. The moles are unchanged, which is exactly why M₁V₁ = M₂V₂ works.
• "Saturated means concentrated." A saturated solution is one in equilibrium with undissolved solute. Silver chloride's saturated solution is about 10^-5 M, which is extremely dilute and completely saturated at the same time.
• "Endothermic dissolving should not happen." Ammonium nitrate absorbs 25.7 kJ per mole and dissolves readily, because the entropy gain outweighs the enthalpy cost.
• "Mixing a 0.2 M and a 0.1 M solution gives 0.15 M." Only if the volumes are equal. Add moles, add volumes, then divide: 50 mL of 0.2 M with 150 mL of 0.1 M gives 0.125 M.
• "A higher absorbance always means a more accurate reading." Above an absorbance of about 1, very little light reaches the detector and the Beer-Lambert relationship stops being reliably linear. Dilute and read again.

The takeaway

• Converting mass percent to molarity requires the density, because one is per unit mass and the other is per unit volume.
• Molarity is moles per litre of solution; molality is moles per kilogram of solvent and is the temperature-independent choice.
• Dilution conserves moles, so M₁V₁ = M₂V₂, and mixing requires adding moles and volumes separately.
• Enthalpy of solution is lattice energy plus hydration enthalpy, and it can be positive, as in ammonium nitrate at plus 25.7 kJ per mole, or negative, as in calcium chloride at minus 81.3.
• Gas solubility falls with temperature and rises with partial pressure, which is Henry's law.
• Beer-Lambert absorbance is proportional to concentration and path length, so one standard is enough to calibrate an unknown, provided the absorbance stays roughly below 1.
• Colligative properties count particles: a 0.100 m NaCl solution depresses the freezing point twice as much as 0.100 m glucose.

Sources

1. OpenStax. (2019). Molarity (Section 3.3). In Chemistry 2e. Rice University. openstax.org
2. OpenStax. (2019). The dissolution process (Section 11.1). In Chemistry 2e. Rice University. openstax.org
3. OpenStax. (2019). Solubility (Section 11.3). In Chemistry 2e. Rice University. openstax.org
4. OpenStax. (2019). Colligative properties (Section 11.4). In Chemistry 2e. Rice University. openstax.org
5. OpenStax. (2019). Quantitative chemical analysis (Section 4.5). In Chemistry 2e. Rice University. openstax.org

A number that stopped being measured

On 20 May 2019 the mole was redefined. It had been the number of atoms in exactly 12 grams of carbon-12, a quantity that had to be determined experimentally and carried an uncertainty. Since that date, one mole contains exactly 6.02214076 times 10²³ elementary entities, by definition. The Avogadro constant is no longer measured; it is fixed, and the kilogram is now defined in terms of the Planck constant instead.

Practically nothing changed at the bench. Conceptually a great deal did: the mole is now a counting unit like the dozen, with a very large number attached, rather than a quantity tethered to a particular lump of carbon.

What the mole is for

Chemistry has a scale problem. Reactions happen between individual atoms and molecules, which combine in small whole-number ratios: two hydrogen molecules to one oxygen molecule. Laboratories work in grams, which is what a balance reads. The mole connects the two, and every quantitative calculation in the rest of this course passes through it.

Key idea: A balanced equation is a statement about numbers of particles. A balance measures mass. Molar mass is the exchange rate, and it is the only route between them.

Three conversions cover almost everything.

• Mass to moles: divide by the molar mass in grams per mole.
• Moles to particles: multiply by 6.022 times 10²³ per mole.
• Moles to volume of a solution: divide by the molarity, from Lesson 8.

Write every factor as a fraction with its units and cancel them explicitly. If the units of your answer are not the units you wanted, the arithmetic is wrong regardless of what the calculator says, and catching that is the single most valuable habit in quantitative chemistry.

Where molar mass comes from

The atomic mass on a periodic table is not the mass of any single atom. It is a weighted average over the isotopes as they occur naturally on Earth. Chlorine is the standard demonstration.

Isotope	Mass (u)	Abundance	Contribution
Chlorine-35	34.96885	75.76 percent	34.96885 times 0.7576 = 26.492
Chlorine-37	36.96590	24.24 percent	36.96590 times 0.2424 = 8.961
Weighted average			35.45 u

No chlorine atom weighs 35.45 u. Roughly three in four weigh 35 and one in four weighs 37, and 35.45 is what a large sample averages to. This is why a mass spectrum of chlorine gas shows peaks rather than a single line, and why the periodic table value is quoted to four significant figures while an isotope mass is quoted to seven.

Percent composition, worked

Problem. What percentage of iron(III) oxide, by mass, is iron?

Step 1. Molar mass. Two irons at 55.845 give 111.69. Three oxygens at 15.999 give 48.00. Total 159.69 g per mole.

Step 2. Divide the part by the whole. 111.69 divided by 159.69 = 0.6994, or 69.94 percent iron.

That number is worth a moment. An ore body assayed at 60 percent Fe₂O₃ contains 0.60 times 69.94, or about 42 percent iron by mass, and the difference between quoting the oxide and quoting the metal is the difference between two very different valuations of the same rock.

From percentages back to a formula

Percent composition can be run in reverse to find a formula. The procedure is fixed.

1. Assume exactly 100 g of compound, so each percentage becomes a mass in grams.
2. Convert each mass to moles by dividing by that element's molar mass.
3. Divide every mole value by the smallest of them.
4. If the results are not close to whole numbers, multiply all of them by the smallest integer that makes them so.

Worked. A compound is 40.00 percent carbon, 6.71 percent hydrogen and 53.29 percent oxygen by mass. Its molar mass is 180.16 g per mole.

Step 1 and 2. In 100 g there are 40.00 g C, 6.71 g H and 53.29 g O.

• C: 40.00 divided by 12.011 = 3.330 mol
• H: 6.71 divided by 1.008 = 6.657 mol
• O: 53.29 divided by 15.999 = 3.331 mol

Step 3. Divide by 3.330: C gives 1.000, H gives 1.999, O gives 1.000. The empirical formula is CH₂O.

Step 4. Scale to the molecular formula. The empirical formula mass is 12.011 + 2(1.008) + 15.999 = 30.03 g per mole. Divide the true molar mass by it: 180.16 divided by 30.03 = 6.00. Multiply every subscript by 6 to get C₆H₁₂O₆, which is glucose.

Remember: An empirical formula is the simplest whole-number ratio and cannot be found from percentages alone beyond that. Formaldehyde CH₂O, acetic acid C₂H₄O₂ and glucose C₆H₁₂O₆ all give exactly the same percent composition. You need a separate measurement of the molar mass to choose between them.

Combustion analysis, which measures what it cannot see

The cleverest version of this calculation never weighs the elements at all. Burn a known mass of an organic compound in excess oxygen, trap the carbon dioxide and the water separately, and weigh those. Every carbon in the sample ends up in the CO₂; every hydrogen ends up in the H₂O.

Problem. A 0.2500 g sample of a hydrocarbon burns completely, producing 0.7485 g of CO₂ and 0.4086 g of H₂O. Find its empirical formula.

Step 1. Carbon. Moles of CO₂ = 0.7485 g divided by 44.01 g per mol = 0.017008 mol. Each CO₂ holds one carbon, so moles of C = 0.017008. Mass of C = 0.017008 times 12.011 = 0.2043 g.

Step 2. Hydrogen. Moles of H₂O = 0.4086 g divided by 18.02 g per mol = 0.022675 mol. Each water holds two hydrogens, so moles of H = 0.045350. Mass of H = 0.045350 times 1.008 = 0.0457 g.

Step 3. Check for oxygen in the original sample. Carbon plus hydrogen accounts for 0.2043 + 0.0457 = 0.2500 g, which is the entire sample. There is no oxygen in the compound. This step is not optional: if the masses had fallen short, the difference would be oxygen, and skipping the check is the most common way this calculation goes wrong.

Step 4. Ratio. 0.017008 to 0.045350. Divide both by the smaller: 1.000 to 2.666. Multiply by 3: 3 to 8. The empirical formula is C₃H₈, propane.

The core of it: The oxygen that appears in the products came from the air, not from the sample. That is exactly why you cannot weigh the oxygen in the products and call it the sample's oxygen; you have to find it by difference.

Hydrates and a second use for mass difference

Copper(II) sulfate crystals are blue because each formula unit carries five water molecules, written CuSO₄.5H₂O. Heat them and the water leaves as vapour, and the residue is white anhydrous CuSO₄. Weighing before and after gives the water content directly.

Suppose 2.500 g of the blue hydrate leaves 1.597 g of white solid. Water lost = 0.903 g, which is 0.903 divided by 18.02 = 0.0501 mol. Anhydrous CuSO₄ has a molar mass of 159.6 g per mole, so 1.597 divided by 159.6 = 0.01001 mol. The ratio of water to salt is 0.0501 divided by 0.01001 = 5.00, confirming the pentahydrate.

Common misconceptions

• "A mole is a unit of mass." It is a count. One mole of hydrogen atoms weighs about 1 g and one mole of lead atoms weighs 207 g; the number of atoms is identical.
• "The atomic mass on the periodic table is the mass of one atom." It is an abundance-weighted average over isotopes. No chlorine atom has a mass of 35.45 u.
• "Percent composition gives you the molecular formula." It gives the empirical formula only. CH₂O and C₆H₁₂O₆ have identical percent compositions, and a molar mass measurement is required to distinguish them.
• "In combustion analysis, the oxygen in the CO₂ came from the compound." It came from the excess oxygen you burned the sample in. Any oxygen in the sample is found by subtracting the carbon and hydrogen masses from the sample mass.
• "Avogadro's number is measured to more decimal places every year." Since 2019 it is exact by definition, at 6.02214076 times 10²³ per mole. What used to be measured is now fixed.
• "Moles of molecules equals moles of atoms." One mole of CO₂ contains one mole of carbon atoms and two moles of oxygen atoms, three moles of atoms in total.

Looking back

• Since 2019 the mole has been exactly 6.02214076 times 10²³ entities by definition, not by measurement.
• Molar mass is the exchange rate between the particle counts a balanced equation describes and the masses a balance reads.
• Periodic table masses are abundance-weighted isotope averages; chlorine's 35.45 comes from 75.76 percent chlorine-35 and 24.24 percent chlorine-37.
• Percent composition yields an empirical formula through four fixed steps; a molar mass is then needed to reach the molecular formula.
• Combustion analysis routes every carbon into CO₂ and every hydrogen into H₂O, and any oxygen in the sample must be found by mass difference.
• Carry the units through every conversion. Wrong units mean wrong arithmetic, every time.

Sources

1. OpenStax. (2019). Formula mass and the mole concept (Section 3.1). In Chemistry 2e. Rice University. openstax.org
2. OpenStax. (2019). Determining empirical and molecular formulas (Section 3.2). In Chemistry 2e. Rice University. openstax.org
3. National Institute of Standards and Technology. (n.d.). SI redefinition: the mole. U.S. Department of Commerce. nist.gov
4. National Institute of Standards and Technology. (n.d.). CODATA value: Avogadro constant. U.S. Department of Commerce. physics.nist.gov
5. National Center for Biotechnology Information. (n.d.). PubChem periodic table of elements. National Library of Medicine. pubchem.ncbi.nlm.nih.gov

Two reactants, and only one of them will be gone at the end

Put 5.00 g of aluminium foil into a beaker with 25.0 g of copper(II) chloride dissolved in water. The blue colour fades, the solution warms, and reddish copper metal settles out. When it stops, one reactant has been used up entirely and the other is still sitting there. Which one, and how much is left over, can be worked out before you touch a beaker.

That calculation is the subject of this lesson, and it is the one that industry cares about most. A reagent you buy and do not consume is money on the floor.

The map, and why every route goes through moles

A balanced equation counts particles. Grams cannot be compared directly to grams, because different substances weigh different amounts per particle. So every stoichiometry problem takes the same route:

mass of A, to moles of A, to moles of B, to mass of B

Only the middle step uses the balanced equation. The first and last steps use molar masses. Miss the balancing and every number afterwards is wrong, which is why the first thing to do with any stoichiometry problem is to check that the equation is balanced before reading the rest of the question.

