Equilibrium

A sealed flask holds 10 g of limestone heated to 1073 K. It decomposes until the carbon dioxide pressure settles at 1.16 atm.

Now open it, tip in another 90 g of limestone, reseal, and reheat to the same temperature. Ten times the solid. What is the new pressure?

Almost everyone says higher. Some say ten times higher.

The answer is 1.16 atm. Exactly what it was before.

$\mathrm{CaC}{\mathrm{O}}_3(\mathrm{s})\rightleftharpoons \mathrm{CaO}(\mathrm{s})+\mathrm{C}{\mathrm{O}}_2(\mathrm{g}),K_p=p_{\mathrm{C}{\mathrm{O}}_2}$

A pure solid has a fixed density and a fixed molar concentration. Doubling the pile does not make it more concentrated, so it cannot appear in the expression, so it cannot move the answer. The flask does not know how much limestone is in it.

That is the shape of this whole chapter. $K$ is fixed by temperature and nothing else, so when you disturb the system, the composition is what has to move.

This chapter is really two chapters wearing one name. The first half is general equilibrium and Le Chatelier's principle. The second half is ionic equilibrium: acids, bases, buffers, solubility. They look unrelated and they are the same idea applied twice.

Le Chatelier's principle is not a separate law. It is the observation that $K$ refuses to move when you disturb the system, so the composition must move instead until $Q$ equals $K$ again.

Hold that and the second half becomes bookkeeping. A buffer, a solubility product and a hydrolysis calculation are the same equilibrium expression with different names attached.

1. Dynamic Equilibrium

Equilibrium. The state in which the forward and reverse rates have become equal, so concentrations stop changing.

Concentrations stop changing. Nothing stops happening. Those are very different claims and the chapter turns on the difference.

The evidence that both directions keep running is direct, not inferred. Replace the water above a saturated salt solution with heavy water. Deuterium eventually shows up inside the crystals, although the mass of solid never changed by a milligram.

Equilibrium also requires a closed system. An open beaker never reaches liquid-vapour equilibrium, because the vapour leaves before the reverse rate can catch up.

Physical equilibria

The same idea covers phase changes and dissolution, none of which involve a chemical reaction at all.

Henry's law. The partial pressure of a gas above a solution is proportional to its mole fraction in solution.

$p=K_{H}x$

A larger $K_H$ means a less soluble gas, since $K_H$ sits on the pressure side. Gas solubility falls with rising temperature, which is why a warm fizzy drink goes flat faster than a cold one. The Solutions chapter works this quantitatively.

2. The Equilibrium Constant

For $aA+bB\rightleftharpoons cC+dD$, the law of chemical equilibrium gives

$K_c=\frac{[C{]}^c[D{]}^d}{[A{]}^a[B{]}^b}$

with every concentration measured at equilibrium. For gases, partial pressures give $K_p$, and the two are related by

$K_p=K_c(RT{)}^{\Delta n_g}$

where $\Delta n_g$ is the change in moles of gas, products minus reactants. When $\Delta n_g=0$ the two constants are numerically equal.

Trap. $\Delta n_g$ counts gases only. Solids and liquids in the equation contribute nothing to it, exactly as they contribute nothing to the expression itself.

Illustration 1

At 1073 K, $K_p=1.16$ atm for $\mathrm{CaC}{\mathrm{O}}_3(\mathrm{s})\rightleftharpoons \mathrm{CaO}(\mathrm{s})+\mathrm{C}{\mathrm{O}}_2(\mathrm{g})$. Find the mass of carbon dioxide in a 10.0 L flask at equilibrium, and state what changes if the mass of limestone is increased tenfold.

Only the gas appears, so the equilibrium pressure is 1.16 atm directly.

$n=\frac{pV}{RT}=\frac{1.16\times 10.0}{0.0821\times 1073}=\frac{11.6}{88.1}=0.132\text{mol}$

$\text{mass}=0.132\times 44=5.8\text{g}$

Increasing the limestone changes nothing at all. Not the pressure, not the mass of gas, not $K_p$. A pure solid has a fixed molar concentration set by its density, so adding more of it leaves every term in the expression untouched.

