H2 Chemistry · JC 1 and 2
Chemical Equilibria
Many reactions do not go to completion but reach a state where reactants and products coexist in fixed amounts. In H2 Chemistry you learn to describe this dynamic equilibrium, write and calculate equilibrium constants, and predict how changing conditions shifts the position of equilibrium. These ideas are the foundation for acid and base equilibria and solubility, and they explain the compromise conditions chosen for industrial processes such as the Haber process.
What the syllabus expects
- Explain the features of a dynamic equilibrium in a closed system
- Write expressions for Kc and Kp for homogeneous and heterogeneous equilibria, and deduce their units
- Calculate Kc, Kp and equilibrium amounts using initial, change and equilibrium (ICE) tables
- Use Le Chatelier's principle to predict the effect of changes in concentration, pressure, temperature and the presence of a catalyst
- Explain why only a change in temperature changes the value of the equilibrium constant, and relate this to the sign of ΔH
- Apply equilibrium and rate ideas to industrial processes such as the Haber process
Dynamic equilibrium
A reversible reaction in a closed system reaches dynamic equilibrium when the rate of the forward reaction equals the rate of the backward reaction. Both reactions are still happening, which is what dynamic means, but because they happen at the same rate the concentrations of all species stay constant. The concentrations are constant, not equal, and the equilibrium mixture may contain mostly reactants or mostly products. The same equilibrium can be reached from either direction. A closed system is needed so that no reactant or product can escape, otherwise the reaction would keep moving in one direction. The position of equilibrium describes the relative amounts of reactants and products, and it is this position that Le Chatelier's principle predicts.
Writing Kc and Kp
For aA + bB ⇌ cC + dD, the equilibrium constant in terms of concentration is Kc = [C]^c[D]^d ÷ ([A]^a[B]^b), where square brackets mean equilibrium concentration in mol dm⁻³. For gaseous equilibria you can also use partial pressures: Kp = p(C)^c p(D)^d ÷ (p(A)^a p(B)^b). The partial pressure of a gas is its mole fraction multiplied by the total pressure, and the partial pressures add up to the total pressure. For heterogeneous equilibria, pure solids and pure liquids are left out of the expression, because their concentrations are effectively constant. So for CaCO₃(s) ⇌ CaO(s) + CO₂(g), Kp = p(CO₂). Units come from substituting the units into the expression, so for N₂ + 3H₂ ⇌ 2NH₃, Kc has units of mol⁻² dm⁶. A large K means the position of equilibrium lies to the right, favouring products. A small K means it lies to the left.
Le Chatelier's principle
Le Chatelier's principle states that when a system at equilibrium is subjected to a change, the position of equilibrium shifts so as to reduce the effect of that change. Adding a reactant shifts the position to the right to use some of it up, and removing a product does the same. Increasing the total pressure of a gaseous system shifts the position towards the side with fewer moles of gas, and if both sides have the same number of moles of gas, pressure has no effect on the position. Adding an inert gas at constant volume does not change any partial pressure, so the position does not change. A catalyst speeds up the forward and backward reactions equally, so equilibrium is reached faster but its position is unchanged. You can also reason with the reaction quotient, Q. If a change makes Q less than K, the system reacts forward until Q equals K again.
Temperature, K and the Haber process
Changes in concentration, pressure and catalysts change the position of equilibrium but not the value of K. Only temperature changes K. Raising the temperature favours the endothermic direction. So for an exothermic forward reaction, K decreases as temperature increases, and for an endothermic forward reaction, K increases. In the Haber process, N₂ + 3H₂ ⇌ 2NH₃ with ΔH = −92 kJ mol⁻¹, a low temperature would give a higher equilibrium yield of ammonia but the rate would be too slow. A temperature of around 450 °C with an iron catalyst is a compromise between yield and rate. A high pressure of around 200 atm favours the side with fewer gas moles (4 on the left, 2 on the right) and increases the rate, but higher pressures are costly to build and run safely. Unreacted gases are recycled.
Worked example
1.00 mol of hydrogen and 1.00 mol of iodine are sealed in a 1.00 dm³ flask and heated to a temperature at which H₂(g) + I₂(g) ⇌ 2HI(g) has Kc = 64. Calculate the equilibrium concentrations of all three gases.