The point: The coefficients in a balanced equation are mole ratios. They are never mass ratios and never volume ratios, except for gases at the same temperature and pressure, which Lesson 11 explains.

Worked: the aluminium and copper chloride problem

Balanced equation. 2 Al(s) + 3 CuCl₂(aq) gives 2 AlCl₃(aq) + 3 Cu(s).

Step 1. Moles of each reactant.

• Al: 5.00 g divided by 26.98 g per mol = 0.1853 mol
• CuCl₂: 25.0 g divided by 134.45 g per mol = 0.1860 mol

The two are almost identical in moles, which is exactly the situation where guessing fails. The equation does not consume them one for one.

Step 2. Find the limiting reagent by asking what each could do alone. If all the CuCl₂ reacted, it would need 0.1860 mol times (2 mol Al divided by 3 mol CuCl₂) = 0.1240 mol of aluminium. You have 0.1853 mol, which is more than enough. So the aluminium is in excess and the copper(II) chloride is limiting.

Step 3. Theoretical yield of copper. Run the calculation from the limiting reagent only. 0.1860 mol CuCl₂ times (3 mol Cu divided by 3 mol CuCl₂) = 0.1860 mol Cu. Mass = 0.1860 times 63.55 g per mol = 11.8 g of copper.

Step 4. Excess remaining. Aluminium used = 0.1240 mol. Aluminium left = 0.1853 minus 0.1240 = 0.0613 mol, which is 0.0613 times 26.98 = 1.65 g of aluminium foil still in the beaker.

Step 5. Check. The masses of everything consumed should account for the masses of everything produced. Aluminium consumed 3.35 g plus copper chloride 25.0 g gives 28.35 g in; copper 11.8 g plus aluminium chloride (0.1240 mol times 133.34 = 16.5 g) gives 28.3 g out. Mass is conserved to the precision of the data.

The same reaction, one number changed

Now start with 1.00 g of aluminium and the same 25.0 g of copper(II) chloride. Nothing about the chemistry changes. Everything about the answer does.

• Al: 1.00 divided by 26.98 = 0.03707 mol
• CuCl₂: still 0.1860 mol

How much CuCl₂ would the aluminium need? 0.03707 times (3 divided by 2) = 0.05561 mol, and you have 0.1860 mol available. This time the aluminium is limiting. Copper produced = 0.03707 times (3 divided by 2) = 0.05561 mol, which is 3.53 g. Copper chloride left over = 0.1860 minus 0.05561 = 0.1304 mol, or 17.5 g.

Why this matters: The limiting reagent is not the one you have less of by mass, and not the one with the smaller coefficient. It is whichever one runs out first once the mole ratio is applied, and the only way to know is to do the comparison.

Theoretical, actual and percent yield

The 11.8 g of copper above is the theoretical yield: what a perfect reaction with perfect recovery would give. Real laboratories do worse. Product sticks to glassware, side reactions consume starting material, the reaction does not run to completion, and some product is lost in filtering and drying.

percent yield = (actual yield divided by theoretical yield) times 100

If the filtered, dried copper weighed 10.4 g, the percent yield is (10.4 divided by 11.8) times 100 = 88.1 percent.

A percent yield above 100 is not a triumph, it is a diagnosis. Almost always it means the product was not fully dried and you weighed solvent along with it, or an impurity precipitated with the product.

Stoichiometry in solution: the titration calculation

When the reagents are dissolved, moles come from concentration times volume rather than from mass divided by molar mass. Everything else is identical.

Problem. 25.00 mL of hydrochloric acid of unknown concentration requires 18.65 mL of 0.1050 M sodium hydroxide to reach the endpoint. Find the acid's concentration.

Step 1. Equation. HCl(aq) + NaOH(aq) gives NaCl(aq) + H₂O(l). The ratio is 1 to 1.

Step 2. Moles of the known. 0.01865 L times 0.1050 mol per L = 1.958 times 10^-3 mol NaOH.

Step 3. Mole ratio. One to one, so moles of HCl = 1.958 times 10^-3.

Step 4. Divide by the acid's volume. 1.958 times 10^-3 mol divided by 0.02500 L = 0.07833 M.

Change the acid to sulfuric and step 3 changes with it: H₂SO₄ plus 2 NaOH gives Na₂SO₄ plus 2 H₂O, so moles of acid are half the moles of base, and the same titre would give 0.03917 M. Forgetting the 2 here is the most common error in titration arithmetic, and it produces an answer exactly twice too large.

Why industry cares about the excess

Deliberately running one reactant in excess is normal practice, and it is not waste when it is chosen well. In the industrial synthesis of ammonia, hydrogen and nitrogen are fed in near their stoichiometric 3 to 1 ratio and recycled, because both are expensive. In esterification, chemists often flood the reaction with the cheaper alcohol to push an equilibrium toward the product, a use of excess you will be able to justify quantitatively after Lesson 15.

The measure that captures this is atom economy: the mass of the desired product divided by the total mass of all reactants, expressed as a percentage. A reaction can have a 95 percent yield and a terrible atom economy if most of the mass of the reagents ends up in a by-product. Percent yield asks how well the reaction was run; atom economy asks whether the reaction was worth choosing.

Common misconceptions

• "The limiting reagent is the one with the smaller mass." In the first example, 5.00 g of aluminium was in excess against 25.0 g of copper chloride. Compare moles after applying the mole ratio, never masses.
• "The limiting reagent is the one with the smaller coefficient." Coefficients say how the reactants are consumed relative to each other, not how much of each you happen to have.
• "Coefficients are mass ratios." They are mole ratios. In 2 H₂ plus O₂, two moles of hydrogen weigh about 4 g while one mole of oxygen weighs 32 g.
• "A yield over 100 percent means the reaction over-performed." It means the product is wet or impure. Mass cannot be created, so the extra mass came from somewhere it should not have.
• "Theoretical yield is calculated from the excess reagent." It is calculated from the limiting reagent alone; the excess stops mattering the moment the limiting reagent is used up.
• "High percent yield means an efficient process." Not on its own. Atom economy can still be poor if most of the reactant mass leaves as by-product.

What to remember

• Every stoichiometry problem runs mass to moles to moles to mass, and only the middle step uses the balanced equation.
• Find the limiting reagent by asking how much of the other reactant each one would require, then comparing with what you actually have.
• Theoretical yield is always calculated from the limiting reagent, and the excess is found by subtracting what was consumed.
• Percent yield is actual over theoretical times 100; a result above 100 signals a wet or impure product.
• In solution, moles come from molarity times volume, and the acid to base mole ratio must come from the balanced equation, not assumed to be 1 to 1.
• Percent yield measures how well a reaction was run; atom economy measures how much of the reactant mass could ever have reached the product.

Sources

1. OpenStax. (2019). Writing and balancing chemical equations (Section 4.1). In Chemistry 2e. Rice University. openstax.org
2. OpenStax. (2019). Reaction stoichiometry (Section 4.3). In Chemistry 2e. Rice University. openstax.org
3. OpenStax. (2019). Reaction yields (Section 4.4). In Chemistry 2e. Rice University. openstax.org
4. OpenStax. (2019). Quantitative chemical analysis (Section 4.5). In Chemistry 2e. Rice University. openstax.org
5. Trost, B. M. (1991). The atom economy: a search for synthetic efficiency. Science, 254(5037), 1471-1477.

A J-shaped tube and a plug of trapped air

In 1662 Robert Boyle sealed the short arm of a J-shaped glass tube, trapping a column of air, and poured mercury into the long arm. Every time he doubled the pressure on the trapped air, its volume halved. The product of pressure and volume stayed constant, and it has been called Boyle's law ever since.

Three more relationships followed, and together they collapse into one equation that covers all of them.

The four laws and the one that contains them

Law	Relationship	Held constant	Everyday version
Boyle	P times V is constant	n and T	Squeeze a syringe and the plunger resists.
Charles	V over T is constant	n and P	A balloon shrinks in the freezer.
Gay-Lussac	P over T is constant	n and V	A sealed aerosol can bursts in a fire.
Avogadro	V over n is constant	P and T	Blowing into a balloon makes it bigger.

Combine all four and you get the ideal gas law:

PV = nRT

with R the gas constant. Its value depends only on your units: 0.08206 L atm per mol per K when pressure is in atmospheres and volume in litres, or 8.314 J per mol per K in SI units. Choose the version that matches the units in the question and half the errors in this topic disappear. Temperature is always in kelvin, without exception, because the laws involve ratios of temperature and only an absolute scale makes a ratio meaningful.

Worked. What volume does 2.50 mol of any ideal gas occupy at 25.0 degrees Celsius and 1.00 atm? Convert first: T = 298.15 K. Then V = nRT divided by P = (2.50 times 0.08206 times 298.15) divided by 1.00 = 61.2 L. Notice that the identity of the gas never entered the calculation.

Molar volume, and a definition that changed

At 0 degrees Celsius and 1 atm, one mole of an ideal gas occupies 22.41 L. That number is worth knowing, and it is worth knowing that it comes with a caveat. IUPAC changed the standard pressure in 1982 from 1 atm to exactly 100 kPa, which is about 1.3 percent lower, and the molar volume under that definition is 22.71 L. Older textbooks and some examination boards still use 22.4; newer sources use 22.7. Neither is wrong, and quoting either without saying which standard you mean is.

Remember: Molar volume is a consequence of PV = nRT, not an independent fact. If you know the equation you can always regenerate the number for whatever conditions the question actually gives, and you should, because most questions are not at any standard state at all.

Density and molar mass from gas measurements

Rearranging the ideal gas law with n replaced by mass over molar mass gives a route from a density measurement to an unknown gas's identity:

M = dRT divided by P, where d is the density in grams per litre.

Worked. An unknown gas has a density of 1.80 g per litre at 25.0 degrees Celsius and 1.00 atm. M = (1.80 times 0.08206 times 298.15) divided by 1.00 = 44.0 g per mole. Carbon dioxide has a molar mass of 44.01, and so do propane and dinitrogen oxide, so this measurement narrows the field without closing it.

Dalton's law, and the water you did not intend to collect

In a mixture of gases that do not react, each gas exerts the pressure it would exert alone, and the total is the sum. So P_total equals the sum of the partial pressures, and each partial pressure equals the mole fraction times the total.

This matters immediately in the laboratory. Gases are often collected by displacing water from an inverted tube, and the gas that arrives is not pure: it is saturated with water vapour. At 25.0 degrees Celsius, water's vapour pressure is 23.8 torr, and that contribution has to be removed.

Worked. Hydrogen is collected over water at 25.0 degrees Celsius. The total pressure inside the tube, once levelled with the outside, is 752 torr. What is the pressure of the hydrogen alone?

P(H₂) = 752 minus 23.8 = 728 torr. Skipping this correction overstates the amount of hydrogen by about 3 percent, and the error grows sharply with temperature because water's vapour pressure roughly doubles for every 10 degrees.

Gas stoichiometry: the airbag

A driver's side airbag inflates with nitrogen produced by detonating sodium azide: 2 NaN₃(s) gives 2 Na(s) + 3 N₂(g). How much azide is needed to fill 65.0 L at 25.0 degrees Celsius and 1.00 atm?

Step 1. Moles of nitrogen required. n = PV divided by RT = (1.00 times 65.0) divided by (0.08206 times 298.15) = 65.0 divided by 24.47 = 2.657 mol.

Step 2. Mole ratio. 2.657 mol N₂ times (2 mol NaN₃ divided by 3 mol N₂) = 1.771 mol NaN₃.

Step 3. Mass. 1.771 times 65.01 g per mol = 115 g of sodium azide.

The whole calculation is Lesson 10's map with one substitution: the gas volume replaced a mass at one end, so PV = nRT replaced a molar mass there. Everything in the middle is unchanged.