Reading the value

$K$ is a verdict on how far the reaction goes, and its magnitude is readable at a glance.

$K$ depends only on temperature. It does not shift with concentration, with pressure, or with a catalyst. This single fact is what makes Le Chatelier's principle work.

Three manipulations recur, and all three follow from the algebra of the expression.

Illustration 2

For ${\mathrm{N}}_2(\mathrm{g})+{\mathrm{O}}_2(\mathrm{g})\rightleftharpoons 2NO(\mathrm{g})$ at 298 K, $K=4.1\times 10^{-31}$. Find $K$ for $\frac{1}{2}{\mathrm{N}}_2+\frac{1}{2}{\mathrm{O}}_2\rightleftharpoons \mathrm{NO}$ and for $2NO\rightleftharpoons {\mathrm{N}}_2+{\mathrm{O}}_2$.

Halving the equation raises $K$ to the power one half.

$K^{\prime }={\left(4.1\times 10^{-31}\right)}^{1/2}=6.4\times 10^{-16}$

Reversing the original inverts it.

$K^{\prime \prime }=\frac{1}{4.1\times 10^{-31}}=2.4\times 10^{30}$

Read the last number. Nitrogen and oxygen sit together in the air at room temperature and essentially none of them combines, while nitric oxide decomposes almost completely if only it can find a route. The exhaust of a car engine contains NO not because equilibrium wants it, but because the gases were briefly at 2000 K and then cooled too fast to come back.

Illustration 3

1.00 mol of $\mathrm{PC}{\mathrm{l}}_5$ is placed in a 2.00 L vessel and 40.0 per cent of it dissociates at 500 K. Find $K_c$ and $K_p$ for $\mathrm{PC}{\mathrm{l}}_5\rightleftharpoons \mathrm{PC}{\mathrm{l}}_3+\mathrm{C}{\mathrm{l}}_2$.

Set out the amounts. Dissociated: 0.400 mol. Remaining $\mathrm{PC}{\mathrm{l}}_5$: 0.600 mol. Formed: 0.400 mol each of $\mathrm{PC}{\mathrm{l}}_3$ and $\mathrm{C}{\mathrm{l}}_2$.

Divide by 2.00 L for concentrations: 0.300, 0.200 and 0.200 M.

$K_c=\frac{(0.200)(0.200)}{0.300}=0.133\mathrm{mol}{\mathrm{L}}^{-1}$

Gas moles rise from 1 to 2, so $\Delta n_g=+1$.

$K_p=K_c(RT{)}^1=0.133\times 0.0821\times 500=5.47\text{atm}$

The units are not decoration. $K_c$ carries mol L${}^{-1}$ and $K_p$ carries atm precisely because $\Delta n_g$ is not zero, and a question that asks you to convert is testing whether you noticed.

Heterogeneous equilibria

Pure solids and pure liquids are omitted from the expression, because their concentrations are fixed properties of the substance and do not change as the reaction proceeds.

$\mathrm{CaC}{\mathrm{O}}_3(\mathrm{s})\rightleftharpoons \mathrm{CaO}(\mathrm{s})+\mathrm{C}{\mathrm{O}}_2(\mathrm{g})\Rightarrow K_p=p_{\mathrm{C}{\mathrm{O}}_2}$

The amount of limestone is irrelevant to the equilibrium pressure. That is the hook this chapter opened on, and it is a favourite exam point precisely because it feels wrong.

The reaction quotient

$Q$ has exactly the same form as $K$, but uses the concentrations at whatever moment you care about, not only at equilibrium.

This comparison answers every direction-of-shift question in the chapter without any chemical intuition at all, which is exactly why it is more reliable than intuition.

Illustration 4

For ${\mathrm{N}}_2{\mathrm{O}}_4(\mathrm{g})\rightleftharpoons 2N{\mathrm{O}}_2(\mathrm{g})$, $K_c=4.6\times 10^{-3}$ at 298 K. A vessel is found to contain $[{\mathrm{N}}_2{\mathrm{O}}_4]=0.500$ M and $[\mathrm{N}{\mathrm{O}}_2]=0.0500$ M. Which way does the reaction go?

