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Chemical Kinetics and Nuclear Chemistry question

2007 · Shift 0 · Q26
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Chemical Kinetics and Nuclear Chemistry question

2007 · Shift 0 · Q26

JEE MainChemistryChemical Kinetics and Nuclear ChemistryMCQ+4 / −1
The energies of activation for forward and reverse reactions for A2A_2A2​ + B2B_2B2​ ⇋\leftrightharpoons⇋ 2ABABAB are 180 kJ mol−1 and 200 kJ mol−1 respectively. The presence of catalyst lowers the activation energy of both (forward and reverse) reactions by 100 kJ mol−1. The enthalpy change of the reaction ( A2A_2A2​ + B2B_2B2​ →\to→ 2ABABAB) in the presence of catalyst will be (in kJ mol−1)
  1. A
    300
  2. B
    120
  3. C
    200
  4. D
    20
View written solutionFree

Correct answer: D

  1. For a reaction, the relation between forward and reverse activation energies and enthalpy change is:
ΔH=Eaforward−Eareverse\Delta H = E_a^{\text{forward}} - E_a^{\text{reverse}}ΔH=Eaforward​−Eareverse​

for the reaction written in the forward direction.

  1. Given:
Eaforward=180 kJ mol−1E_a^{\text{forward}} = 180\ \text{kJ mol}^{-1}Eaforward​=180 kJ mol−1 Eareverse=200 kJ mol−1E_a^{\text{reverse}} = 200\ \text{kJ mol}^{-1}Eareverse​=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 forward reaction is exothermic by 20 kJ mol−120\ \text{kJ mol}^{-1}20 kJ mol−1.

  1. A catalyst lowers the activation energy of both forward and reverse reactions by the same amount. Here, both are lowered by 100 kJ mol−1100\ \text{kJ mol}^{-1}100 kJ mol−1:
Ea,catforward=180−100=80 kJ mol−1E_{a,\text{cat}}^{\text{forward}} = 180 - 100 = 80\ \text{kJ mol}^{-1}Ea,catforward​=180−100=80 kJ mol−1 Ea,catreverse=200−100=100 kJ mol−1E_{a,\text{cat}}^{\text{reverse}} = 200 - 100 = 100\ \text{kJ mol}^{-1}Ea,catreverse​=200−100=100 kJ mol−1

Now the enthalpy change in presence of catalyst is:

ΔHcat=80−100=−20 kJ mol−1\Delta H_{\text{cat}} = 80 - 100 = -20\ \text{kJ mol}^{-1}ΔHcat​=80−100=−20 kJ mol−1
  1. Therefore, the enthalpy change remains unchanged in presence of catalyst:
−20 kJ mol−1\boxed{-20\ \text{kJ mol}^{-1}}−20 kJ mol−1​
  1. Since the options are given as positive values, the magnitude is 20 kJ mol−120\ \text{kJ mol}^{-1}20 kJ mol−1, corresponding to option D.
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