Kinetic molecular theory: why the equation works

The ideal gas law is empirical. Kinetic molecular theory explains it from five assumptions: gas particles are in constant random motion; their volume is negligible compared with the container; they exert no forces on one another; collisions are perfectly elastic; and average kinetic energy is proportional to absolute temperature.

That last assumption is the load-bearing one. Since kinetic energy is one half m v squared, particles with the same average kinetic energy but different masses must have different speeds. The root mean square speed works out to

u_rms = the square root of (3RT divided by M), with M the molar mass in kilograms per mole and R in J per mol per K.

Worked. Nitrogen at 298 K: M is 0.028014 kg per mole, so u_rms = the square root of (3 times 8.314 times 298 divided by 0.028014) = the square root of 265320 = about 515 m per second. That is faster than the speed of sound, which is what you would expect given that sound propagates by molecular collision.

Because speed goes as the inverse square root of molar mass, lighter gases escape through a pinhole faster. Graham's law states the ratio directly: the rate of effusion is inversely proportional to the square root of molar mass. Helium (4.00 g per mole) against argon (39.95) gives a ratio of the square root of (39.95 divided by 4.00), which is 3.16. Helium effuses more than three times as fast, which is why a helium balloon deflates in a day and an air-filled one lasts a week.

Where the ideal gas law fails

Two of the five assumptions are false, and they fail under opposite conditions. Define the compressibility factor Z = PV divided by nRT. An ideal gas has Z = 1 exactly, at every pressure. Real gases do not.

• At moderate pressures and low temperatures, Z falls below 1. Molecules do attract each other, so they strike the walls a little less hard than an ideal gas would, and the measured pressure is lower than predicted. This dominates when molecules are slow enough for attractions to matter.
• At high pressures, Z rises above 1. Molecules occupy real volume, so the space actually available to move in is less than the container volume, and the gas resists compression more than predicted.

The van der Waals equation corrects both effects with two constants per gas: a for the strength of intermolecular attraction, and b for the volume the molecules themselves occupy. Reading a table of them is a direct check on Lesson 7.

Gas	a (L2 atm per mol2)	b (L per mol)	Dominant intermolecular force
He	0.0342	0.0237	Very weak dispersion only
N2	1.39	0.0391	Dispersion, small molecule
CO2	3.59	0.0427	Stronger dispersion, larger molecule
H2O	5.46	0.0305	Hydrogen bonding

Water has the largest a in the table and one of the smallest b values: a small molecule with unusually strong attractions, exactly as Lesson 7 predicted. Helium sits at the other extreme, which is why it behaves nearly ideally down to very low temperatures.

The upshot: A gas approaches ideal behaviour at low pressure and high temperature, because both corrections shrink. Room conditions are close enough to ideal for most calculations, and a gas near its condensation point is not.

Common misconceptions

• "You can use degrees Celsius in gas law problems if you are consistent." You cannot. Doubling from 10 to 20 degrees Celsius does not double the volume; doubling from 283 K to 566 K does. Every gas law temperature is in kelvin.
• "Molar volume is 22.4 L per mole." Only at 0 degrees Celsius and exactly 1 atm. Under the IUPAC standard of 100 kPa it is 22.71, and at room temperature and pressure it is about 24.5.
• "Heavier gases have lower kinetic energy at the same temperature." Average kinetic energy depends only on temperature. Heavier molecules move more slowly at the same kinetic energy, which is why they effuse more slowly.
• "Real gases deviate because the ideal gas law is an approximation for all conditions equally." The two deviations act in opposite directions and dominate under opposite conditions: attraction pulls Z below 1, molecular volume pushes it above 1.
• "Gas collected over water is pure." It is saturated with water vapour, which contributes 23.8 torr at 25 degrees Celsius and must be subtracted from the total pressure.
• "In a mixture, the heavier gas settles to the bottom." At ordinary conditions, molecular motion keeps a gas mixture uniform. Air is not stratified into layers of nitrogen and oxygen in a room.

Recap

• PV = nRT contains Boyle, Charles, Gay-Lussac and Avogadro as special cases, and temperature is always in kelvin.
• Choose R to match your units: 0.08206 L atm per mol per K, or 8.314 J per mol per K.
• Molar mass follows from gas density by M = dRT over P, and molar volume is a consequence of the equation rather than a separate fact.
• Partial pressures add, and gas collected over water carries a water vapour partial pressure that must be subtracted.
• Kinetic molecular theory gives u_rms as the square root of 3RT over M, which produces Graham's law: lighter gases effuse faster by the inverse square root of molar mass.
• Real gases deviate because molecules attract, which lowers Z below 1 at moderate pressure, and because they occupy volume, which raises Z above 1 at high pressure. Ideality improves at low pressure and high temperature.

Sources

1. OpenStax. (2019). Relating pressure, volume, amount, and temperature: the ideal gas law (Section 9.2). In Chemistry 2e. Rice University. openstax.org
2. OpenStax. (2019). Stoichiometry of gaseous substances, mixtures, and reactions (Section 9.3). In Chemistry 2e. Rice University. openstax.org
3. OpenStax. (2019). The kinetic-molecular theory (Section 9.5). In Chemistry 2e. Rice University. openstax.org
4. OpenStax. (2019). Non-ideal gas behavior (Section 9.6). In Chemistry 2e. Rice University. openstax.org
5. National Institute of Standards and Technology. (n.d.). CODATA value: molar gas constant. U.S. Department of Commerce. physics.nist.gov

A polystyrene cup and a thermometer

Put 100.0 g of water in a polystyrene cup at 22.0 degrees Celsius. Stir in 5.00 g of sodium hydroxide pellets and watch the thermometer climb to 34.7 before it starts to fall again. That 12.7 degree rise is a measurement of the enthalpy of solution of sodium hydroxide, and getting from one to the other takes three lines of arithmetic.

Step 1. Heat absorbed by the solution. Treat the total mass of the solution, 105.0 g, as having water's specific heat capacity of 4.184 J per gram per degree.

q = m c (delta T) = 105.0 g times 4.184 J per g per degree times 12.7 degrees = 5579 J, or 5.58 kJ.

Step 2. Assign the sign. The solution gained that energy, so the chemical system lost it: q for the reaction is minus 5.58 kJ.

Step 3. Per mole. 5.00 g of NaOH, molar mass 40.00, is 0.125 mol. So the enthalpy of solution is minus 5.58 kJ divided by 0.125 mol = minus 44.6 kJ per mole. The accepted value is about minus 44.5 kJ per mole, so a cup, a thermometer and a balance got within half a percent.

Systems, surroundings, and getting the sign right

The system is the chemicals you are studying. The surroundings is everything else, and in a coffee cup experiment that means the water, the cup and the thermometer. Energy released by the system is absorbed by the surroundings, and the two have opposite signs.

• Exothermic: the system releases energy, the surroundings warm up, delta H is negative.
• Endothermic: the system absorbs energy, the surroundings cool down, delta H is positive.

Key idea: The thermometer measures the surroundings. Every sign error in this topic comes from reporting the surroundings' change as if it were the system's.

Two calorimeters, two different quantities

	Coffee cup calorimeter	Bomb calorimeter
Held constant	Pressure, open to the atmosphere	Volume, a sealed steel vessel
Measures	Enthalpy change, delta H	Internal energy change, delta U
Heat capacity used	Specific heat capacity of the solution	Calibrated heat capacity of the whole apparatus, in kJ per degree
Typical use	Neutralisation, dissolution, displacement in solution	Combustion, food energy content

The distinction matters because a reaction that produces gas does work pushing back the atmosphere. In an open cup that work happens and is not counted as heat, so the heat measured is delta H. In a sealed bomb no expansion is possible, no work is done, and the heat measured is delta U. The two differ by the work term, which for reactions involving gases can be several kilojoules per mole.

Worked, bomb calorimeter. Burning 1.500 g of glucose in a bomb calorimeter with a heat capacity of 3.72 kJ per degree raises the temperature by 6.32 degrees. The heat released is 3.72 times 6.32 = 23.5 kJ. Glucose has a molar mass of 180.16, so 1.500 g is 0.008326 mol, and the energy released per mole is 23.5 divided by 0.008326 = 2823 kJ per mole. The accepted enthalpy of combustion of glucose is close to minus 2803 kJ per mole.

Enthalpy is a state function, and that is the whole trick

Enthalpy depends only on the current state of a system, not on how it got there. Two consequences follow, and every remaining calculation in this lesson uses one of them.

1. Reversing a reaction reverses the sign of delta H.
2. Multiplying an equation by a factor multiplies delta H by the same factor.

Put those together and you can add equations like algebra. That is Hess's law: if a reaction can be written as the sum of others, its enthalpy change is the sum of theirs.

Worked. Find the standard enthalpy of formation of ethane, 2 C(s) + 3 H₂(g) giving C₂H₆(g), from three measurable combustions.

• (i) C(s) + O₂(g) gives CO₂(g), delta H = minus 393.5 kJ
• (ii) H₂(g) + half O₂(g) gives H₂O(l), delta H = minus 285.8 kJ
• (iii) C₂H₆(g) + three and a half O₂(g) gives 2 CO₂(g) + 3 H₂O(l), delta H = minus 1560 kJ

Step 1. Decide what each equation must do. The target has 2 C on the left, so take (i) twice. It has 3 H₂ on the left, so take (ii) three times. It has C₂H₆ on the right, so reverse (iii).

Step 2. Add the enthalpies with those operations applied.

2 times (minus 393.5) + 3 times (minus 285.8) + (plus 1560) = minus 787.0 minus 857.4 plus 1560 = minus 84.4 kJ per mole.

Step 3. Check the bookkeeping. Adding the manipulated equations gives 2 C + 3 H₂ + 3.5 O₂ on the left and C₂H₆ + 3.5 O₂ on the right. The oxygen and the carbon dioxide and water cancel completely, leaving exactly the target reaction. The tabulated enthalpy of formation of ethane is about minus 84.0 kJ per mole.

The shortcut: standard enthalpies of formation

Rather than hunting for equations to add, chemists tabulate one number per compound: the standard enthalpy of formation, delta H_f standard, which is the enthalpy change when one mole of the compound forms from its elements in their standard states. Elements in their standard states have a value of exactly zero, by definition. Then

delta H reaction = the sum of (n times delta H_f) for products, minus the sum of (n times delta H_f) for reactants.

Worked: the combustion of methane. CH₄(g) + 2 O₂(g) gives CO₂(g) + 2 H₂O(l). Using delta H_f values of minus 74.6 for CH₄, minus 393.5 for CO₂, minus 285.8 for liquid water, and zero for O₂:

Products: minus 393.5 + 2 times (minus 285.8) = minus 965.1 kJ. Reactants: minus 74.6 + 0 = minus 74.6 kJ. Difference: minus 965.1 minus (minus 74.6) = minus 890.5 kJ per mole of methane.

Note the state symbol. If the water leaves as vapour instead, using minus 241.8 for H₂O(g) gives minus 802.5 kJ. The 88 kJ difference is the energy needed to vaporise two moles of water, and it is why a condensing boiler recovers more energy from the same fuel than an older one that sends the water out of the flue as steam.

Bond enthalpies: a cruder route, and a check

Breaking bonds costs energy; forming bonds releases it. So

delta H is approximately the sum of the bonds broken, minus the sum of the bonds formed.

Run methane combustion this way, with the water as vapour so that only gas-phase bonds are involved. Bonds broken: 4 C-H at 413 kJ each, giving 1652, plus 2 O=O at 498 each, giving 996; total 2648 kJ. Bonds formed: 2 C=O at 799 each, giving 1598, plus 4 O-H at 463 each, giving 1852; total 3450 kJ. Then delta H = 2648 minus 3450 = minus 802 kJ.

Compare that with the minus 802.5 kJ the formation enthalpies gave for the same reaction with gaseous water. Two entirely independent data sets, agreeing to within a kilojoule.

In short: The bond enthalpy method is an estimate, because tabulated bond enthalpies are averages over many compounds and the true C-H bond energy in methane differs from the one in ethanol. It works only for gas-phase species, since it accounts for nothing that happens between molecules.

