Question

For a reversible reaction \( A \rightleftharpoons B \), the \( \Delta H _{\text {forward reaction }}=20 kJ mol ^{-1} \). The activation energy of the uncatalysed forward reaction is \( 300 kJ mol ^{-1} \). When the reaction is catalysed keeping the reactant concentration same, the rate of the catalysed forward reaction at \( 27^{\circ} C \) is found to be same as that of the uncatalysed reaction at \( 327^{\circ} C \). 

Answer 1

Using the Arrhenius equation:

\(k = Ae ^{(-E_a / RT)}\)

where k is the rate constant, A is the pre-exponential factor, E_a is the activation energy, R is the gas constant, and T is the temperature in Kelvin.

The equation is used for both the uncatalyzed forward reaction and the catalyzed backward reaction. 

By setting the two equations equal to each other and cancelling out the pre-exponential factor, we get:

\(e^{\frac{300 \times 10^3}{600 \times R}}=e^{\frac{-E a}{300 \times R}}\)

Simplifying the equation, we get:

\(\frac{300 \times 10^3}{600 \times R}=\frac{E_a}{300 \times R}\)

Solving for E_a, we get:

\(E_a=\frac{10^3}{2} \times 300=150 \times 10^3 \mathrm{~J} \mathrm{~mol}^{-1}=150 \mathrm{~kJ} \mathrm{~mol}^{-1}\)

This gives us the activation energy for the uncatalyzed forward reaction.

To find the activation energy for the catalyzed backward reaction, we use the relationship:

\(E_{\text {rev,catalysed }} \quad=E_{\text {fwd,uncat }} \quad-\Delta H_{\text {forward reaction }}\)

Substituting the given values, we get:

\(E_{\text {rev,catalysed }}=150 \mathrm{~kJ} \mathrm{~mol}^{-1}-20 \mathrm{~kJ} \mathrm{~mol}^{-1}=130 \mathrm{~kJ} \mathrm{~mol}^{-1}\)

Therefore, the activation energy for the catalyzed backward reaction is \(130 \mathrm{~kJ} \mathrm{~mol}^{-1}.\)