Question

The rate constant is given by Arrhenius equation,
$k=A{e}^{-E_a/RT}$
Calculate the value of $lo{g}_{10}\frac{k_{ca}}{k_{un}}$ at ${27}^{\circ }C$.
where,
$k_{ca}$ and $k_{un}$ are the rate constants for catalysed and uncatalysed reaction respectively.
The energy of activation of catalysed and uncatalysed reactions are $162kJmo{l}^{-1}$ and $350kJmo{l}^{-1}$ respectively.

• A. $32.72$
• B. $12.44$
• C. $0.85$
• D. $1$

Answer 1

The correct option is A $32.72$

Let $k_{ca}$ and $k_{un}$ be the rate constants for catalysed and uncatalysed reactions.
$2.303lo{g}_{10}{k}_{ca}=2.303lo{g}_{10}A-\frac{162\times {10}^3}{RT}...\left(i\right)$

and $2.303lo{g}_{10}{k}_{un}=2.303lo{g}_{10}A-\frac{350\times {10}^3}{RT}...\left(ii\right)$

Subtracting eq. (ii) from eq. (i)
$lo{g}_{10}\frac{k_{ca}}{k_{un}}=\frac{{10}^3}{2.303RT}(350-162)$
$lo{g}_{10}\frac{k_{ca}}{k_{un}}=\frac{188\times {10}^3}{2.303\times 8.314\times 300}\approx 32.72$