Common misconceptions

• "Exothermic means the reaction gets hot." The surroundings get hot. The system is releasing energy, so delta H for the system is negative, and confusing the two flips every sign in the problem.
• "Breaking bonds releases energy." Breaking bonds always costs energy; forming bonds always releases it. A combustion is exothermic because the bonds formed in the products are stronger than the bonds broken in the reactants.
• "A bomb calorimeter measures delta H." It measures delta U, because the sealed vessel prevents any expansion work. A coffee cup, open to the atmosphere, measures delta H.
• "Delta H depends on the route the reaction takes." It does not, and Hess's law is the direct consequence. A catalyst changes the route and leaves delta H untouched.
• "Elements have an enthalpy of formation you look up." Elements in their standard states are defined as zero. Graphite is zero and diamond is not, because graphite is the standard state of carbon.
• "Bond enthalpy calculations should give the exact answer." They are estimates built from averages and apply only to gases. Agreement within a few kilojoules is a good result, not a guarantee.

Putting it together

• q = m c (delta T) converts a measured temperature change into an energy, and dividing by moles converts that into an enthalpy per mole.
• The thermometer reads the surroundings, so the system's sign is the opposite of the temperature change.
• A coffee cup at constant pressure measures delta H; a bomb at constant volume measures delta U.
• Enthalpy is a state function, so reactions can be reversed, scaled and added, which is Hess's law.
• Standard enthalpies of formation give a one-line route: products minus reactants, with elements at zero.
• Bond enthalpies estimate a gas-phase reaction enthalpy as bonds broken minus bonds formed, and agree with the formation route to within a few kilojoules.

Sources

1. OpenStax. (2019). Energy basics (Section 5.1). In Chemistry 2e. Rice University. openstax.org
2. OpenStax. (2019). Calorimetry (Section 5.2). In Chemistry 2e. Rice University. openstax.org
3. OpenStax. (2019). Enthalpy (Section 5.3). In Chemistry 2e. Rice University. openstax.org
4. OpenStax. (2019). Standard thermodynamic properties for selected substances (Appendix G). In Chemistry 2e. Rice University. openstax.org
5. National Institute of Standards and Technology. (n.d.). NIST Chemistry WebBook, SRD 69. U.S. Department of Commerce. webbook.nist.gov

A cold pack that should not work

Snap an instant cold pack and it falls to a few degrees above freezing while sitting in a room at 22 degrees Celsius. Inside, ammonium nitrate is dissolving and absorbing 25.7 kJ per mole from its surroundings, including from your hand. Nothing pushed it. It happened on its own, against the flow of energy that enthalpy alone would predict.

Lesson 12 gave you the energy accounting and no way to tell whether a reaction will actually happen. Something else is deciding, and this lesson names it.

Entropy is a count, not a mess

Entropy, symbol S, is often glossed as disorder, and that gloss causes more trouble than it saves. The precise version is that entropy measures the number of distinguishable microscopic arrangements, called microstates, that produce the same observable state. Ludwig Boltzmann's formulation says entropy is proportional to the logarithm of that number.

Take four gas molecules in a box divided into two halves. There is exactly one arrangement with all four on the left, and six arrangements with two on each side. Nothing prefers the middle; there are simply more ways to be there. Scale that to 10²³ molecules and the number of ways to be spread out exceeds the number of ways to be bunched by a factor so vast that the gas never spontaneously collects in one half.

Why this matters: Entropy increases not because nature likes disorder, but because there are overwhelmingly more arrangements that look spread out than arrangements that look sorted. It is a counting argument, and that is why it is inviolable at scale.

Reading the sign of delta S off an equation

Three rules cover almost every case, in order of how much they matter.

1. Count moles of gas. Gases have far more accessible arrangements than liquids or solids, so any change in the number of gas molecules dominates. 2 KClO₃(s) giving 2 KCl(s) + 3 O₂(g) goes from zero moles of gas to three: strongly positive.
2. Watch the phase. Solid to liquid to gas is increasing entropy at every step, and the second step is much larger than the first. Water's standard molar entropy is 70.0 J per mol per K as a liquid and 188.8 as a vapour.
3. Count particles and complexity. More particles, or more complex particles with more ways to vibrate and rotate, means more entropy. Dissolving an ionic solid usually raises entropy because two ions in solution have more freedom than one formula unit in a lattice.

Standard molar entropies differ from formation enthalpies in one important way: they are absolute, not relative. The third law of thermodynamics fixes the entropy of a perfect crystal at absolute zero as exactly zero, so every substance has a real, positive S value, elements included. Graphite is 5.7 J per mol per K, nitrogen gas is 191.6, and carbon dioxide is 213.8.

Two entropies, one universe

The second law says a process is spontaneous when the entropy of the universe increases. The universe here means the system plus its surroundings, and the surroundings' share depends on how much heat the system gave them and at what temperature:

delta S surroundings = minus delta H system divided by T

The temperature in the denominator is why an exothermic reaction is more decisive in the cold. Releasing 50 kJ into surroundings at 200 K raises their entropy far more than releasing it at 1000 K, in the same way that a shout matters more in a library than in a stadium.

Multiply the total entropy change of the universe by minus T and you get a quantity that refers only to the system, which is the whole point:

delta G = delta H minus T (delta S)

Gibbs free energy is negative exactly when the universe's entropy increases. It lets you judge spontaneity from properties of the system alone, without ever calculating anything about the surroundings.

delta H	delta S	Spontaneous when	Example
Negative	Positive	Always, at every temperature	Combustion of octane
Positive	Negative	Never, at any temperature	Ozone forming from oxygen at room temperature
Negative	Negative	At low temperature	Water freezing
Positive	Positive	At high temperature	Ice melting, limestone decomposing

Worked: why ice melts at exactly 0 degrees Celsius

Melting ice absorbs 6.01 kJ per mole and increases entropy by 22.0 J per mol per K. Put those into the Gibbs equation at three temperatures, keeping units consistent by working in joules.

• At 263 K (minus 10 degrees Celsius): delta G = 6010 minus (263 times 22.0) = 6010 minus 5786 = plus 224 J. Positive, so ice does not melt.
• At 273.15 K (0 degrees Celsius): delta G = 6010 minus (273.15 times 22.0) = 6010 minus 6009 = about zero. Neither direction is favoured, so ice and water coexist. This is what a melting point is.
• At 298 K (25 degrees Celsius): delta G = 6010 minus (298 times 22.0) = 6010 minus 6556 = minus 546 J. Negative, so ice melts.

The melting point is not a property that had to be looked up. It is the temperature at which delta H divided by delta S equals T, and 6010 divided by 22.0 is 273 K.

Worked: the crossover temperature of a lime kiln

Calcium carbonate decomposes to quicklime and carbon dioxide: CaCO₃(s) gives CaO(s) + CO₂(g). Using formation enthalpies of minus 1207.6, minus 635.1 and minus 393.5 kJ per mole, and standard entropies of 92.9, 38.1 and 213.8 J per mol per K:

delta H = (minus 635.1 minus 393.5) minus (minus 1207.6) = plus 179.0 kJ. Strongly endothermic.

delta S = (38.1 + 213.8) minus 92.9 = plus 159.0 J per K. Positive, because a mole of gas appeared where there was none.

At 298 K: delta G = 179000 minus (298 times 159.0) = 179000 minus 47382 = plus 131.6 kJ. Limestone is entirely stable on a building at room temperature, as cathedrals demonstrate.

Crossover: set delta G to zero. T = delta H divided by delta S = 179000 divided by 159.0 = 1126 K, or about 853 degrees Celsius. Industrial lime kilns run above 900 degrees Celsius, comfortably past the crossover, which is exactly what the calculation says they must do.

Worked: the tension inside the Haber process

N₂(g) + 3 H₂(g) gives 2 NH₃(g), with delta H of minus 92.2 kJ. Entropies: NH₃ 192.8, N₂ 191.6, H₂ 130.7 J per mol per K.

delta S = 2(192.8) minus [191.6 + 3(130.7)] = 385.6 minus 583.7 = minus 198.1 J per K. Four moles of gas became two, so the sign is no surprise.

delta G at 298 K = minus 92200 minus (298 times minus 198.1) = minus 92200 plus 59034 = minus 33.2 kJ. Favourable at room temperature.

Crossover: T = minus 92200 divided by minus 198.1 = 465 K, about 192 degrees Celsius. Above that the reaction becomes non-spontaneous. Yet industrial plants run at roughly 400 to 500 degrees Celsius, far above the crossover. They do it because at 192 degrees the reaction is thermodynamically favourable and kinetically hopeless, so slow that no ammonia would be made. The plant trades away yield to buy rate, then claws yield back with 150 to 300 atmospheres of pressure and an iron catalyst. Lesson 14 supplies the rate half of that argument and Lesson 15 the pressure half.

Spontaneous does not mean fast

Diamond converting to graphite has a free energy change of about minus 2.9 kJ per mole at room temperature. Every diamond in existence is thermodynamically unstable and is, very slowly, becoming pencil lead. Nobody notices, because the conversion requires breaking a covalent network and the activation barrier is enormous.

Remember: Thermodynamics tells you whether a change is downhill. It says nothing whatever about how long the journey takes. A mixture of hydrogen and oxygen has a hugely negative delta G and will sit unchanged for years until a spark arrives.

Common misconceptions

• "Entropy is disorder." Entropy counts microstates. The disorder gloss fails on cases like a protein folding, where the system becomes more ordered while the water around it gains far more entropy than the protein loses.
• "Entropy always increases." The entropy of the universe increases in any spontaneous process. A system's entropy falls all the time, as when water freezes; the surroundings gain more than the system loses.
• "Exothermic reactions are spontaneous." Usually but not always. Water freezing at 25 degrees Celsius is exothermic and not spontaneous, because the entropy loss outweighs the enthalpy gain at that temperature.
• "Spontaneous means fast." Diamond to graphite is spontaneous and takes geological time. Spontaneity is about direction; rate is a separate question with a separate theory.
• "Elements have zero entropy like they have zero formation enthalpy." Standard entropies are absolute values anchored at absolute zero by the third law, so every substance has a positive S. Graphite is 5.7 J per mol per K, not zero.
• "A negative delta G means the reaction goes to completion." It means the reaction proceeds until equilibrium, which for a modestly negative delta G may leave plenty of reactant behind.

The short version

• Entropy counts the microscopic arrangements consistent with an observed state; changes in the number of moles of gas dominate the sign.
• Standard entropies are absolute, fixed by the third law, so elements have positive values.
• Spontaneity requires the entropy of the universe to increase, and delta G = delta H minus T delta S repackages that condition using system properties only.
• The four sign combinations give always, never, low temperature only, and high temperature only.
• Setting delta G to zero gives the crossover temperature, T = delta H over delta S: 273 K for ice melting, 1126 K for limestone decomposing, 465 K for ammonia synthesis.
• Thermodynamic favourability says nothing about rate: diamond to graphite is spontaneous and effectively never happens.

Sources

1. OpenStax. (2019). Spontaneity (Section 16.1). In Chemistry 2e. Rice University. openstax.org
2. OpenStax. (2019). Entropy (Section 16.2). In Chemistry 2e. Rice University. openstax.org
3. OpenStax. (2019). The second and third laws of thermodynamics (Section 16.3). In Chemistry 2e. Rice University. openstax.org
4. OpenStax. (2019). Free energy (Section 16.4). In Chemistry 2e. Rice University. openstax.org
5. OpenStax. (2019). Standard thermodynamic properties for selected substances (Appendix G). In Chemistry 2e. Rice University. openstax.org

Three experiments, one rate law

Nitrogen monoxide reacts with hydrogen: 2 NO(g) + 2 H₂(g) gives N₂(g) + 2 H₂O(g). Run it three times at the same temperature, changing only the starting concentrations, and measure how fast nitrogen appears at the very beginning of each run.