$Q=\frac{[\mathrm{N}{\mathrm{O}}_2{]}^2}{[{\mathrm{N}}_2{\mathrm{O}}_4]}=\frac{(0.0500{)}^2}{0.500}=\frac{2.50\times 10^{-3}}{0.500}=5.00\times 10^{-3}$

$Q=5.00\times 10^{-3}$ against $K=4.6\times 10^{-3}$, so $Q>K$ and the reaction runs backward, consuming $\mathrm{N}{\mathrm{O}}_2$.

That margin is 9 per cent. No amount of staring at "0.5 M of one and 0.05 M of the other" would have told you which side of $K$ it fell on. Compute $Q$; never estimate it.

The thermodynamic link

$\Delta G^{\circ }=-RT\ln K$

A negative standard Gibbs energy change corresponds to $K>1$. This is the bridge to Chemical Thermodynamics, and it is also the reason $K$ depends on temperature: the enthalpy term inside $\Delta G^{\circ }$ is effectively divided by $T$.

Illustration 5

Use $K_c=4.6\times 10^{-3}$ for ${\mathrm{N}}_2{\mathrm{O}}_4\rightleftharpoons 2N{\mathrm{O}}_2$ at 298 K to find $\Delta G^{\circ }$.

$\Delta G^{\circ }=-(8.314)(298)\ln \left(4.6\times 10^{-3}\right)=-(2478)(-5.38)=+13.3\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$

Positive, as a $K$ below 1 requires. But notice what the positive sign does not mean: the reaction is not forbidden. It ran until $\mathrm{N}{\mathrm{O}}_2$ reached 4.6 per cent of the way there, and a sealed tube of ${\mathrm{N}}_2{\mathrm{O}}_4$ is visibly brown. $\Delta G^{\circ }$ describes the standard state; $\Delta G$ is what reaches zero at equilibrium.

3. Le Chatelier's Principle

Le Chatelier's principle. A system at equilibrium, when disturbed, shifts in the direction that partly counteracts the disturbance.

Concentration. Adding a reactant makes $Q<K$, so the system moves forward. Removing a product does the same, which is why ammonia is continuously condensed out of the Haber loop.

Pressure. Compression shifts towards the side with fewer moles of gas. When both sides have equal gas moles, pressure has no effect whatsoever.

Temperature. The only disturbance that touches $K$ itself. Heating favours the endothermic direction, so for an exothermic forward reaction, heating reduces $K$ and lowers the yield.

Catalyst. It speeds both directions equally. Equilibrium arrives sooner and sits in exactly the same place. A catalyst has never improved a yield in the history of chemistry.

Illustration 6

${\mathrm{N}}_2{\mathrm{O}}_4\rightleftharpoons 2N{\mathrm{O}}_2$ sits at equilibrium in a sealed vessel. Argon is added, (a) holding the volume constant, (b) holding the total pressure constant. Predict each case.

(a) Constant volume. Argon takes part in nothing and occupies no space that was not already there. The partial pressures of ${\mathrm{N}}_2{\mathrm{O}}_4$ and $\mathrm{N}{\mathrm{O}}_2$ are each $n_{i}RT/V$, and neither $n_i$ nor $V$ has changed. So $Q$ is unchanged, and nothing happens. The total pressure rises, and that rise is entirely argon.

(b) Constant pressure. To keep total pressure fixed while adding argon, the vessel must expand. Now $V$ has increased, so every partial pressure falls, and the system shifts towards the side with more gas moles to push back: forward, making more $\mathrm{N}{\mathrm{O}}_2$.

Trap. "Adding an inert gas increases the pressure, so the equilibrium shifts" is wrong at constant volume and right at constant pressure. The phrase that decides it is which quantity is held fixed, and questions bury that phrase deliberately.

The industrial compromise

The Haber process, ${\mathrm{N}}_2+3{\mathrm{H}}_2\rightleftharpoons 2N{\mathrm{H}}_3$, is exothermic and cuts gas moles from 4 to 2.

