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Thermodynamics question

2005 · Shift 0 · Q4
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Thermodynamics question

2005 · Shift 0 · Q4

JEE MainChemistryThermodynamicsMCQ+4 / −1
Consider an endothermic reaction, X →\to→ Y with the activation energies Eb and Ef for the backward and forward reactions, respectively. In general :
  1. A
    Eb < Ef
  2. B
    Eb > Ef
  3. C
    Eb = Ef
  4. D
    There is no definite relation between Eb and Ef
View written solutionFree

Correct answer: A

  1. Relate activation energies to enthalpy change

For a reaction X→YX \rightarrow YX→Y let:

  • EfE_fEf​ = activation energy of the forward reaction
  • EbE_bEb​ = activation energy of the backward reaction

If the reaction is endothermic, then the products are at higher energy than the reactants. So, ΔH=HY−HX>0\Delta H = H_Y - H_X > 0ΔH=HY​−HX​>0

  1. Energy profile relation

Let the transition state energy be E‡E^{\ddagger}E‡. Then, Ef=E‡−HXE_f = E^{\ddagger} - H_XEf​=E‡−HX​ Eb=E‡−HYE_b = E^{\ddagger} - H_YEb​=E‡−HY​

Subtracting, Ef−Eb=(E‡−HX)−(E‡−HY)=HY−HX=ΔHE_f - E_b = (E^{\ddagger} - H_X) - (E^{\ddagger} - H_Y) = H_Y - H_X = \Delta HEf​−Eb​=(E‡−HX​)−(E‡−HY​)=HY​−HX​=ΔH

Thus, Ef−Eb=ΔHE_f - E_b = \Delta HEf​−Eb​=ΔH

Since the reaction is endothermic, ΔH>0\Delta H > 0ΔH>0 Therefore, Ef>EbE_f > E_bEf​>Eb​ which means Eb<EfE_b < E_fEb​<Ef​

  1. Check options
  • A: Eb<EfE_b < E_fEb​<Ef​ ✅ Correct
  • B: Eb>EfE_b > E_fEb​>Ef​ ❌ Incorrect
  • C: Eb=EfE_b = E_fEb​=Ef​ ❌ Incorrect
  • D: There is no definite relation ❌ Incorrect

Therefore, the correct option is A.

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