Experiment	[NO] (M)	[H2] (M)	Initial rate (M per s)
1	0.100	0.100	1.23 times 10-3
2	0.100	0.200	2.46 times 10-3
3	0.200	0.100	4.92 times 10-3

Step 1. Isolate hydrogen. Between experiments 1 and 2, [NO] is fixed and [H₂] doubles. The rate doubles, from 1.23 to 2.46. Doubling a concentration and doubling the rate means the exponent is 1: the reaction is first order in H₂.

Step 2. Isolate nitrogen monoxide. Between 1 and 3, [H₂] is fixed and [NO] doubles. The rate quadruples, from 1.23 to 4.92. Doubling and quadrupling means the exponent is 2: second order in NO.

Step 3. Write the rate law and find k. Rate = k[NO]²[H₂]. Substituting experiment 1: 1.23 times 10^-3 = k times (0.100)² times (0.100) = k times 1.00 times 10^-3, so k = 1.23.

Step 4. Units. Rate is M per s and the concentration term is M³, so k must carry M^-2 s^-1 to make the units balance. Always derive the units of k rather than memorising them; the derivation is a free check on the order.

The point: The exponents in a rate law come from experiment, never from the coefficients in the balanced equation. Here the coefficient of H₂ is 2 and the order in H₂ is 1, and no amount of staring at the equation would have told you that.

Watching one reaction over time

Initial rates compare separate runs. To follow a single run, you need an integrated rate law, which comes from integrating the rate expression with respect to time. Three cases cover almost everything.

Order	Integrated form	Straight line when you plot	Half-life	Units of k
Zero	[A] = [A]0 minus kt	[A] against t	[A]0 divided by 2k	M s-1
First	ln[A] = ln[A]0 minus kt	ln[A] against t	0.693 divided by k	s-1
Second	1 over [A] = 1 over [A]0 plus kt	1 over [A] against t	1 divided by k[A]0	M-1 s-1

The third column is how order is determined in practice. Take concentration-time data, make all three plots, and see which one is a straight line. That plot's slope gives k, and the plot that worked names the order.

Worked, first order. A reaction is first order with k = 0.0250 s^-1. How long does the concentration take to fall from 0.500 M to 0.125 M?

Route one, half-lives: t_1/2 = 0.693 divided by 0.0250 = 27.7 s. Going from 0.500 to 0.250 is one half-life; 0.250 to 0.125 is another. Two half-lives is 55.4 s.

Route two, the equation: ln(0.500 divided by 0.125) = kt, so ln 4 = 1.386 = 0.0250 t, giving t = 55.5 s. Both routes agree, and the second works for any ratio, not just powers of two.

What matters here: Half-life is independent of starting concentration only for a first-order reaction. That is why radioactive decay, which is strictly first order, has a fixed half-life, and it is why a constant half-life across a data set is diagnostic of first-order kinetics.

Why temperature matters so much

A rough rule says a reaction roughly doubles in rate for every 10 degree rise. The reason is not that molecules move about 2 percent faster, which is all the 10 degrees buys them. It is the shape of the Maxwell-Boltzmann distribution of molecular energies. Only molecules arriving with at least the activation energy, E_a, can react, and those sit in the thin high-energy tail of the distribution. Warming the sample shifts the whole curve right, and the fraction of the area beyond a fixed threshold grows disproportionately.

The Arrhenius equation makes this quantitative: k = A times e raised to the power of minus E_a over RT, where A collects the collision frequency and the fraction of collisions with the right orientation.

Worked: find the activation energy. A reaction has k = 2.0 times 10^-3 at 300 K and 8.0 times 10^-3 at 320 K. Using the two-point form, ln(k₂ over k₁) = (E_a over R) times (1 over T₁ minus 1 over T₂):

• ln(8.0 divided by 2.0) = ln 4 = 1.386
• 1 over 300 minus 1 over 320 = 20 divided by 96000 = 2.083 times 10^-4 per K
• E_a = 1.386 times 8.314 divided by 2.083 times 10^-4 = 55.3 kJ per mole

A 20 degree rise, from 300 to 320 K, quadrupled the rate. That is a 6.7 percent change in absolute temperature producing a 300 percent change in rate, and it is entirely the exponential term doing the work.

Mechanisms: what the rate law is really telling you

Almost no reaction happens in a single collision of all its reactants. A mechanism is the sequence of elementary steps that actually occurs, and for an elementary step, and only for an elementary step, the coefficients are the orders. A mechanism is acceptable only if it satisfies two tests: the steps must add to the overall equation, and the predicted rate law must match the measured one.

Return to the nitrogen monoxide reaction, whose measured rate law was k[NO]²[H₂]. Here is a proposed mechanism.

1. 2 NO gives N₂O₂, fast and reversible
2. N₂O₂ + H₂ gives N₂O + H₂O, slow
3. N₂O + H₂ gives N₂ + H₂O, fast

Test 1, do the steps add up? Cancelling N₂O₂ and N₂O, which appear once on each side, leaves 2 NO + 2 H₂ giving N₂ + 2 H₂O. It matches.

Test 2, does the rate law come out right? The slow step is rate determining, so rate = k₂[N₂O₂][H₂]. But N₂O₂ is an intermediate and cannot appear in a rate law, because you cannot control its concentration in a flask. Step 1 is a fast equilibrium, so [N₂O₂] is proportional to [NO]². Substituting gives rate proportional to [NO]²[H₂], which is exactly what was measured.

Note carefully what this does and does not prove. Agreement makes the mechanism plausible; it never makes it certain, because another mechanism might predict the same rate law. Mechanisms are supported by evidence and discarded by evidence, and they are never proved.

Intermediates and catalysts differ in their order of appearance. An intermediate is produced in an early step and consumed later, so it appears first on a right-hand side. A catalyst is consumed early and regenerated later, so it appears first on a left-hand side. Neither shows up in the overall equation.

Catalysis

A catalyst provides an alternative pathway with a lower activation energy. It does not lower the energy of the reactants or products, so it changes neither delta H nor the equilibrium constant. It speeds the forward and reverse reactions by the same factor, which means it gets you to the same equilibrium position faster and never to a better one.

The stratospheric ozone case shows the mechanism plainly. A chlorine atom, released when ultraviolet light breaks up a chlorofluorocarbon, reacts with ozone: Cl + O₃ gives ClO + O₂. Then ClO + O gives Cl + O₂. The chlorine atom is back, ready to do it again, and the net reaction is O₃ + O giving 2 O₂. Chlorine appears on the left of the first step and the right of the second: a catalyst, and one chlorine atom can destroy a very large number of ozone molecules before it is finally removed from the cycle.

Common misconceptions

• "Rate law exponents come from the balanced equation." They come from experiment. In 2 NO + 2 H₂, the order in H₂ is 1 despite a coefficient of 2. Only for a single elementary step do the coefficients equal the orders.
• "Half-life is always independent of concentration." Only for first order. A second-order half-life is 1 over k[A]₀, so it doubles every time the concentration halves.
• "A catalyst shifts the equilibrium toward products." It accelerates both directions equally. The equilibrium position is thermodynamics; the catalyst only changes how quickly you arrive.
• "Raising the temperature by 10 degrees doubles the rate because molecules move twice as fast." Average speed rises by under 2 percent. What doubles is the fraction of molecules in the high-energy tail that exceed the activation energy.
• "An intermediate can appear in the rate law." It cannot, because its concentration is not something an experimenter sets. It must be eliminated by substituting from a preceding fast equilibrium.
• "A mechanism that predicts the right rate law is proved correct." It is consistent with the data, which is the most any mechanism ever is. A different mechanism may predict the same law.

What to carry forward

• Orders are found experimentally, usually by the method of initial rates: change one concentration at a time and watch what the rate does.
• Derive the units of k from the rate law rather than memorising them; the units confirm the overall order.
• Zero, first and second order each give a different linear plot, and whichever plot is straight identifies the order and hands you k as its slope.
• A constant half-life across a run is the signature of first-order kinetics.
• The Arrhenius equation makes rate exponentially sensitive to temperature; two rate constants at two temperatures give the activation energy.
• A mechanism must add to the overall equation and reproduce the measured rate law, with intermediates eliminated through fast pre-equilibria.
• A catalyst lowers the activation energy of both directions, so it changes rate and never changes delta H or K.

Sources

1. OpenStax. (2019). Chemical reaction rates (Section 12.1). In Chemistry 2e. Rice University. openstax.org
2. OpenStax. (2019). Rate laws (Section 12.3). In Chemistry 2e. Rice University. openstax.org
3. OpenStax. (2019). Integrated rate laws (Section 12.4). In Chemistry 2e. Rice University. openstax.org
4. OpenStax. (2019). Reaction mechanisms (Section 12.6). In Chemistry 2e. Rice University. openstax.org
5. OpenStax. (2019). Catalysis (Section 12.7). In Chemistry 2e. Rice University. openstax.org

Two flasks, two starting points, one destination

Seal 1.000 mol of hydrogen and 1.000 mol of iodine vapour in a one litre flask at 448 degrees Celsius. The purple colour of the iodine fades, but it never disappears. Now seal 2.000 mol of colourless hydrogen iodide in an identical flask at the same temperature. Purple appears. Leave both long enough and the two flasks become indistinguishable: in each, 0.2196 M hydrogen, 0.2196 M iodine and 1.561 M hydrogen iodide.

Both numbers are calculated later in this lesson, from opposite directions, and they come out the same. That is the central fact about equilibrium: the destination does not depend on which end you start from.

What is happening at the molecular level

Equilibrium is not a reaction stopping. Both directions continue at full speed; they simply proceed at equal rates, so the concentrations stop changing. Label a few iodine atoms with a heavier isotope and you would find them redistributed through the hydrogen iodide within minutes, in a flask whose measured concentrations have not moved at all.

Key idea: Equilibrium means equal rates, not equal amounts. A system with 99 percent products and 1 percent reactants is at equilibrium if those percentages are stable.

The equilibrium constant

For a general reaction aA + bB giving cC + dD, the equilibrium constant is

K = ([C]^c[D]^d) divided by ([A]^a[B]^b)

Products on top, reactants on the bottom, each raised to its coefficient. Two rules govern what goes in.

• Pure solids and pure liquids are omitted. Their concentration does not change: doubling the amount of solid calcium carbonate in a flask does not make it twice as concentrated, because it is still solid calcium carbonate. So for CaCO₃(s) giving CaO(s) + CO₂(g), K is simply the carbon dioxide term.
• The solvent in a dilute solution is omitted for the same reason.

Gases can be expressed as partial pressures instead of concentrations, giving K_p, and the two are related by K_p = K_c(RT) raised to the power of the change in moles of gas. When that change is zero, as in H₂ + I₂ giving 2 HI, the two constants are numerically identical.

The size of K tells you where the reaction sits. A K of 10⁶ means essentially complete conversion; a K of 10^-6 means barely any reaction; a K near 1 means a genuine mixture at equilibrium.

Q, and knowing which way a system will move

The reaction quotient Q uses the same expression as K but with whatever concentrations you have right now, at equilibrium or not. Comparing them gives the direction.

Comparison	Meaning	Direction of net change
Q less than K	Too few products relative to equilibrium	Forward, toward products
Q equals K	At equilibrium	No net change
Q greater than K	Too many products	Reverse, toward reactants

Worked. For H₂ + I₂ giving 2 HI, K_c = 50.5 at 448 degrees Celsius. A flask contains 0.500 M H₂, 0.500 M I₂ and 1.000 M HI. Which way does it go? Q = (1.000)² divided by (0.500 times 0.500) = 1.00 divided by 0.250 = 4.00. Since 4.00 is less than 50.5, Q is below K, and the reaction runs forward, making more hydrogen iodide until Q climbs to 50.5.

The ICE table, worked from the reactant side

Problem. 1.000 mol each of H₂ and I₂ in a 1.00 L flask at 448 degrees Celsius, K_c = 50.5. Find all three equilibrium concentrations.

	H2	I2	HI
Initial	1.000	1.000	0
Change	minus x	minus x	plus 2x
Equilibrium	1.000 minus x	1.000 minus x	2x

Step 1. Substitute. 50.5 = (2x)² divided by (1.000 minus x)².

