Chemical equilibria
At AS you predicted which way an equilibrium moves. Now you put numbers on it. The equilibrium constants Kc and Kp measure exactly where an equilibrium lies, you can calculate them from experimental data, and they explain why changing the temperature behaves so differently from changing concentration, pressure or adding a catalyst.
The equilibrium constant Kc
In a closed system a reversible reaction settles into dynamic equilibrium: the forward and backward reactions are still happening, but at the same rate, so the concentrations of everything stop changing. For a given reaction at a given temperature, the equilibrium concentrations always fit one fixed ratio, whatever amounts you started with. That ratio is the equilibrium constant.
For the general reaction \( m\mathrm{A} + n\mathrm{B} \rightleftharpoons p\mathrm{C} + q\mathrm{D} \):
\[ K_c = \frac{[\mathrm{C}]^p\,[\mathrm{D}]^q}{[\mathrm{A}]^m\,[\mathrm{B}]^n} \]- Square brackets mean the concentration at equilibrium, in mol dm−3.
- Products go on top, reactants underneath, and each concentration is raised to the power of its number in the balanced equation.
- Kc belongs to an equation as written. Write the equation backwards and the new Kc is the reciprocal of the old one; double every number in the equation and Kc is squared. Always quote the equation with the value.
Homogeneous and heterogeneous equilibria
In a homogeneous equilibrium everything is in the same phase, for example all gases, or all dissolved in one solution. In a heterogeneous equilibrium there is more than one phase, such as a solid in contact with a gas or a solution. A pure solid (or a pure liquid that forms its own separate layer) has a fixed concentration however much of it is there, so it is left out of the Kc expression.
| Equilibrium | Type | Kc |
|---|---|---|
| H2(g) + I2(g) ⇌ 2HI(g) | homogeneous | \( \dfrac{[\mathrm{HI}]^2}{[\mathrm{H_2}][\mathrm{I_2}]} \) |
| CH3COOH(l) + C2H5OH(l) ⇌ CH3COOC2H5(l) + H2O(l) | homogeneous (one liquid mixture) | \( \dfrac{[\mathrm{CH_3COOC_2H_5}][\mathrm{H_2O}]}{[\mathrm{CH_3COOH}][\mathrm{C_2H_5OH}]} \) |
| CaCO3(s) ⇌ CaO(s) + CO2(g) | heterogeneous | \( [\mathrm{CO_2}] \) |
| Ag+(aq) + Fe2+(aq) ⇌ Ag(s) + Fe3+(aq) | heterogeneous | \( \dfrac{[\mathrm{Fe^{3+}}]}{[\mathrm{Ag^+}][\mathrm{Fe^{2+}}]} \) |
Water is left out only when it is the solvent, present in huge excess. In the esterification above, water is one of four liquids mixed in a single solution and is formed in similar amounts to the ester, so it does appear in Kc.
Working out the units of Kc
Kc has units only when the powers on the top and bottom do not cancel. Put mol dm−3 in for every concentration and simplify.
Write the expression: \( K_c = \dfrac{[\mathrm{NH_3}]^2}{[\mathrm{N_2}][\mathrm{H_2}]^3} \).
Substitute the unit: \( \dfrac{(\mathrm{mol\,dm^{-3}})^2}{(\mathrm{mol\,dm^{-3}})(\mathrm{mol\,dm^{-3}})^3} = \dfrac{(\mathrm{mol\,dm^{-3}})^2}{(\mathrm{mol\,dm^{-3}})^4} \).
Cancel: \( (\mathrm{mol\,dm^{-3}})^{-2} = \mathrm{mol^{-2}\,dm^{6}} \).
A quick check: there are 2 molecules on top and 4 underneath, so the unit is (mol dm−3)2−4.
For H2 + I2 ⇌ 2HI there are 2 on top and 2 underneath, so Kc has no units. For 2SO2 + O2 ⇌ 2SO3 it is (mol dm−3)2−3 = dm3 mol−1.
