Chemical equilibrium

Most reactions do not go to completion. Long before all the reactants are used up, the products begin turning back into reactants, and the two opposing changes settle down to the same speed. Nothing appears to happen after that, yet both reactions are still running. This balanced, unfinished state is chemical equilibrium, and it decides how much ammonia a Haber plant can make and how much sulphur trioxide a contact plant can make.

8.1 Physical and chemical equilibrium

A reversible change is one that can go in both directions under the same conditions; it is written with a double half-arrow, ⇌. When the two opposing changes go at the same rate in a closed system, the system is in equilibrium.

Physical equilibrium involves only a change of physical state — no new substance is formed.

Physical equilibrium	Example	Condition
Solid ⇌ Liquid	H₂O(s) ⇌ H₂O(l)	melting point, 0 °C at 1 atm
Liquid ⇌ Vapour	H₂O(l) ⇌ H₂O(g)	any closed vessel; fixes the vapour pressure
Solid ⇌ Vapour	I₂(s) ⇌ I₂(g)	sublimation, as in camphor or naphthalene
Solute ⇌ Solution	sugar(s) ⇌ sugar(aq)	a saturated solution
Gas ⇌ Solution	CO₂(g) ⇌ CO₂(aq)	a sealed bottle of soda

Chemical equilibrium involves a chemical reaction, so new substances are formed. Examples:

• N₂(g) + 3H₂(g) ⇌ 2NH₃(g)
• 2SO₂(g) + O₂(g) ⇌ 2SO₃(g)
• H₂(g) + I₂(g) ⇌ 2HI(g)
• CH₃COOH(l) + C₂H₅OH(l) ⇌ CH₃COOC₂H₅(l) + H₂O(l)
• CaCO₃(s) ⇌ CaO(s) + CO₂(g) (in a closed vessel)

8.2 Dynamic nature of chemical equilibrium

At the start of a reaction only reactants are present, so the forward rate is large and the backward rate is zero. As reactants are consumed the forward rate falls; as products accumulate the backward rate rises. Sooner or later the two rates become equal.

Equilibrium is dynamic, not static. Its characteristics are:

1. It is reached only in a closed system at constant temperature.
2. The forward and backward reactions never stop; they merely go at equal rates. Isotopic labelling proves this — label the iodine as I-131 in 2HI and the label soon appears in I₂.
3. All observable properties (concentration, colour, pressure, density) become constant with time.
4. The same equilibrium state can be reached from either direction — from pure reactants or from pure products — provided the temperature and total composition are the same.
5. A catalyst does not change the position of equilibrium; it only makes the system reach it faster, because it speeds up both reactions equally.

8.3 Law of mass action

The molar concentration of a substance is called its active mass, written in square brackets: [A] in mol L⁻¹. For the general reversible reaction

aA + bB ⇌ cC + dD

the law of mass action gives the two rates as

8.4 Expression for equilibrium constant and its importance

At equilibrium the two rates are equal, r_f = r_b:

K_c is the equilibrium constant in terms of concentration. If all species are gases we may use partial pressures instead:

The units of K_c are (\text{mol L}^{-1})^{\Delta n} and of K_p are (\text{atm})^{\Delta n}, where \Delta n = (c+d) - (a+b) is the change in the number of gaseous moles. When \Delta n = 0 both are dimensionless.

Reaction	K_c expression	\Delta n	Units of K_c
H₂ + I₂ ⇌ 2HI	[HI]²/([H₂][I₂])	0	none
N₂ + 3H₂ ⇌ 2NH₃	[NH₃]²/([N₂][H₂]³)	−2	mol⁻² L²
PCl₅ ⇌ PCl₃ + Cl₂	[PCl₃][Cl₂]/[PCl₅]	+1	mol L⁻¹
2SO₂ + O₂ ⇌ 2SO₃	[SO₃]²/([SO₂]²[O₂])	−1	mol⁻¹ L
CaCO₃(s) ⇌ CaO(s) + CO₂(g)	[CO₂]	+1	mol L⁻¹

Importance of the equilibrium constant

1. Extent of reaction. A large K (say >10^{3}) means the equilibrium mixture is almost all products; a small K (<10^{-3}) means the reaction hardly proceeds. K \approx 1 means comparable amounts of both.
2. Direction of reaction. Compute the reaction quotient Q using the same expression but with the actual concentrations. If Q < K the reaction moves forward; if Q > K it moves backward; if Q = K it is already at equilibrium.
3. Calculation of equilibrium concentrations and of the yield of a product.
4. K depends only on temperature — not on pressure, volume, initial concentrations or the presence of a catalyst.
5. Reversing a reaction replaces K by 1/K; multiplying the equation through by n replaces K by K^{n}.