Step 2. Exploit the perfect square. Both sides are squares, so take the square root of each: 7.106 = 2x divided by (1.000 minus x). This avoids the quadratic formula entirely, and it works whenever the expression is a ratio of squares.

Step 3. Solve. 7.106 minus 7.106x = 2x, so 9.106x = 7.106 and x = 0.7804.

Step 4. Read off the concentrations. [H₂] = [I₂] = 1.000 minus 0.7804 = 0.2196 M, and [HI] = 2 times 0.7804 = 1.561 M.

Step 5. Check. (1.561)² divided by (0.2196)² = 2.437 divided by 0.04822 = 50.5. Always substitute back; it catches arithmetic slips and sign errors in one step.

The same equilibrium, approached from the other side

Now start with 2.000 M HI and nothing else. The reaction is 2 HI giving H₂ + I₂, whose constant is the reciprocal: 1 divided by 50.5 = 0.0198.

The ICE table gives HI as 2.000 minus 2x, with H₂ and I₂ each at x. So 0.0198 = x² divided by (2.000 minus 2x)². Taking square roots, 0.1407 = x divided by (2.000 minus 2x), giving 0.2814 minus 0.2814x = x, so 1.2814x = 0.2814 and x = 0.2196.

Therefore [H₂] = [I₂] = 0.2196 M and [HI] = 2.000 minus 0.4392 = 1.561 M. Identical to the previous answer, to four figures, from a completely different starting point.

Remember: Reversing a reaction inverts K. Doubling the coefficients squares it. Adding two reactions multiplies their constants. These follow from the same state-function logic as Hess's law.

Le Chatelier's principle, and what it does not do

A system at equilibrium that is disturbed shifts in the direction that partly offsets the disturbance. Four disturbances matter, and only one of them changes K.

Change	Shift	Does K change?
Add a reactant	Toward products	No
Remove a product	Toward products	No
Decrease volume, raising pressure	Toward the side with fewer moles of gas	No
Add inert gas at constant volume	No shift	No
Add a catalyst	No shift	No
Raise the temperature	In the endothermic direction	Yes

The pressure row needs a moment. Compressing the flask raises every partial pressure, but the reaction can relieve some of that by reducing the total number of gas molecules. For H₂ + I₂ giving 2 HI, both sides have two moles of gas, so squeezing the flask changes nothing about the position of equilibrium. Adding argon at constant volume changes the total pressure but not the partial pressure of any reacting species, so it also does nothing.

Temperature is different in kind. Treat heat as a reactant or a product: for an exothermic reaction, heat is a product, so warming the system pushes it backward. Because the constant itself is temperature dependent through delta G equals minus RT ln K, this is the only listed change that gives a genuinely different K rather than a new position under the same K.

The industrial compromise, finally complete

Return to ammonia synthesis, N₂ + 3 H₂ giving 2 NH₃, exothermic at minus 92.2 kJ and going from four moles of gas to two. Le Chatelier says: high pressure helps, because the product side has fewer gas molecules; low temperature helps, because the forward direction is exothermic.

Lesson 13 showed that low temperature makes the rate hopeless. Lesson 14 showed why. So a plant runs at roughly 400 to 500 degrees Celsius for rate, accepts the poorer equilibrium constant that follows, buys back conversion with 150 to 300 atmospheres of pressure, uses an iron catalyst to improve rate without touching K, and continuously condenses ammonia out of the stream, which is the Le Chatelier move of removing a product to keep pulling the equilibrium forward. Every one of those five decisions is a line from this module.

Linking K to free energy

The bridge between Lesson 13 and this one is a single equation: delta G standard = minus RT ln K. For the hydrogen iodide system at 448 degrees Celsius, which is 721 K, with K = 50.5:

delta G = minus 8.314 times 721 times ln(50.5) = minus 8.314 times 721 times 3.922 = minus 23.5 kJ per mole.

Note how modest that is. A free energy change of only minus 23.5 kJ produces a K of 50, which is decisively product-favoured but leaves 22 percent of the starting material unreacted. Large equilibrium constants do not require large free energy changes, because the relationship is logarithmic.

Common misconceptions

• "At equilibrium the reaction has stopped." Both directions continue at equal rates. Isotope labelling shows atoms exchanging in a flask whose concentrations are constant.
• "At equilibrium the amounts of reactants and products are equal." Only if K happens to be near 1. A K of 50.5 leaves seven times more product than reactant here.
• "Adding more solid shifts the equilibrium." Pure solids and liquids do not appear in K, because their concentrations are fixed. Adding limestone to a decomposition equilibrium changes nothing.
• "A catalyst increases the yield." It reaches the same equilibrium faster. Yield at equilibrium is set by K, which the catalyst does not touch.
• "Any change in pressure shifts any gaseous equilibrium." Only when the two sides differ in moles of gas. And adding an inert gas at constant volume shifts nothing at all, because no reacting partial pressure changes.
• "Temperature shifts equilibrium the same way other changes do." Temperature is the only one that changes K itself. Everything else moves the system to a new position consistent with the same K.

Pulling it together

• Equilibrium is equal forward and reverse rates, and the final state is independent of the starting side.
• K is products over reactants, each raised to its coefficient, with pure solids, pure liquids and the solvent omitted.
• Q uses the same expression with current concentrations: Q below K drives the reaction forward, Q above K drives it back.
• An ICE table converts a starting composition into an equation in x, and a ratio of perfect squares can be solved by taking square roots instead of using the quadratic formula.
• Always substitute the answer back into the K expression as a check.
• Le Chatelier predicts the direction of a shift; only a temperature change alters the value of K.
• Delta G standard equals minus RT ln K, and because the relationship is logarithmic, a modest free energy change gives a large K.

Sources

1. OpenStax. (2019). Chemical equilibria (Section 13.1). In Chemistry 2e. Rice University. openstax.org
2. OpenStax. (2019). Equilibrium constants (Section 13.2). In Chemistry 2e. Rice University. openstax.org
3. OpenStax. (2019). Shifting equilibria: Le Chatelier's principle (Section 13.3). In Chemistry 2e. Rice University. openstax.org
4. OpenStax. (2019). Equilibrium calculations (Section 13.4). In Chemistry 2e. Rice University. openstax.org
5. OpenStax. (2019). Free energy (Section 16.4). In Chemistry 2e. Rice University. openstax.org

A number your body will not let go of

Human arterial blood is held between pH 7.35 and 7.45. Drift much outside that band and enzymes stop working; below about 6.8 or above about 7.8, sustained, a person dies. Meanwhile a resting adult produces on the order of fifteen moles of carbon dioxide a day, and carbon dioxide dissolved in water makes carbonic acid. Fifteen moles of acid generated daily, and a pH band 0.1 units wide.

The mechanism that manages this is a buffer, and by the end of this lesson you will be able to calculate exactly how one behaves.

Brønsted and Lowry: a transfer, not a substance

An acid donates a proton; a base accepts one. Nothing is intrinsically an acid, because the definition describes a role in a particular reaction. When an acid gives up its proton, what remains is a base, ready to take one back, and the two are a conjugate pair.

In HF + H₂O giving F^- + H₃O⁺, hydrogen fluoride is the acid and fluoride is its conjugate base; water is the base and the hydronium ion is its conjugate acid. Water appears as a base here and as an acid when ammonia is dissolved in it, which makes it amphiprotic.

The strength relationship inside a pair is inverse and quantitative. Multiply the Ka of an acid by the Kb of its conjugate base and you always get Kw:

K_a times K_b = K_w = 1.0 times 10^-14 at 25 degrees Celsius

A strong acid therefore has an extremely weak conjugate base. Chloride does essentially nothing in water, which is why sodium chloride solutions are neutral.

pH, and the fact that 7 is not magic

Water self-ionises slightly, and the product of the two ion concentrations is fixed at any given temperature. At 25 degrees Celsius, Kw is 1.0 times 10^-14, so pure water has 10^-7 M of each ion and a pH of 7.00, with pH plus pOH equal to 14.00.

That last sentence is true only at 25 degrees Celsius. Self-ionisation is endothermic, so warming water increases Kw. At 37 degrees Celsius, body temperature, Kw is roughly 2.4 times 10^-14, and neutral water has a pH near 6.8. Blood at pH 7.4 is therefore more basic relative to neutral than the number 7.4 makes it look.

What matters here: Neutral means equal concentrations of hydronium and hydroxide, not pH 7. The number 7 is a consequence of Kw at one particular temperature.

Strong acids: no equilibrium to solve

The six common strong acids are HCl, HBr, HI, HNO₃, H₂SO₄ and HClO₄; the strong bases are the hydroxides of the group 1 metals and of calcium, strontium and barium. These ionise essentially completely, so the calculation is a single step.

Worked. 0.010 M HCl. Complete ionisation gives [H₃O⁺] = 0.010 M, so pH = minus log(0.010) = 2.00.

Weak acids: an ICE table in disguise

A weak acid ionises only partly, and how far is set by Ka. Acetic acid, the acid in vinegar, has Ka = 1.8 times 10^-5.

Problem. Find the pH and percent ionisation of 0.100 M acetic acid, written HA for brevity.

Step 1. ICE. HA starts at 0.100 and ends at 0.100 minus x. H₃O⁺ and A^- both start at essentially zero and end at x.

Step 2. Substitute. 1.8 times 10^-5 = x² divided by (0.100 minus x).

Step 3. Approximate, then justify. Because Ka is small, x will be small compared with 0.100, so take the denominator as 0.100. Then x² = 1.8 times 10^-6 and x = 1.34 times 10^-3.

Step 4. Check the approximation. x divided by the initial concentration = 1.34 times 10^-3 divided by 0.100 = 1.34 percent. The usual rule is that the approximation is acceptable below 5 percent, so it holds comfortably here. If it had failed, you would solve the full quadratic.

Step 5. pH. minus log(1.34 times 10^-3) = 2.87, and the acid is 1.34 percent ionised.

Compare that with 0.100 M hydrochloric acid, which has a pH of 1.00. Same concentration, and the hydronium concentration differs by a factor of about 75. Concentration and strength are independent variables.

Dilution does something counterintuitive

Repeat the calculation at 0.0100 M acetic acid. x² = 1.8 times 10^-5 times 0.0100 = 1.8 times 10^-7, so x = 4.24 times 10^-4 and pH = 3.37.

The solution is less acidic, as expected. But the percent ionisation has risen from 1.34 to 4.24 percent. Diluting a weak acid increases the fraction that ionises, because dilution shifts the ionisation equilibrium toward the side with more particles. Le Chatelier again, in the place where students least expect it.

Salts are not spectators

Dissolve a salt and check whether either ion is the conjugate of a weak partner.

Salt	Cation from	Anion from	Solution
NaCl	Strong base	Strong acid	Neutral
NaCH3COO	Strong base	Weak acid	Basic, because acetate takes a proton from water
NH4Cl	Weak base	Strong acid	Acidic, because ammonium gives a proton to water
NH4CH3COO	Weak base	Weak acid	Depends on which K is larger

Buffers, and the calculation that shows how well they work

A buffer is a solution containing appreciable amounts of both members of a conjugate pair. Added acid is absorbed by the base member; added base is absorbed by the acid member. Rearranging the Ka expression and taking logarithms gives the Henderson-Hasselbalch equation:

pH = pK_a + log([A^-] divided by [HA])

Worked, part 1: the buffer at rest. One litre of solution containing 0.100 mol of acetic acid and 0.100 mol of sodium acetate. pKa = minus log(1.8 times 10^-5) = 4.74. The ratio is 1, its logarithm is zero, so pH = 4.74. When the two members are equal, the pH equals the pKa exactly.

Worked, part 2: hit it with base. Add 0.010 mol of sodium hydroxide. It converts 0.010 mol of acetic acid into acetate, so the acid falls to 0.090 mol and the base rises to 0.110 mol.

pH = 4.74 + log(0.110 divided by 0.090) = 4.74 + log(1.222) = 4.74 + 0.087 = 4.83.