Calculating Kc from experimental data
A question usually gives the starting amounts of the reactants and one piece of information about the equilibrium mixture: the amount of one substance, or the percentage of a reactant that has reacted. The balanced equation then tells you how everything else changed. Lay it out in a table with three rows, often called an ICE table: Initial moles, Change, Equilibrium moles. Then divide by the volume to get concentrations.
- Reactant at equilibrium = moles at the start − moles that reacted.
- Product at equilibrium = moles at the start + moles that formed.
- The changes are always in the ratio of the numbers in the balanced equation.
0.500 mol of H2 and 0.500 mol of I2 are heated in a sealed 2.00 dm3 flask. At equilibrium the flask contains 0.786 mol of HI. Find Kc.
HI formed = 0.786 mol. The equation says 2 HI form for every 1 H2 and 1 I2 used, so 0.393 mol of each reactant reacted.
| H2 | I2 | HI | |
|---|---|---|---|
| Initial / mol | 0.500 | 0.500 | 0 |
| Change / mol | −0.393 | −0.393 | +0.786 |
| Equilibrium / mol | 0.107 | 0.107 | 0.786 |
| Concentration / mol dm−3 | 0.0535 | 0.0535 | 0.393 |
\( K_c = \dfrac{[\mathrm{HI}]^2}{[\mathrm{H_2}][\mathrm{I_2}]} = \dfrac{0.393^2}{0.0535 \times 0.0535} = 54.0 \), with no units.
Because the units cancel, the volumes cancel too: putting the equilibrium moles straight in gives the same answer. That shortcut only works when Kc has no units.
2.00 mol of SO2 and 1.20 mol of O2 are mixed in a 5.00 dm3 vessel with a catalyst. At equilibrium there are 1.60 mol of SO3. Find Kc.
1.60 mol of SO3 formed, so 1.60 mol of SO2 and 0.800 mol of O2 reacted (ratio 2 : 1 : 2).
Equilibrium moles: SO2 2.00 − 1.60 = 0.400; O2 1.20 − 0.800 = 0.400; SO3 1.60.
Divide by 5.00 dm3: [SO2] = 0.0800, [O2] = 0.0800, [SO3] = 0.320 mol dm−3.
\( K_c = \dfrac{[\mathrm{SO_3}]^2}{[\mathrm{SO_2}]^2[\mathrm{O_2}]} = \dfrac{0.320^2}{0.0800^2 \times 0.0800} = 200\ \mathrm{dm^3\,mol^{-1}} \)
Here the volume does not cancel. Forgetting to divide by 5.00 would give 40.0, a wrong answer.
Practical: finding Kc for an esterification
Ethanoic acid and ethanol react slowly to make ethyl ethanoate and water. A little strong acid speeds the reaction up so that equilibrium is reached in about a week at room temperature:
CH3COOH(l) + C2H5OH(l) ⇌ CH3COOC2H5(l) + H2O(l)
Method
- Use burettes to measure known volumes of ethanoic acid, ethanol and a dilute acid catalyst into a boiling tube. Stopper it, swirl and leave it for a week.
- Find how much acid the catalyst alone contributes by titrating a separate sample of the same volume of catalyst with sodium hydroxide.
- Wash the equilibrium mixture into a 250 cm3 volumetric flask with distilled water, make it up to the mark and invert it to mix.
- Pipette 25.0 cm3 portions into conical flasks and titrate with standard sodium hydroxide using phenolphthalein, until you have concordant titres.
- Starting amounts of the acid and the alcohol come from their volumes: mass = density × volume, then moles = mass ÷ Mr.
- The titration measures all the acid present: the ethanoic acid left at equilibrium plus the catalyst. Subtract the catalyst's share to find the ethanoic acid.
- Diluting with water does not spoil the result because the titration is done quickly, and the uncatalysed reaction is far too slow to shift much in that time.
- The pink end-point can fade: as the hydroxide removes ethanoic acid, the equilibrium slowly shifts left, hydrolysing some ester and releasing more acid. Take the first permanent pink.
- To check that a week was long enough, set up identical mixtures and analyse them after different times. If Kc stops changing, equilibrium had been reached.
The mixture started with 0.200 mol of ethanoic acid and 0.200 mol of ethanol. Each 25.0 cm3 portion of the diluted mixture needed 21.40 cm3 of 0.500 mol dm−3 NaOH. The catalyst alone contains 0.0400 mol of H+.