8.5 Relationship between K_p and K_c

For an ideal gas, PV = nRT, so the partial pressure of a gas is related to its molar concentration by

Substituting this for every species in the expression for K_p:

Three cases follow at once:

\Delta n	Relation	Example
\Delta n = 0	K_p = K_c	H₂ + I₂ ⇌ 2HI
\Delta n > 0	K_p > K_c	PCl₅ ⇌ PCl₃ + Cl₂
\Delta n < 0	K_p < K_c	N₂ + 3H₂ ⇌ 2NH₃

8.6 Le Chatelier's Principle

Change applied	Direction of shift	Reason
Increase concentration of a reactant	forward	system consumes the added reactant
Remove a product as it forms	forward	system replaces what was removed
Increase total pressure (decrease volume)	towards the side with fewer gaseous moles	that reduces the pressure again
Decrease pressure	towards the side with more gaseous moles	—
Increase temperature	in the endothermic direction	that absorbs the added heat
Decrease temperature	in the exothermic direction	—
Add a catalyst	no shift	both rates are speeded up equally
Add an inert gas at constant volume	no shift	partial pressures are unchanged

Only a change of temperature changes the value of K. Concentration and pressure changes move the position of equilibrium but leave K the same.

Application 1 — Haber process. N₂(g) + 3H₂(g) ⇌ 2NH₃(g); ΔH = −92.4 kJ mol⁻¹

Since \Delta n = -2, a high pressure (about 200 atm) pushes the equilibrium towards ammonia. Since the forward reaction is exothermic, a low temperature would give a better yield — but at low temperature the reaction is far too slow, so a compromise optimum temperature of about 450–500 °C is used, with a finely divided iron catalyst promoted by molybdenum. Ammonia is condensed out continuously, which pulls the equilibrium further forward.

Application 2 — Contact process. 2SO₂(g) + O₂(g) ⇌ 2SO₃(g); ΔH = −196 kJ mol⁻¹

Again \Delta n = -1 and the forward reaction is exothermic, so high pressure and low temperature favour SO₃. In practice the equilibrium yield is already above 95 % at 1–2 atm, so high pressure is not worth its cost; the plant runs at 400–450 °C over a V₂O₅ catalyst with an excess of air (Le Chatelier applied to concentration).

Application 3 — physical equilibrium. Ice ⇌ water is accompanied by a decrease in volume, so increasing the pressure melts ice. This is why a wire loaded at both ends cuts slowly through a block of ice — the phenomenon of regelation.

Chapter summary

• Chemical equilibrium is the state of a reversible reaction in a closed system where the forward and backward rates are equal. It is dynamic, reachable from either side, and unaffected in position by a catalyst.
• Law of mass action: rate ∝ product of active masses (molar concentrations), each raised to its stoichiometric coefficient.
• For aA + bB ⇌ cC + dD, K_c = \dfrac{[C]^{c}[D]^{d}}{[A]^{a}[B]^{b}} and K_p = \dfrac{p_C^{c}p_D^{d}}{p_A^{a}p_B^{b}}; pure solids and pure liquids are omitted.
• K_p = K_c(RT)^{\Delta n} with \Delta n the change in gaseous moles; K_p = K_c when \Delta n = 0.
• K measures the extent of reaction and depends only on temperature; comparing the reaction quotient Q with K predicts the direction of change.
• Le Chatelier: a disturbed equilibrium shifts so as to oppose the disturbance. High pressure favours the side with fewer gas moles; a temperature rise favours the endothermic direction.
• Haber: N₂ + 3H₂ ⇌ 2NH₃, ΔH = −92.4 kJ mol⁻¹, run at ≈200 atm and 450–500 °C over promoted iron. Contact: 2SO₂ + O₂ ⇌ 2SO₃, ΔH = −196 kJ mol⁻¹, at 400–450 °C over V₂O₅.

Practice questions

Group A — Multiple choice (1 mark each)

1. For the reaction N₂(g) + 3H₂(g) ⇌ 2NH₃(g), the relation between K_p and K_c is [1]aK_p = K_cbK_p = K_c RTcK_p = K_c(RT)^{-2}dK_p = K_c(RT)^{2}
2. Which of the following will not change the position of an equilibrium? [1]aadding a catalystbchanging the temperaturecchanging the pressure when \Delta n \ne 0dremoving a product
3. The value of the equilibrium constant depends on [1]ainitial concentrationsbpressurectemperaturedthe catalyst used
4. In the reaction CaCO₃(s) ⇌ CaO(s) + CO₂(g), the equilibrium constant K_c is [1]a[CaO][CO₂]/[CaCO₃]b[CO₂]c1/[CO₂]d[CaO][CO₂]
5. If Q > K_c for a reacting mixture, the reaction will [1]ago forwardbgo backwardcalready be at equilibriumdstop completely

Answers to Group A

1. (c) — \Delta n = 2 - 4 = -2.