Worked, part 3: the control. Add the same 0.010 mol of sodium hydroxide to one litre of pure water at pH 7.00. [OH^-] becomes 0.010 M, so pOH = 2.00 and pH = 12.00.

The buffer moved 0.09 pH units. Water moved 5.00. That factor of more than fifty is what a buffer buys you, and notice that the buffer did not prevent the change; it absorbed most of it.

The upshot: A buffer works well while both members remain in reasonable supply, which in practice means a ratio between about 1 to 10 and 10 to 1, or a pH within one unit of the pKa. Choose a buffer by finding an acid whose pKa is near the pH you want.

Polyprotic acids and the blood buffer

Phosphoric acid loses three protons, each harder than the last, with pKa values near 2.1, 7.2 and 12.3. Pulling a proton off an already negative ion costs more than pulling it off a neutral molecule, so the constants fall by several orders of magnitude at each step. The middle pKa near 7.2 is why phosphate buffers are standard in biology laboratories.

Blood uses carbonic acid and hydrogen carbonate. Under the conditions in blood, the effective pKa is about 6.1, and the physiological ratio of hydrogen carbonate to dissolved carbon dioxide is about 20 to 1. Then

pH = 6.1 + log(20) = 6.1 + 1.30 = 7.4

which is the number the lesson opened with. What makes this buffer unusually powerful is that it is open: the lungs adjust how fast carbon dioxide leaves, and the kidneys adjust hydrogen carbonate. A closed buffer in a beaker eventually exhausts itself; this one is continuously topped up from both ends.

Common misconceptions

• "Strong acid means concentrated acid." Strength is the degree of ionisation; concentration is how much is dissolved. 0.100 M acetic acid at pH 2.87 and 0.100 M hydrochloric acid at pH 1.00 are equally concentrated and not equally strong.
• "pH 7 is always neutral." Neutral means equal hydronium and hydroxide concentrations. At 37 degrees Celsius that occurs near pH 6.8, because Kw rises with temperature.
• "Diluting a weak acid decreases the fraction ionised." It increases it. Acetic acid is 1.34 percent ionised at 0.100 M and 4.24 percent at 0.0100 M.
• "A buffer keeps pH constant." It resists change. The acetate buffer above still moved 0.09 units; it simply moved fifty times less than water would have.
• "pH cannot be negative or above 14." Both happen. 2 M hydrochloric acid has a pH below zero, and concentrated sodium hydroxide exceeds 14. The 0 to 14 range is a convenience, not a limit.
• "All salt solutions are neutral." Only when both ions come from a strong acid and a strong base. Ammonium chloride is acidic and sodium acetate is basic.

Summing up

• Acids donate protons and bases accept them, so every acid has a conjugate base and Ka times Kb equals Kw.
• pH 7 is neutral only at 25 degrees Celsius; Kw and therefore neutral pH shift with temperature.
• A strong acid calculation is one step; a weak acid calculation is an ICE table with a small-x approximation that must be checked against the 5 percent rule.
• 0.100 M acetic acid has a pH of 2.87 and is 1.34 percent ionised; diluting it tenfold raises the ionised fraction to 4.24 percent.
• Salt solutions are acidic, basic or neutral depending on which parent acid and base were weak.
• The Henderson-Hasselbalch equation gives buffer pH as pKa plus the logarithm of the base to acid ratio, and pH equals pKa when they are equal.
• A buffer is effective within about one pH unit of its pKa, and it resists rather than prevents change.

Sources

1. OpenStax. (2019). Bronsted-Lowry acids and bases (Section 14.1). In Chemistry 2e. Rice University. openstax.org
2. OpenStax. (2019). pH and pOH (Section 14.2). In Chemistry 2e. Rice University. openstax.org
3. OpenStax. (2019). Relative strengths of acids and bases (Section 14.3). In Chemistry 2e. Rice University. openstax.org
4. OpenStax. (2019). Buffers (Section 14.6). In Chemistry 2e. Rice University. openstax.org
5. OpenStax. (2019). Ionization constants of weak acids (Appendix H). In Chemistry 2e. Rice University. openstax.org

Six pH units in a fifth of a millilitre

Put 25.00 mL of 0.100 M hydrochloric acid in a flask and add 0.100 M sodium hydroxide from a burette. For the first 24.90 mL the pH climbs slowly, from 1.00 to about 3.70. Then, between 24.90 and 25.10 mL, one fifth of a single millilitre, it leaps to 10.30. Six and a half pH units for two drops.

That cliff is why titration works as an analytical method. The endpoint is not a gentle transition you have to judge; it is a near-vertical wall, and almost any indicator that changes colour anywhere in that range will find it to within a drop.

Why the cliff is there

pH is logarithmic, so a vertical jump on the curve is an arithmetic statement about ratios. At 24.90 mL the flask holds 0.010 mmol of excess acid in 49.90 mL, which is 2.0 times 10^-4 M, giving pH 3.70. At 25.10 mL it holds 0.010 mmol of excess base in 50.10 mL, giving pOH 3.70 and pH 10.30. The absolute amounts are tiny and identical; the pH swing is enormous because it measures the logarithm of a concentration that has crossed from one side of 10^-7 to the other.

For a strong acid with a strong base, the equivalence point sits at pH 7.00, because the salt formed, sodium chloride here, has no acid-base character at all.

The weak acid curve, point by point

Now titrate 25.00 mL of 0.100 M acetic acid with 0.100 M sodium hydroxide. The same total amount of base is needed, 2.50 mmol, but the curve looks entirely different. Four points define it.

Volume of NaOH	What is in the flask	pH
0.00 mL	Weak acid alone	2.87
12.50 mL	Equal acid and conjugate base	4.74
25.00 mL	Acetate alone, 0.0500 M	8.72
30.00 mL	Acetate plus excess hydroxide	11.96

At 0.00 mL, this is Lesson 16's weak acid calculation: x² over 0.100 equals 1.8 times 10^-5, so [H₃O⁺] is 1.34 times 10^-3 and pH is 2.87. The curve starts almost two units higher than the strong acid did.

At 12.50 mL, exactly half the acid has been converted. The flask is a buffer with equal amounts of both members, so pH equals pK_a = 4.74. This is the half-equivalence point, and it is the single most useful feature of the curve: read the pH at half the equivalence volume and you have measured the pKa of an unknown acid.

At 25.00 mL, all the acid has become acetate, now diluted into 50.00 mL, so its concentration is 0.0500 M. Acetate is a weak base with K_b = K_w divided by K_a = 1.0 times 10^-14 divided by 1.8 times 10^-5 = 5.6 times 10^-10. Then [OH^-] is the square root of (5.6 times 10^-10 times 0.0500) = the square root of 2.8 times 10^-11 = 5.3 times 10^-6, so pOH = 5.28 and pH = 8.72.

Key idea: The equivalence point of a weak acid with a strong base is basic, not neutral, because the solution at that point is a solution of the acid's conjugate base. Only a strong acid with a strong base gives pH 7.00 at equivalence.

At 30.00 mL, there is 0.500 mmol of excess hydroxide in 55.00 mL, or 9.09 times 10^-3 M. pOH = 2.04, so pH = 11.96. Past equivalence, the weak and strong curves converge, because excess strong base swamps everything else.

Choosing an indicator

An indicator is itself a weak acid whose acid and base forms have different colours, and it changes over roughly two pH units centred near its own pKa. The rule is that the indicator's range must lie inside the vertical section of the curve.

Titration	pH at equivalence	Suitable indicator
Strong acid with strong base	7.00	Almost any, since the cliff spans about 3.7 to 10.3
Weak acid with strong base	Above 7, here 8.72	Phenolphthalein, changing about 8.3 to 10.0
Strong acid with weak base	Below 7	Methyl red, changing about 4.4 to 6.2

Using phenolphthalein for a strong acid titrated with a weak base would report the endpoint far too late, because the curve has already flattened out well below the indicator's range.

Solubility as an equilibrium

CHEM 100 gave you solubility rules: nitrates always dissolve, most chlorides dissolve except silver, lead and mercury, and so on. Those rules are a summary of a quantitative fact. Every ionic solid dissolves to some extent, and the extent is an equilibrium constant.

For AgCl(s) giving Ag⁺(aq) + Cl^-(aq), the solid does not appear in the expression, so

K_sp = [Ag⁺][Cl^-]

Worked: molar solubility of silver chloride. Take K_sp as 1.8 times 10^-10, noting that published values for this salt range from about 1.6 to 1.8 times 10^-10 depending on the source, so quote the source with the number. If s mol per litre dissolves, then both ion concentrations are s, and s² = 1.8 times 10^-10, giving s = 1.34 times 10^-5 M. In grams, that is 1.34 times 10^-5 times 143.32 = 1.9 times 10^-3 g per litre, about two milligrams. "Insoluble" turns out to mean two milligrams per litre rather than zero.

Worked: a salt of the form MX₂. Suppose K_sp = 4.0 times 10^-11. Dissolving s moles gives [M²⁺] = s and [X^-] = 2s, so K_sp = s times (2s)² = 4s³. Then s³ = 1.0 times 10^-11 and s = 2.15 times 10^-4 M.

In short: You cannot compare solubilities of two salts by comparing their Ksp values unless they have the same formula type. A 1 to 1 salt and a 1 to 2 salt convert Ksp into solubility through different equations.

The common ion effect

Dissolve silver chloride in 0.10 M sodium chloride instead of pure water. The chloride is already there, so the equilibrium is pushed back toward the solid. With [Cl^-] fixed at essentially 0.10 M,

[Ag⁺] = K_sp divided by [Cl^-] = 1.8 times 10^-10 divided by 0.10 = 1.8 times 10^-9 M.

The solubility fell from 1.34 times 10^-5 to 1.8 times 10^-9 M, a factor of about 7400. This is Le Chatelier applied to a dissolution equilibrium, and it is the working principle behind washing a precipitate with a dilute solution of one of its own ions rather than with pure water.

Will a precipitate form?

Compare the ion product Q with K_sp, exactly as Q was compared with K in Lesson 15. If Q exceeds K_sp, the solution is supersaturated and solid appears until Q falls back to K_sp.

Worked. Mix 100.0 mL of 1.0 times 10^-3 M silver nitrate with 100.0 mL of 1.0 times 10^-3 M sodium chloride. Mixing doubles the volume, so each concentration halves to 5.0 times 10^-4 M. Then Q = (5.0 times 10^-4)² = 2.5 times 10^-7. Since 2.5 times 10^-7 is far greater than 1.8 times 10^-10, a precipitate forms.

Do not forget the dilution step. Combining two solutions always dilutes both, and omitting that is the commonest error in this calculation.

Why acid rain dissolves cathedrals

If the anion of a sparingly soluble salt is the conjugate base of a weak acid, adding acid removes it from solution and drags the dissolution equilibrium forward. Calcium carbonate is the case that matters. Carbonate reacts with hydronium to make hydrogen carbonate and eventually carbon dioxide and water, so every proton added takes a carbonate ion out of the equilibrium and more limestone dissolves.

That single fact accounts for limestone caves, for the pitting of marble statues in polluted air, and for the fact that seashells and coral skeletons dissolve more readily as ocean water absorbs carbon dioxide and its pH falls. Silver chloride, by contrast, is unaffected by acid, because chloride is the conjugate base of a strong acid and does not react with hydronium at all.

Worth holding on to: A salt whose anion is a weak base becomes more soluble as pH falls. A salt whose anion comes from a strong acid does not.

