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Chemical Kinetics and Nuclear Chemistry question

2025 · 4 Apr · Shift 1 · Q1
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Chemical Kinetics and Nuclear Chemistry question

2025 · 4 Apr · Shift 1 · Q1

JEE MainChemistryChemical Kinetics and Nuclear ChemistryMCQ+4 / −1
For A2+B2⇌2ABEa\mathrm{A}_2+\mathrm{B}_2 \rightleftharpoons 2 \mathrm{AB}\mathrm{E}_{\mathrm{a}}A2​+B2​⇌2ABEa​ for forward and backward reaction are 180 and 200 kJ mol−1200 \mathrm{~kJ} \mathrm{~mol}^{-1}200 kJ mol−1 respectively If catalyst lowers Ea\mathrm{E}_{\mathrm{a}}Ea​ for both reaction by 100 kJ mol−1100 \mathrm{~kJ} \mathrm{~mol}^{-1}100 kJ mol−1. Which of the following statement is correct?
  1. A
    Catalyst does not alter the Gibbs energy change of a reaction.
  2. B
    The enthalpy change for the reaction is +20 kJ mol−1+20 \mathrm{~kJ} \mathrm{~mol}^{-1}+20 kJ mol−1.
  3. C
    Catalyst can cause non-spontaneous reactions to occur.
  4. D
    The enthalpy change for the catalysed reaction is different from that of uncatalysed reaction.
View written solutionFree

Correct answer: A

  1. Use the relation between activation energies and enthalpy change

For a reversible reaction,

ΔH=Ea(forward)−Ea(backward)\Delta H = E_a(\text{forward}) - E_a(\text{backward})ΔH=Ea​(forward)−Ea​(backward)

Given:

Eaforward=180 kJ mol−1,Eabackward=200 kJ mol−1E_a^{\text{forward}} = 180\ \text{kJ mol}^{-1}, \qquad E_a^{\text{backward}} = 200\ \text{kJ mol}^{-1}Eaforward​=180 kJ mol−1,Eabackward​=200 kJ mol−1

So,

ΔH=180−200=−20 kJ mol−1\Delta H = 180 - 200 = -20\ \text{kJ mol}^{-1}ΔH=180−200=−20 kJ mol−1

Thus, the reaction is exothermic.

  1. Effect of catalyst on activation energy

A catalyst lowers the activation energy of both forward and backward reactions by the same amount.

Here, both are lowered by 100 kJ mol−1100\ \text{kJ mol}^{-1}100 kJ mol−1:

Eaforward,catalysed=180−100=80 kJ mol−1E_a^{\text{forward,catalysed}} = 180 - 100 = 80\ \text{kJ mol}^{-1}Eaforward,catalysed​=180−100=80 kJ mol−1 Eabackward,catalysed=200−100=100 kJ mol−1E_a^{\text{backward,catalysed}} = 200 - 100 = 100\ \text{kJ mol}^{-1}Eabackward,catalysed​=200−100=100 kJ mol−1

Then,

ΔHcatalysed=80−100=−20 kJ mol−1\Delta H_{\text{catalysed}} = 80 - 100 = -20\ \text{kJ mol}^{-1}ΔHcatalysed​=80−100=−20 kJ mol−1

So catalyst does not change the enthalpy change.

  1. Now evaluate each option

Option A: Catalyst does not alter the Gibbs energy change of a reaction.

This is true.

A catalyst changes only the reaction pathway and lowers activation energy. It does not change state functions like:

  • Gibbs free energy change ΔG\Delta GΔG
  • Enthalpy change ΔH\Delta HΔH
  • Equilibrium constant

So A is correct.

Option B: The enthalpy change for the reaction is +20 kJ mol−1+20\ \text{kJ mol}^{-1}+20 kJ mol−1.

From calculation,

ΔH=−20 kJ mol−1\Delta H = -20\ \text{kJ mol}^{-1}ΔH=−20 kJ mol−1

So this is false.

Option C: Catalyst can cause non-spontaneous reactions to occur.

A catalyst cannot make a non-spontaneous reaction spontaneous. It only increases the rate of a reaction that is thermodynamically feasible.

So this is false.

Option D: The enthalpy change for the catalysed reaction is different from that of uncatalysed reaction.

We found:

ΔHuncatalysed=−20 kJ mol−1,ΔHcatalysed=−20 kJ mol−1\Delta H_{\text{uncatalysed}} = -20\ \text{kJ mol}^{-1}, \qquad \Delta H_{\text{catalysed}} = -20\ \text{kJ mol}^{-1}ΔHuncatalysed​=−20 kJ mol−1,ΔHcatalysed​=−20 kJ mol−1

So this is false.

  1. Final answer

Only Option A is correct.

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