2. (a) — a catalyst speeds up both directions equally.

3. (c) — K is a function of temperature alone.

4. (b) — pure solids have constant active mass and are omitted.

5. (b) — there is too much product, so the reverse reaction dominates until Q falls to K_c.

Group B — Short answer (5 marks each)

1. State the law of mass action and use it to derive the expression for the equilibrium constant K_c of the reaction aA + bB ⇌ cC + dD. [5]
2. Give any five characteristics of dynamic chemical equilibrium. [5]
3. Derive the relationship K_p = K_c(RT)^{\Delta n} and state when K_p = K_c. [5]
4. 1 mol of ethanoic acid and 1 mol of ethanol were mixed and allowed to reach equilibrium at 298 K. At equilibrium 2/3 mol of ester had formed. Calculate K_c for the esterification. [5]
5. For 2SO₂(g) + O₂(g) ⇌ 2SO₃(g), K_c = 4.0\ \text{mol}^{-1}\text{L} at 1000 K. Calculate K_p at the same temperature. [5]

Answers to Group B

1. Statement as in §8.3, then equate k_f[A]^a[B]^b = k_r[C]^c[D]^d at equilibrium and rearrange to K_c = k_f/k_r = [C]^c[D]^d/([A]^a[B]^b).

2. Any five from §8.2: closed system; equal forward and backward rates; reactions continue (dynamic); measurable properties constant; attainable from either direction; catalyst does not shift it; K fixed at a given temperature.

3. As derived in §8.5, using p_i = [i]RT. K_p = K_c when \Delta n = 0.

4. CH₃COOH + C₂H₅OH ⇌ CH₃COOC₂H₅ + H₂O with x = 2/3. Equilibrium amounts: acid = 1/3, alcohol = 1/3, ester = 2/3, water = 2/3. Because \Delta n = 0, the volume V cancels:

K_c = \frac{(2/3)(2/3)}{(1/3)(1/3)} = \frac{4/9}{1/9} = 4

5. \Delta n = 2 - 3 = -1, T = 1000\ \text{K}, RT = 0.0821 \times 1000 = 82.1. K_p = K_c(RT)^{-1} = 4.0/82.1 = 0.0487\ \text{atm}^{-1}.

Group C — Long answer (8 marks each)

1. (a) What is meant by dynamic chemical equilibrium? Distinguish it from physical equilibrium with one example of each. [3](b) State Le Chatelier's principle and apply it to the manufacture of ammonia by the Haber process, justifying the pressure, the temperature and the use of a catalyst. [5]
2. (a) Derive K_p = K_c(RT)^{\Delta n}. [3](b) 4 mol of PCl₅ were heated in a 4 L closed vessel at 250\ ^{\circ}\text{C}. At equilibrium 50 % had dissociated into PCl₃ and Cl₂. Using an ICE table, calculate K_c and K_p. [5]

Answers to Group C

1. (a) See §8.1 and §8.2 — physical equilibrium involves only a change of state (ice ⇌ water), chemical equilibrium forms new substances (N₂ + 3H₂ ⇌ 2NH₃). (b) Statement of the principle, then: \Delta n = -2 so high pressure (200 atm) increases the yield; the forward reaction is exothermic (ΔH = −92.4 kJ mol⁻¹) so a low temperature increases the yield but lowers the rate, hence the compromise at 450–500 °C; the iron catalyst does not change the yield, only the time taken to reach it.

2(b). Dissociated = 0.50 \times 4 = 2\ \text{mol}.

	PCl₅	PCl₃	Cl₂
Initial (mol)	4	0	0
Change (mol)	−2	+2	+2
Equilibrium (mol)	2	2	2

Dividing by 4 L: [PCl_5] = [PCl_3] = [Cl_2] = 0.5\ \text{mol L}^{-1}.

K_c = \frac{(0.5)(0.5)}{0.5} = 0.5\ \text{mol L}^{-1}

With \Delta n = +1 and T = 523\ \text{K}, K_p = 0.5 \times 0.0821 \times 523 = 21.5\ \text{atm}.