- Initial concentrations: [H₂] = 1.00, [I₂] = 1.00 and [HI] = 0 mol dm⁻³.
- Let x mol dm⁻³ of H₂ react. Then I₂ also falls by x and HI rises by 2x, from the 1 : 1 : 2 ratio.
- Equilibrium: [H₂] = 1.00 − x, [I₂] = 1.00 − x, [HI] = 2x.
- Kc = (2x)² ÷ (1.00 − x)² = 64. Take the square root of both sides: 2x ÷ (1.00 − x) = 8.
- So 2x = 8 − 8x, giving 10x = 8 and x = 0.80.
- Check: [HI]² ÷ ([H₂][I₂]) = 1.60² ÷ (0.20 × 0.20) = 2.56 ÷ 0.0400 = 64, which matches Kc.
Answer: [H₂] = 0.20 mol dm⁻³, [I₂] = 0.20 mol dm⁻³ and [HI] = 1.60 mol dm⁻³. Kc has no units here because the moles of gas are equal on both sides.
Common mistakes
- Including solids, or water as a solvent, in a Kc expression. Leave out pure solids and liquids, whose concentrations are effectively constant.
- Saying a change in pressure or concentration changes K. It changes the position of equilibrium. Only temperature changes K.
- Saying a catalyst increases the equilibrium yield. It speeds up both reactions equally and only shortens the time taken to reach equilibrium.
- Putting moles straight into Kc when the volume is not 1 dm³. Divide by the volume first, which matters whenever the total moles on each side differ.
- Thinking that equilibrium means equal concentrations of reactants and products. It means constant concentrations.
- Predicting a shift when an inert gas is added at constant volume. The partial pressures of the reacting gases do not change, so neither does the position.
Exam tips
- Set out every equilibrium calculation as an ICE table with initial, change and equilibrium rows, and state the volume used to convert moles to concentrations.
- In Le Chatelier answers, name the change, state the direction of the shift and give the reason in terms of reducing that change, such as 'to the side with fewer moles of gas'.
- Always check whether a question asks about the position of equilibrium, the value of K or the rate. These are three separate things and markers expect you to distinguish them.
Practice questions
1. Easy: Write the Kc expression and its units for N₂(g) + 3H₂(g) ⇌ 2NH₃(g).
Show answer
Kc = [NH₃]² ÷ ([N₂][H₂]³). Units: (mol dm⁻³)² ÷ (mol dm⁻³)⁴ = (mol dm⁻³)⁻² = mol⁻² dm⁶.
2. Medium: 1.00 mol of N₂O₄ is allowed to reach equilibrium, N₂O₄(g) ⇌ 2NO₂(g). 20% of the N₂O₄ dissociates and the total pressure at equilibrium is 1.20 atm. Calculate Kp.
Show answer
Kp = 0.20 atm. At equilibrium, N₂O₄ = 0.80 mol and NO₂ = 0.40 mol, total 1.20 mol. Partial pressures: p(N₂O₄) = 0.80/1.20 × 1.20 = 0.80 atm and p(NO₂) = 0.40/1.20 × 1.20 = 0.40 atm. Kp = 0.40² ÷ 0.80 = 0.16 ÷ 0.80 = 0.20 atm.
3. Harder: For 2SO₂(g) + O₂(g) ⇌ 2SO₃(g), ΔH is negative. State and explain the effect on the equilibrium yield of SO₃ and on Kc of (a) halving the volume of the container, (b) raising the temperature, (c) adding a V₂O₅ catalyst.
Show answer
(a) Yield increases, Kc unchanged. Halving the volume doubles every concentration, so Q = [SO₃]² ÷ ([SO₂]²[O₂]) becomes 4 ÷ (4 × 2) = 1/2 of Kc. Since Q is less than Kc, the system reacts forward, towards fewer gas moles. (b) Yield decreases and Kc decreases, because the forward reaction is exothermic, so raising the temperature favours the endothermic backward reaction. (c) No change to yield or Kc, but equilibrium is reached faster.
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