Question

Two reactions of the same order have equal exponential factors but their activation energies differ by $24.9kJ/mol$. Calculate the ratio of the rate constants of this reaction at ${27}^{\circ }C$. $(R=8.314J{K}^{-1}mo{l}^{-1})$

• A. $2.1\times {10}^4$
• B. $2.1\times {10}^3$
• C. $2.1$
• D. None of these

Answer 1

The correct option is A $2.1\times {10}^4$

We know that,
$lo{g}_{10}k=lo{g}_{10}A-\frac{E}{2.303RT}$

So, $lo{g}_{10}{k}_2=lo{g}_{10}A-\frac{E_2}{2.303RT}....eqn\left(1\right)$

and $lo{g}_{10}{k}_1=lo{g}_{10}A-\frac{E_1}{2.303RT}.....eqn\left(2\right)$

Substracting equation (2) from (1), we get,

$lo{g}_{10}\left(\frac{k_2}{k_1}\right)=\frac{(E_1-E_2)}{2.303RT}$

$lo{g}_{10}\left(\frac{k_2}{k_1}\right)=\frac{24.9\times 1000}{2.303\times 8.314\times 300}$

$\frac{k_2}{k_1}=2.14\times {10}^4$
Hence, (a) is correct.