Common misconceptions

• "The equivalence point is always pH 7." Only for a strong acid with a strong base. A weak acid titrated with a strong base finishes basic, here at 8.72, because the flask contains the conjugate base.
• "The endpoint and the equivalence point are the same thing." The equivalence point is where the moles match; the endpoint is where the indicator changes. A well-chosen indicator makes them nearly coincide, and a badly chosen one does not.
• "A larger Ksp always means a more soluble salt." Only within the same formula type. Converting Ksp to molar solubility uses a different equation for a 1 to 1 salt than for a 1 to 2 salt.
• "Insoluble means none dissolves." Silver chloride dissolves to about two milligrams per litre. Insoluble is a practical label, not a statement that the concentration is zero.
• "Adding a common ion dissolves more solid." It dissolves less. Adding chloride pushes the silver chloride equilibrium back toward the solid, cutting solubility by a factor of thousands.
• "pH does not affect solubility." It strongly affects any salt whose anion is a weak base. Calcium carbonate dissolves in acid; silver chloride does not.

What you now know

• The vertical section of a titration curve is a logarithmic artefact of tiny excesses crossing from acid to base, which is what makes the endpoint sharp.
• A strong-strong titration reaches equivalence at pH 7.00; a weak acid with a strong base reaches it above 7, at 8.72 in the acetic acid case.
• At half the equivalence volume, pH equals pKa, which is how a titration curve measures an unknown acid's strength.
• An indicator must change colour within the vertical section, so phenolphthalein suits a weak acid with a strong base and methyl red suits the reverse.
• Ksp is an equilibrium constant with the solid omitted, and converting it into molar solubility depends on the formula type.
• A common ion suppresses solubility sharply, and comparing Q with Ksp predicts whether mixing two solutions produces a precipitate.
• Lowering the pH increases the solubility of salts whose anions are weak bases, which is why acid attacks limestone and not silver chloride.

Sources

1. OpenStax. (2019). Acid-base titrations (Section 14.7). In Chemistry 2e. Rice University. openstax.org
2. OpenStax. (2019). Precipitation and dissolution (Section 15.1). In Chemistry 2e. Rice University. openstax.org
3. OpenStax. (2019). Solubility products (Appendix J). In Chemistry 2e. Rice University. openstax.org
4. OpenStax. (2019). Hydrolysis of salts (Section 14.4). In Chemistry 2e. Rice University. openstax.org
5. National Center for Biotechnology Information. (n.d.). Silver chloride, PubChem compound summary CID 24561. National Library of Medicine. pubchem.ncbi.nlm.nih.gov

The same reaction, twice, one of them useful

Drop a strip of zinc into copper(II) sulfate solution. The blue fades, reddish copper coats the zinc, and the beaker gets warm. All the energy leaves as heat, and you cannot use any of it.

Now separate the two halves. Zinc in zinc sulfate in one beaker, copper in copper sulfate in another, a wire between the metals and a salt bridge between the solutions. The same reaction runs, but the electrons now have to travel through the wire to get from the zinc to the copper ions, and on the way they will do work. At standard conditions the cell pushes them with 1.10 volts.

Separating oxidation from reduction in space is the whole idea of electrochemistry, and it is the difference between a warm beaker and a battery.

Balancing by half-reactions

Redox equations rarely balance by inspection, because both mass and charge must come out right. Split the reaction into an oxidation half and a reduction half and the procedure becomes mechanical. In acidic solution:

1. Split into two half-reactions and balance every element except oxygen and hydrogen.
2. Balance oxygen by adding H₂O.
3. Balance hydrogen by adding H⁺.
4. Balance charge by adding electrons to the more positive side.
5. Multiply each half-reaction so the electron counts match, then add and cancel.

Worked. Permanganate oxidising iron(II) in acid.

• Reduction half: MnO₄^- gives Mn²⁺. Add 4 H₂O on the right for the oxygens, then 8 H⁺ on the left for the hydrogens. Charge is now plus 7 on the left and plus 2 on the right, so add 5 electrons to the left: MnO₄^- + 8 H⁺ + 5 e^- gives Mn²⁺ + 4 H₂O.
• Oxidation half: Fe²⁺ gives Fe³⁺ + e^-.
• Multiply the oxidation by 5 and add: MnO₄^- + 5 Fe²⁺ + 8 H⁺ gives Mn²⁺ + 5 Fe³⁺ + 4 H₂O.

Check both: 1 Mn, 5 Fe, 4 O and 8 H on each side, and total charge is plus 17 on both sides. This reaction is also a titration in disguise, since permanganate is deep purple and its products are almost colourless, so it is its own indicator.

Anatomy of a galvanic cell

Oxidation happens at the anode and reduction at the cathode, always, in every cell of either kind. In a galvanic cell, which runs spontaneously, the anode is the negative terminal.

The salt bridge is not decorative. As zinc dissolves, its beaker accumulates positive charge; as copper deposits, its beaker loses positive charge. Within moments that imbalance would stop the reaction entirely. The salt bridge lets inert ions migrate to neutralise both sides, and removing it stops the current instantly.

Cell notation compresses all of this into a line, anode on the left: Zn(s) | Zn²⁺(1 M) || Cu²⁺(1 M) | Cu(s), where a single bar is a phase boundary and the double bar is the salt bridge.

Standard reduction potentials

No single electrode's potential can be measured alone, because a voltmeter needs two connections. So everything is measured against one arbitrary reference, the standard hydrogen electrode, defined as exactly 0.000 V. Every tabulated potential is a reduction, at 1 M and 1 atm and 25 degrees Celsius.

Half-reaction (reduction)	E standard (V)
F2 + 2 e- gives 2 F-	+2.87
MnO4- + 8 H+ + 5 e- gives Mn2+ + 4 H2O	+1.51
Ag+ + e- gives Ag	+0.80
Cu2+ + 2 e- gives Cu	+0.34
2 H+ + 2 e- gives H2	0.000
Fe2+ + 2 e- gives Fe	minus 0.44
Zn2+ + 2 e- gives Zn	minus 0.76
Li+ + e- gives Li	minus 3.04

Read the table as a ranking of appetite for electrons. Fluorine at plus 2.87 is the strongest oxidising agent in ordinary chemistry; lithium metal at minus 3.04 is the strongest reducing agent, which is exactly why lithium is in your phone.

E standard for the cell = E standard of the cathode minus E standard of the anode. For the zinc-copper cell: 0.34 minus (minus 0.76) = 1.10 V. A positive cell potential means the reaction as written is spontaneous.

Why this matters: Potential is intensive. If you multiply a half-reaction by five to balance electrons, its potential does not change. Voltage is energy per unit charge, and scaling the reaction scales both.

Three quantities, one reaction

Cell potential, free energy and the equilibrium constant are three descriptions of the same thermodynamics, linked by two equations.

delta G standard = minus nFE standard, where n is the moles of electrons transferred and F is the Faraday constant, 96485 coulombs per mole of electrons.

Worked. For the zinc-copper cell, n is 2 and E is 1.10 V, so delta G = minus 2 times 96485 times 1.10 = minus 212267 J, or minus 212 kJ per mole. Strongly spontaneous, as the warm beaker suggested.

Combining with delta G = minus RT ln K gives, at 25 degrees Celsius, log K = nE standard divided by 0.0592. So log K = 2 times 1.10 divided by 0.0592 = 37.2, and K is about 1.6 times 10³⁷. Zinc metal and copper ions do not coexist in any meaningful amount.

Non-standard conditions: the Nernst equation

Tabulated potentials assume 1 M solutions. Real cells drift away from that as they discharge, and the Nernst equation tracks it. At 25 degrees Celsius:

E = E standard minus (0.0592 divided by n) times log Q

Worked. The same cell with [Cu²⁺] at 0.010 M and [Zn²⁺] at 1.0 M. Q = [Zn²⁺] divided by [Cu²⁺] = 1.0 divided by 0.010 = 100.

E = 1.10 minus (0.0592 divided by 2) times log(100) = 1.10 minus 0.0296 times 2 = 1.10 minus 0.0592 = 1.04 V.

Depleting the product-side reactant a hundredfold cost only 59 millivolts, because the relationship is logarithmic. It also explains what a flat battery is: as the reaction proceeds, Q rises toward K, and when Q equals K the potential is exactly zero. A dead battery is a cell at equilibrium.

Corrosion, and why a zinc coating protects steel

Iron rusts because Fe²⁺/Fe sits at minus 0.44 V while oxygen reduction in moist air sits well above it, so the overall process is spontaneous. Coating steel with zinc, called galvanising, protects it in a way paint cannot. Zinc's potential of minus 0.76 V is more negative than iron's, so zinc is oxidised preferentially. Scratch the coating and the exposed iron still does not rust, because the surrounding zinc continues to give up electrons on its behalf. This is sacrificial protection, and the same principle puts blocks of magnesium at minus 2.37 V on ship hulls and buried pipelines.

Electrolysis: paying to run it backwards

Apply an external voltage larger than a cell's own and you can force a non-spontaneous reaction. The definitions do not change: oxidation is still at the anode and reduction still at the cathode, but now the anode is the positive terminal, because the external supply is pulling electrons out of it.

The quantitative law is Faraday's. Charge is current times time, and moles of electrons is charge divided by F.

Worked: silver plating. A current of 2.00 A runs for 30.0 minutes through a silver nitrate solution. What mass of silver deposits?

1. Charge. Q = I times t = 2.00 A times 1800 s = 3600 C.
2. Moles of electrons. 3600 divided by 96485 = 0.03731 mol.
3. Moles of silver. Ag⁺ + e^- gives Ag, so one electron per atom: 0.03731 mol.
4. Mass. 0.03731 times 107.87 = 4.03 g.

Run identical current and time through copper(II) sulfate instead and you get 0.03731 divided by 2 = 0.01866 mol of copper, or 1.19 g, because Cu²⁺ needs two electrons per atom. Same charge, different mass, and the difference is entirely the charge on the ion.

Remember: Electrolysing molten sodium chloride gives sodium metal and chlorine. Electrolysing sodium chloride solution gives hydrogen and chlorine, because water is easier to reduce than sodium ions. The solvent is a competitor in every aqueous electrolysis, and ignoring it produces predictions that are simply wrong.

Common misconceptions

• "The anode is always negative." Oxidation defines the anode. It is negative in a galvanic cell and positive in an electrolytic cell, because in electrolysis an external supply drives the electrons.
• "Multiplying a half-reaction multiplies its potential." Potential is energy per unit charge and does not scale. Multiply the half-reaction by 5 and E is unchanged, though n in delta G equals minus nFE does change.
• "The salt bridge carries the electrons." Electrons travel through the external wire. The salt bridge carries ions, keeping each compartment electrically neutral so the reaction can continue.
• "A dead battery has run out of electrons." It has reached equilibrium. Q has risen to equal K, so the Nernst equation gives E of exactly zero.
• "Electrolysing brine produces sodium metal." It produces hydrogen at the cathode, because reducing water is easier than reducing sodium ions. Sodium metal requires the molten salt.
• "Galvanising works by sealing the surface." It works electrochemically. Zinc is oxidised in preference to iron, so a scratched galvanised surface still protects the exposed steel.

The takeaway

• Balance redox equations by half-reactions, adding water for oxygen, hydrogen ions for hydrogen, and electrons for charge.
• Oxidation is at the anode and reduction at the cathode in both cell types; only the sign of the terminals changes.
• Standard cell potential is the cathode value minus the anode value, and a positive result means a spontaneous reaction.
• Potentials are intensive and never scale with the size of the reaction.
• Delta G standard equals minus nFE standard, and log K equals nE standard divided by 0.0592 at 25 degrees Celsius.
• The Nernst equation corrects for non-standard concentrations, and a battery is flat when Q has risen to K and E has fallen to zero.
• Faraday's law converts current and time into mass deposited, dividing by the charge on the ion.

Sources

1. OpenStax. (2019). Review of redox chemistry (Section 17.1). In Chemistry 2e. Rice University. openstax.org
2. OpenStax. (2019). Galvanic cells (Section 17.2). In Chemistry 2e. Rice University. openstax.org
3. OpenStax. (2019). Electrode and cell potentials (Section 17.3). In Chemistry 2e. Rice University. openstax.org
4. OpenStax. (2019). Electrolysis (Section 17.7). In Chemistry 2e. Rice University. openstax.org
5. National Institute of Standards and Technology. (n.d.). CODATA value: Faraday constant. U.S. Department of Commerce. physics.nist.gov