Le Chatelier therefore wants low temperature and high pressure. Industry runs at roughly 700 K and 200 atm.

The pressure obeys the principle. The temperature flatly disobeys it.

At the temperature Le Chatelier recommends, the rate is so slow that equilibrium would take years to arrive. A magnificent yield that never materialises is worth nothing. So the plant accepts a worse equilibrium in exchange for reaching it, recovers the ammonia by condensation, and recycles the unreacted gases.

This is the standard illustration that thermodynamics and kinetics can point in opposite directions, and examiners ask it in exactly that form.

4. Acids and Bases

Three definitions, each broader than the last.

Arrhenius fails outside water. Bronsted-Lowry handles ammonia acting as a base with no hydroxide anywhere in sight. Lewis covers $\mathrm{B}{\mathrm{F}}_3$ combining with $\mathrm{N}{\mathrm{H}}_3$, where no proton moves at all.

Conjugate pair. Two species differing by exactly one proton.

A strong acid has a weak conjugate base, and the relationship is quantitative rather than a slogan.

$K_a\times K_b=K_w\Rightarrow \mathrm{p}{K}_a+\mathrm{p}{K}_b=14\text{at }298\text{K}$

Knowing either constant gives the other immediately, which halves the number of tables to remember.

Water and the pH scale

Water self-ionises slightly, and that reaction is itself an equilibrium.

$K_w=[{\mathrm{H}}^{+}][\mathrm{O}{\mathrm{H}}^{-}]=1.0\times 10^{-14}\text{at }298\text{K}$

$\mathrm{pH}=-\log [{\mathrm{H}}^{+}],\mathrm{pH}+\mathrm{pOH}=14$

The 14 is temperature specific, not universal. Self-ionisation is endothermic, so heating shifts it forward and $K_w$ rises to about $5.1\times 10^{-13}$ at 373 K.

Both ion concentrations rise together, to about $7.2\times 10^{-7}$ M, so the pH of pure boiling water is near 6.14.

Trap. That water is still exactly neutral. Neutrality means $[{\mathrm{H}}^{+}]=[\mathrm{O}{\mathrm{H}}^{-}]$, not pH 7. The number 7 is a coincidence of 298 K.

Weak electrolytes

A weak acid ionises only partly, to a degree $\alpha$. Ostwald's dilution law relates the two.

$K_a=\frac{c{\alpha }^2}{1-\alpha }\approx c{\alpha }^2\text{for }\alpha \lesssim 0.05$

$\Rightarrow \alpha =\sqrt{\frac{K_a}{c}},[{\mathrm{H}}^{+}]=c\alpha =\sqrt{K_{a}c},\mathrm{pH}=\frac{1}{2}\left(\mathrm{p}{K}_a-\log c\right)$

So $\alpha$ rises as the solution is diluted while $K_a$ stays fixed. Dilution increases the fraction ionised and decreases the hydrogen ion concentration at the same time.

That pair of statements sounds contradictory. It is not, and the arithmetic settles it in one line.

Illustration 7

Compare 0.10 M and 0.0010 M acetic acid, with $K_a=1.8\times 10^{-5}$.

Dilution by a factor of 100 made the acid ten times more ionised and ten times less acidic.

The resolution: $\alpha$ scales as $c^{-1/2}$ while $[{\mathrm{H}}^{+}]=c\alpha$ scales as $c^{+1/2}$. A larger share of a much smaller amount is still a smaller amount. There is no paradox, only two different questions being answered by two different quantities.

Polyprotic acids ionise in stages

An acid with more than one ionisable proton loses them one at a time, each stage carrying its own constant.

The reason is electrostatic. Removing a proton from a neutral molecule is far easier than pulling a positive charge away from an ion that is already negative, and harder still from a doubly negative one.

The practical consequence: the first ionisation supplies essentially all the hydrogen ions, so the pH of a polyprotic acid can be computed from $K_{a1}$ alone. Later stages matter only for the concentrations of the intermediate anions.

5. Common Ion Effect and Buffers

Common ion effect. The suppression of the ionisation of a weak electrolyte by adding a strong electrolyte that shares one of its ions.

