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NVAMediumJEE 2023Arrhenius Equation & Activation Energy

JEE Chemistry 2023 Question with Solution

For a reversible reaction A⇌B\mathrm{A \rightleftharpoons B}, the ΔHforward reaction=20 kJ mol−1\Delta H_{forward\ reaction} = 20\ \mathrm{kJ\,mol^{-1}}. The activation energy of the uncatalysed forward reaction is 300 kJ mol−1300\ \mathrm{kJ\,mol^{-1}}. When the reaction is catalysed keeping the reactant concentration same, the rate of the catalysed forward reaction at 27∘C27^\circ\mathrm{C} is found to be same as that of the uncatalysed reaction at 327∘C327^\circ\mathrm{C}. The activation energy of the catalysed backward reaction is _____ kJ mol−1\mathrm{kJ\,mol^{-1}}.

Answer

Correct answer:130

Step-by-step solution

Standard Method

Given:

  • Reversible reaction A⇌B\mathrm{A \rightleftharpoons B}
  • ΔHforward=20 kJ mol−1\Delta H_{forward} = 20\ \mathrm{kJ\,mol^{-1}}
  • Uncatalysed forward activation energy =300 kJ mol−1= 300\ \mathrm{kJ\,mol^{-1}}
  • Catalysed forward reaction at 300 K300\ \mathrm{K} has the same rate as uncatalysed forward reaction at 600 K600\ \mathrm{K}

Find: Activation energy of the catalysed backward reaction.

Using Arrhenius equation,

k=Ae−Ea/RTk = A e^{-E_a/RT}

If the two rates are the same, the solution uses

Ea1T1=Ea2T2\frac{E_{a1}}{T_1} = \frac{E_{a2}}{T_2}

For the uncatalysed forward reaction,

Ea=300 kJ mol−1,T1=327∘C=600 KE_a = 300\ \mathrm{kJ\,mol^{-1}}, \qquad T_1 = 327^\circ\mathrm{C} = 600\ \mathrm{K}

For the catalysed forward reaction,

T2=27∘C=300 KT_2 = 27^\circ\mathrm{C} = 300\ \mathrm{K}

Therefore,

300600=Ea(cat, forward)300\frac{300}{600} = \frac{E_a(\text{cat, forward})}{300}

So,

Ea(cat, forward)=150 kJ mol−1E_a(\text{cat, forward}) = 150\ \mathrm{kJ\,mol^{-1}}

For a reversible reaction,

Ea,forward−Ea,backward=ΔHE_{a,forward} - E_{a,backward} = \Delta H

Hence,

Ea,backward=Ea,forward−ΔHE_{a,backward} = E_{a,forward} - \Delta H

Substituting the values,

Ea,backward=150−20=130 kJ mol−1E_{a,backward} = 150 - 20 = 130\ \mathrm{kJ\,mol^{-1}}

Therefore, the activation energy of the catalysed backward reaction is 130 kJ mol−1130\ \mathrm{kJ\,mol^{-1}}.

Energy Difference Shortcut

Given: The catalysed forward rate at 300 K300\ \mathrm{K} matches the uncatalysed forward rate at 600 K600\ \mathrm{K}.

Find: Catalysed backward activation energy.

From the proportionality used in the solution,

300600=Ea(cat, forward)300\frac{300}{600} = \frac{E_a(\text{cat, forward})}{300}

So the catalysed forward activation energy becomes half of 300 kJ mol−1300\ \mathrm{kJ\,mol^{-1}}, i.e.

150 kJ mol−1150\ \mathrm{kJ\,mol^{-1}}

Now use the reversible reaction relation,

Ea,forward−Ea,backward=ΔH=20 kJ mol−1E_{a,forward} - E_{a,backward} = \Delta H = 20\ \mathrm{kJ\,mol^{-1}}

Thus,

Ea,backward=150−20=130 kJ mol−1E_{a,backward} = 150 - 20 = 130\ \mathrm{kJ\,mol^{-1}}

Therefore, the required answer is 130 kJ mol−1130\ \mathrm{kJ\,mol^{-1}}.

Common mistakes

  • Using 2727 and 327327 directly instead of converting to Kelvin is incorrect because Arrhenius relations require absolute temperature. Convert first: 27∘C=300 K27^\circ\mathrm{C} = 300\ \mathrm{K} and 327∘C=600 K327^\circ\mathrm{C} = 600\ \mathrm{K}.

  • Reversing the enthalpy relation is a common error. For the forward reaction given here, the correct relation is Ea,forward−Ea,backward=ΔHE_{a,forward} - E_{a,backward} = \Delta H, so backward activation energy is found by subtracting ΔH\Delta H from the forward activation energy.

  • Assuming a catalyst changes ΔH\Delta H is wrong. A catalyst lowers activation energies for both directions but does not alter the enthalpy change of the reaction. Use the same ΔH=20 kJ mol−1\Delta H = 20\ \mathrm{kJ\,mol^{-1}} after catalysis.

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