Question

Consider the reaction $\text{A} \rightarrow 2 \text{ B} + \text{C,} \, Δ \text{H} = - 1 5 \text{ kcal}$ . The energy of activation of backward reaction is 20 kcal mol^-1. In presence of catalyst, the energy of activation of forward reaction is 3 kcal mol^-1. At 400 K the catalyst causes the rate of the forward reaction to increase by the number of times equal to-

• A. e^3.5
• B. e^2.5
• C. e^-2.5
• D. e^2.303

Answer 1

Answer: (B)

$\mathrm{E}_{\mathrm{a}(\mathrm{f})}-\mathrm{E}_{\mathrm{a}(\mathrm{b})}=\Delta \mathrm{H}=-15 \mathrm{kcal}$

$\Rightarrow \quad E_{a(f)}=-15+20=5 \mathrm{kcal}$

$\frac{\mathrm{k}_{\text {(catalyst) }}}{\mathrm{k}_{\mathrm{a}}}=\mathrm{e}^{\mathrm{E}_{\mathrm{a}}-\mathrm{E}_{\text {a(catalyst) }} / \mathrm{RT}}$

$=\mathrm{e}^{(5-3) \times 10^3 / 2 \times 400}$

$= \text{e}^{2 \text{.} 5}$