At 27^{circ}C in the presence of a catalyst, | vedclass

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(B) According to the Arrhenius equation, $k = A e^{-E_a / RT}$.For the catalysed reaction, $k_c = A e^{-E_{ac} / RT}$ and for the uncatalysed reaction

Vedclass · Accepted Answer

$1.741$

Vedclass · Answer

(B) According to the Arrhenius equation, $k = A e^{-E_a / RT}$.For the catalysed reaction, $k_c = A e^{-E_{ac} / RT}$ and for the uncatalysed reaction, $k_u = A e^{-E_{au} / RT}$.Given that the frequency factor $A$ is the same, the ratio is $\frac{k_c}{k_u} = e^{(E_{au} - E_{ac}) / RT} = e^{\Delta E_a / RT}$.Taking the logarithm on both sides: $\log_{10} \left( \frac{k_c}{k_u} ight) = \frac{\Delta E_a}{2.303 RT}$.Given $\Delta E_a = 10 \ kJ \ mol^{-1} = 10000 \ J \ mol^{-1}$, $R = 8.314 \ J \ K^{-1} \ mol^{-1}$, and $T = 27 + 273 = 300 \ K$.$\log_{10} \left( \frac{k_c}{k_u} ight) = \frac{10000}{2.303 	imes 8.314 	imes 300} = \frac{10000}{5744.14} \approx 1.741$.

At $27^{\circ}C$ in the presence of a catalyst, the activation energy of a reaction is lowered by $10 \ kJ \ mol^{-1}$. The logarithm ratio of $\frac{k(\text{catalysed})}{k(\text{uncatalysed})}$ is .... (Consider that the frequency factor for both the reactions is the same)

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