Chemistry · Chemical kinetics · chapter 3 of 10

Class 12 Chemistry Chapter 3: Chemical Kinetics case study question with answers

One case-based question of the kind the board asks: a passage, then 3 parts worth 4 marks in all, with the answer and the marking scheme for each. Written from the NCERT chapter.

Case study · 4 marks

A student studies a gas-phase reaction A + B → C + D. Reactant molecules must pass through an activated complex before they form products, and the energy needed to reach it is the activation energy. The figure plots potential energy against the reaction coordinate for the reaction alone (solid curve) and for the same reaction with a catalyst added (dashed curve). Both paths start from the same reactants and end at the same products. Read the energies from the figure to answer the questions.

Part 1 · 1 mark

The activation energy of the forward reaction without the catalyst is

(a) 80 kJ mol⁻¹(b) 130 kJ mol⁻¹(c) 110 kJ mol⁻¹(d) 40 kJ mol⁻¹

Show answer

Answer: (a) 80 kJ mol⁻¹

Ea(forward) = peak − reactants = 130 − 50 = 80 kJ mol⁻¹.

Part 2 · 1 mark

The enthalpy change ΔH of the reaction is

(a) −30 kJ mol⁻¹(b) 30 kJ mol⁻¹(c) −80 kJ mol⁻¹(d) −110 kJ mol⁻¹

Show answer

Answer: (a) −30 kJ mol⁻¹

ΔH = products − reactants = 20 − 50 = −30 kJ mol⁻¹; exothermic.

Part 3 · 2 marks

Find the activation energy of the backward reaction C + D → A + B along each path, and state what you notice about how the catalyst changes the forward and backward barriers.

Answer and marking scheme

Uncatalysed: 130 − 20 = 110 kJ mol⁻¹; catalysed: 90 − 20 = 70 kJ mol⁻¹. The forward barrier falls from 80 to 40 kJ mol⁻¹ and the backward from 110 to 70: both are lowered by the same 40 kJ mol⁻¹, so the catalyst speeds both directions alike.

110 and 70 kJ mol⁻¹ (1)
both barriers lowered by the same 40 kJ mol⁻¹ (1)

Using the figure, explain why the catalyst changes neither ΔH nor the equilibrium constant of the reaction.

Answer and marking scheme

Both paths start at the same reactant energy (50) and end at the same product energy (20), so ΔH = −30 kJ mol⁻¹ either way. The catalyst lowers the forward and backward barriers by the same 40 kJ mol⁻¹, speeding both reactions equally, so equilibrium is reached sooner but its position, and K, are unchanged.

same end points, so ΔH unchanged (1)
both barriers lowered equally, so K unchanged (1)

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