Add sodium acetate to acetic acid and the acetate concentration jumps, pushing $Q$ above $K_a$, so the acid ionises less. That is Le Chatelier applied to an ionic equilibrium, and it is the mechanism behind both buffers and controlled precipitation.

Buffers

Buffer. A solution that resists pH change when small amounts of acid or base are added. It holds a weak acid together with its conjugate base, or a weak base with its conjugate acid.

$\mathrm{pH}=\mathrm{p}{K}_a+\log \frac{[\text{salt}]}{[\text{acid}]}$

That is the Henderson-Hasselbalch equation, and three consequences fall straight out of it.

Basic buffers work identically with the roles reversed. Ammonia with ammonium chloride obeys

$\mathrm{pOH}=\mathrm{p}{K}_b+\log \frac{[\text{salt}]}{[\text{base}]}$

and the pH follows by subtracting from 14. Choosing a buffer for a target pH means choosing a weak acid whose $\mathrm{p}{K}_a$ lies near it, which is why phosphate and carbonate buffers dominate biology.

Illustration 8

One litre of buffer holds 0.200 mol acetic acid and 0.200 mol sodium acetate, with $K_a=1.8\times 10^{-5}$. Add 0.0100 mol of HCl. Find the new pH, and compare with adding the same HCl to a litre of pure water.

The starting pH is $\mathrm{p}{K}_a=4.74$, since the ratio is 1.

Added ${\mathrm{H}}^{+}$ is mopped up by the acetate, converting it to acetic acid mole for mole.

$[\text{acid}]=0.200+0.010=0.210,[\text{salt}]=0.200-0.010=0.190$

$\mathrm{pH}=4.74+\log \frac{0.190}{0.210}=4.74-0.043=4.70$

The same 0.0100 mol of HCl in a litre of pure water gives $[{\mathrm{H}}^{+}]=0.0100$ M and a pH of 2.00.

A hundred-fold difference in response, from a solution containing nothing exotic. The acetate is a reservoir that absorbs protons without letting them stay free, and the buffer fails only when that reservoir runs out.

Salt hydrolysis

A salt's solution need not be neutral, because its ions may react with water.

Sodium acetate gives a basic solution and ammonium chloride an acidic one. Both are the conjugate relationship at work: a weak acid has a conjugate base strong enough to matter.

The extent of hydrolysis is governed by a hydrolysis constant, which for the anion of a weak acid is $K_h=K_w/K_a$. Working the equilibrium through gives compact pH expressions.

$\mathrm{pH}=7+\frac{1}{2}\left(\mathrm{p}{K}_a+\log c\right)\text{weak acid with strong base}$

$\mathrm{pH}=7-\frac{1}{2}\left(\mathrm{p}{K}_b+\log c\right)\text{strong acid with weak base}$

$\mathrm{pH}=7+\frac{1}{2}\left(\mathrm{p}{K}_a-\mathrm{p}{K}_b\right)\text{weak acid with weak base}$

Note what vanished from the third one. The concentration has dropped out entirely.

Illustration 9

Find the pH of ammonium acetate solution and of ammonium cyanide solution. Take $\mathrm{p}{K}_a=4.74$ for acetic acid, $\mathrm{p}{K}_a=9.21$ for HCN, and $\mathrm{p}{K}_b=4.74$ for ammonia.

For ammonium acetate the two constants are equal.

$\mathrm{pH}=7+\frac{1}{2}(4.74-4.74)=7.00$

Exactly neutral, and at any concentration you like. Both ions hydrolyse, and they do it to precisely the same extent, so the two effects cancel.

For ammonium cyanide, cyanide is the conjugate base of a very weak acid.

$\mathrm{pH}=7+\frac{1}{2}(9.21-4.74)=7+2.24=9.24$

Distinctly basic. Both salts are made from a weak acid and a weak base, and swapping one partner has moved the solution more than two pH units. In this family the identity of the ions decides everything and the dilution decides nothing.

6. Solubility Product

For a sparingly soluble salt, the solid is in equilibrium with its ions and, being a pure solid, is omitted.