NaOH used per portion = 0.02140 dm3 × 0.500 mol dm−3 = 0.01070 mol, so the 25.0 cm3 portion held 0.01070 mol of acid.
The whole 250 cm3 held ten times as much: 0.1070 mol of acid.
Ethanoic acid at equilibrium = 0.1070 − 0.0400 = 0.0670 mol.
Ethanoic acid that reacted = 0.200 − 0.0670 = 0.133 mol, so 0.133 mol of ethanol reacted, and 0.133 mol each of ester and water formed. Ethanol left = 0.200 − 0.133 = 0.0670 mol.
\( K_c = \dfrac{0.133 \times 0.133}{0.0670 \times 0.0670} = 3.94 \), with no units, so the total volume cancels.
This ignores the water that came in with the dilute catalyst. Including it makes [H2O] slightly bigger and Kc slightly bigger: a known limitation of the method.
Partial pressures and Kp
For reactions between gases it is often easier to measure pressures than concentrations, so a second equilibrium constant, Kp, is written with partial pressures.
- Partial pressure of a gas in a mixture: the pressure it would exert if it alone filled the whole volume of the mixture. It is written pA or p(A).
- Mole fraction: \( x_\mathrm{A} = \dfrac{\text{moles of A}}{\text{total moles of gas}} \). The mole fractions of all the gases add up to 1.
- Partial pressure = mole fraction × total pressure: \( p_\mathrm{A} = x_\mathrm{A}\,P \).
- The partial pressures add up to the total pressure: \( P = p_\mathrm{A} + p_\mathrm{B} + p_\mathrm{C} + \dots \)
For \( m\mathrm{A}(g) + n\mathrm{B}(g) \rightleftharpoons p\mathrm{C}(g) + q\mathrm{D}(g) \):
\[ K_p = \frac{p_\mathrm{C}^{\,p}\; p_\mathrm{D}^{\,q}}{p_\mathrm{A}^{\,m}\; p_\mathrm{B}^{\,n}} \]Only gases appear in Kp. Solids, liquids and dissolved substances are left out, so for CaCO3(s) ⇌ CaO(s) + CO2(g), \( K_p = p_{\mathrm{CO_2}} \), with units of pressure. Units of Kp are worked out like those of Kc, using the pressure unit given in the question (atm or kPa): for N2 + 3H2 ⇌ 2NH3, Kp is in atm−2 (or kPa−2).
1.00 mol of N2O4 is allowed to reach equilibrium. 20.0% of it dissociates, and the total pressure is 1.00 atm. Find Kp.
N2O4 reacted = 0.200 mol, so NO2 formed = 0.400 mol. At equilibrium: 0.800 mol N2O4 and 0.400 mol NO2, 1.200 mol in total.
Mole fractions: \( x_{\mathrm{N_2O_4}} = 0.800/1.200 = \tfrac{2}{3} \); \( x_{\mathrm{NO_2}} = 0.400/1.200 = \tfrac{1}{3} \).
Partial pressures at 1.00 atm: \( p_{\mathrm{N_2O_4}} = \tfrac{2}{3} = 0.667 \) atm, \( p_{\mathrm{NO_2}} = \tfrac{1}{3} = 0.333 \) atm.
\( K_p = \dfrac{p_{\mathrm{NO_2}}^2}{p_{\mathrm{N_2O_4}}} = \dfrac{(1/3)^2}{2/3} = 0.167\ \mathrm{atm} \)
With the rounded values 0.333 and 0.667 you would get 0.166: another reason to keep full figures until the end.
2.00 mol of N2 and 6.00 mol of H2 reach equilibrium at a total pressure of 100 atm. The equilibrium mixture contains 1.00 mol of NH3. Find Kp.
1.00 mol of NH3 formed, so 0.500 mol of N2 and 1.50 mol of H2 reacted. At equilibrium: N2 1.50, H2 4.50, NH3 1.00; total 7.00 mol.
Partial pressures: \( p_{\mathrm{N_2}} = \tfrac{1.50}{7.00} \times 100 = 21.4 \) atm; \( p_{\mathrm{H_2}} = \tfrac{4.50}{7.00} \times 100 = 64.3 \) atm; \( p_{\mathrm{NH_3}} = \tfrac{1.00}{7.00} \times 100 = 14.3 \) atm. Check: they add up to 100 atm.
\( K_p = \dfrac{p_{\mathrm{NH_3}}^2}{p_{\mathrm{N_2}}\,p_{\mathrm{H_2}}^3} = \dfrac{14.3^2}{21.4 \times 64.3^3} = 3.58 \times 10^{-5}\ \mathrm{atm^{-2}} \)
Keep the unrounded partial pressures in your calculator; rounding to three figures first can shift the last digit.
What changes K, and what does not
For a given equation, the value of Kc or Kp depends only on temperature. Changing concentration or pressure moves the position of equilibrium but leaves K unchanged, and a catalyst changes neither.
Concentration
Add more H2 to an equilibrium mixture of H2, I2 and HI. For a moment the fraction \( \frac{[\mathrm{HI}]^2}{[\mathrm{H_2}][\mathrm{I_2}]} \) is smaller than Kc, because its bottom line has grown. The system is no longer at equilibrium. The forward reaction now runs faster than the backward one, using up H2 and I2 and making HI, until the fraction is back up to exactly Kc. The position has moved right; Kc has not changed.
Pressure
Increasing the total pressure changes every partial pressure at once, but not by the same overall factor on the top and bottom of Kp when the two sides have different numbers of gas molecules.
Write each partial pressure as mole fraction × total pressure: \( K_p = \dfrac{(x_{\mathrm{NH_3}}P)^2}{(x_{\mathrm{N_2}}P)(x_{\mathrm{H_2}}P)^3} \).
Collect the P terms: \( K_p = \dfrac{x_{\mathrm{NH_3}}^2}{x_{\mathrm{N_2}}\,x_{\mathrm{H_2}}^3} \times \dfrac{1}{P^2} \).
Double P and the factor \( 1/P^2 \) falls to a quarter. Straight after the change the right-hand side is smaller than Kp, so the mixture is not at equilibrium.
To restore Kp, the mole-fraction part must grow fourfold: more NH3 forms and N2 and H2 are used up. The position shifts right, towards fewer gas molecules, and Kp ends where it started.
If both sides have the same number of gas molecules, the P terms cancel completely and pressure has no effect on the position at all.
Catalysts
A catalyst provides an alternative route with a lower activation energy for both the forward and the backward reaction. Both speed up by the same factor, so equilibrium is reached sooner but its position and the value of K are unchanged.
Temperature
Temperature is the one condition that changes K itself. If the forward reaction is exothermic, raising the temperature makes K smaller: the equilibrium mixture contains less product. If the forward reaction is endothermic, raising the temperature makes K larger. The new value of K then forces the concentrations to adjust, which is why the position also moves. The next section explains why, using entropy.
The size of K, extent of reaction and entropy
The bigger K is, the further the reaction goes before it reaches equilibrium. As a rough guide:
| Value of K | What the equilibrium mixture is like |
|---|---|
| less than about 10−10 | essentially only reactants: the reaction does not happen to any noticeable extent |
| around 0.1 | reactants predominate |
| around 1 | similar amounts of reactants and products |
| around 10 | products predominate |
| more than about 1010 | essentially only products: the reaction goes to completion |
This links to Topic 12. The total entropy change of a reaction and its equilibrium constant are connected by
\[ \Delta S_{\text{total}} = R \ln K \]where R = 8.31 J K−1 mol−1. A positive ΔStotal means K > 1, so products are favoured; a negative ΔStotal means K < 1. A reaction with K = 1010 has ΔStotal = 8.31 × ln 1010 = +191 J K−1 mol−1. Since ΔG = −TΔStotal, the same idea can be written \( \Delta G = -RT\ln K \).
ΔStotal = ΔSsystem + ΔSsurroundings, and \( \Delta S_{\text{surroundings}} = -\dfrac{\Delta H}{T} \).
For an exothermic reaction ΔH is negative, so ΔSsurroundings is positive. As T rises, −ΔH/T gets smaller.