Trap. Comparing $K_{sp}$ values directly is legitimate only within one formula type. A 1:2 salt and a 1:1 salt with the same $K_{sp}$ have very different solubilities, because the expressions carry different powers of $s$.

Precipitation occurs when the ionic product exceeds $K_{sp}$, which is the $Q$ against $K$ comparison again in different clothes.

The common ion effect cuts solubility sharply, which is why silver chloride is far less soluble in dilute hydrochloric acid than in water, and why qualitative analysis separates cations by controlling sulphide concentration with pH.

Illustration 10

100 mL of 0.0100 M $\mathrm{AgN}{\mathrm{O}}_3$ is mixed with 100 mL of 0.0100 M ${\mathrm{K}}_2\mathrm{Cr}{\mathrm{O}}_4$. Will silver chromate precipitate? Take $K_{sp}=1.1\times 10^{-12}$ for $\mathrm{A}{\mathrm{g}}_2\mathrm{Cr}{\mathrm{O}}_4$.

Mixing doubles the volume, so every concentration halves before anything reacts.

$[\mathrm{A}{\mathrm{g}}^{+}]=0.00500\mathrm{M},[\mathrm{Cr}{\mathrm{O}}_4^{2-}]=0.00500\mathrm{M}$

The salt is $\mathrm{A}{\mathrm{g}}_2\mathrm{Cr}{\mathrm{O}}_4$, so the silver term is squared.

$Q=[\mathrm{A}{\mathrm{g}}^{+}{]}^2[\mathrm{Cr}{\mathrm{O}}_4^{2-}]=(5.00\times 10^{-3}{)}^2(5.00\times 10^{-3})=1.25\times 10^{-7}$

$Q$ exceeds $K_{sp}$ by five orders of magnitude, so a precipitate forms copiously.

Two steps decide this question and both are routinely skipped. Halve the concentrations for the dilution on mixing, and square the ion that carries a subscript 2. Miss either and the answer can flip.

Summary

Equilibrium is dynamic. Both directions keep running at equal rates in a closed system, so concentrations become constant without anything stopping.

$K$ depends only on temperature. Concentration, pressure and catalysts leave it untouched, which is precisely why disturbing the system forces the composition to change instead. Reversing an equation inverts $K$, multiplying it by $n$ raises $K$ to the power $n$, and adding equations multiplies their constants.

$K_p=K_c(RT{)}^{\Delta n_g}$ converts between the two forms, counting gases only. Pure solids and liquids are omitted from every expression, so limestone decomposition depends on the carbon dioxide pressure and not at all on how much limestone is present.

Comparing $Q$ with $K$ answers every direction question mechanically, and $\Delta G^{\circ }=-RT\ln K$ connects the chapter to thermodynamics.

Le Chatelier's principle covers concentration, pressure and temperature, with temperature alone changing $K$. An inert gas does nothing at constant volume and acts like expansion at constant pressure. A catalyst never improves a yield, and the Haber conditions are a deliberate compromise between yield and rate.

Bronsted-Lowry extends acids and bases beyond water, Lewis beyond protons entirely, and $K_a{K}_b=K_w$ links any conjugate pair. Neutrality means equal hydrogen and hydroxide concentrations, not pH 7, which is why boiling water is neutral at pH 6.14.

Ostwald's dilution law makes $\alpha$ rise on dilution while $[{\mathrm{H}}^{+}]$ falls, because $\alpha$ scales as $c^{-1/2}$ and $c\alpha$ as $c^{+1/2}$. A weak acid's pH is $\frac{1}{2}(\mathrm{p}{K}_a-\log c)$, and a polyprotic acid's pH comes from $K_{a1}$ alone.

Buffers work by the common ion effect, sit at $\mathrm{pH}=\mathrm{p}{K}_a$ when salt and acid are equal, are unaffected by dilution, and are useful within one pH unit of $\mathrm{p}{K}_a$.

Solubility product predicts precipitation whenever the ionic product exceeds it, a common ion can cut solubility by orders of magnitude, and $K_{sp}$ values are comparable only between salts of the same formula type.