So ΔStotal becomes less positive (or more negative). Because ΔStotal = R ln K, ln K falls and K gets smaller: less product at equilibrium.
For an endothermic reaction ΔSsurroundings is negative, and rising T makes it less negative. ΔStotal increases, so K gets larger.
This is the quantitative reason behind the Le Chatelier rule you learned at AS: heating favours the endothermic direction.
Equilibria in industry: yield against rate
An industrial chemist wants a high yield and a fast rate at a sensible cost. Equilibrium and kinetics often pull in opposite directions, so the conditions chosen are a compromise.
| Haber process N2 + 3H2 ⇌ 2NH3, ΔH = −92 kJ mol−1 | Contact process 2SO2 + O2 ⇌ 2SO3, ΔH = −197 kJ mol−1 | |
|---|---|---|
| Temperature | about 450 °C. Lower would give a bigger K and more ammonia, but the rate would be far too slow. | about 450 °C, for the same reason. |
| Pressure | about 200 atm. Higher pressure favours the side with fewer gas molecules (4 → 2) and speeds the reaction, but costs more to build and run safely. | 1 to 2 atm. K is already so large that the yield is about 99% without high pressure, so the expense is not worth it. |
| Catalyst | iron: equilibrium reached faster, K unchanged. | vanadium(V) oxide, V2O5. |
| Other | ammonia is liquefied and removed, and unreacted gases are recycled. | SO3 is removed by absorbing it, and unreacted gases pass over further catalyst beds. |
When explaining a choice of conditions, give both sides: what the condition does to the equilibrium yield (through K or the position) and what it does to the rate and the cost. A catalyst helps the rate and lets a lower temperature be used, saving energy, but never improves the yield.
Quick checks
What are the units of Kc for 2NO2(g) ⇌ N2O4(g)?
\( K_c = [\mathrm{N_2O_4}]/[\mathrm{NO_2}]^2 \): one concentration on top and two underneath gives (mol dm−3)−1 = dm3 mol−1. mol dm−3 is the unit for the reverse equation.
Which is the correct Kp expression for C(s) + H2O(g) ⇌ CO(g) + H2(g)?
Carbon is a solid, so it does not appear; steam is a gas and does. Products go on top.
A gas mixture contains 0.60 mol of A and 1.40 mol of B at a total pressure of 250 kPa. What is the partial pressure of A?
xA = 0.60 ÷ 2.00 = 0.30, so pA = 0.30 × 250 = 75 kPa. 107 kPa comes from dividing by the moles of B instead of the total.
For an exothermic equilibrium, what happens when the temperature is raised?
Temperature is the only thing that changes K. For an exothermic reaction ΔSsurroundings = −ΔH/T falls as T rises, so ΔStotal and K fall, and the mixture holds less product.
The pressure on the equilibrium 2SO2(g) + O2(g) ⇌ 2SO3(g) is increased at constant temperature. What happens to Kp and to the amount of SO3?
Kp depends only on temperature. With 3 gas molecules on the left and 2 on the right, higher pressure shifts the position right, so the amount of SO3 rises.
A reaction has Kc = 2 × 1012 at 298 K. What can you say about it?
K above about 1010 means essentially complete reaction, and ΔStotal = R ln K is positive. K says nothing about the rate: a reaction can have a huge K and still be very slow.
0.40 mol of PCl5 is heated in a 2.0 dm3 flask: PCl5(g) ⇌ PCl3(g) + Cl2(g). At equilibrium 0.10 mol of Cl2 is present. What is Kc?
Equilibrium: PCl5 0.30 mol, PCl3 0.10 mol, Cl2 0.10 mol. Concentrations 0.15, 0.050 and 0.050 mol dm−3, so Kc = (0.050 × 0.050) ÷ 0.15 = 0.017 mol dm−3. Using moles without dividing by the volume gives 0.033.
Why is the Haber process run at about 450 °C rather than at a lower temperature?
The forward reaction is exothermic, so K is actually larger at lower temperatures. 450 °C is a compromise that gives an acceptable rate with a reasonable yield. A catalyst never changes K.