The rate constant for the reaction,COCl_{2(g) | vedclass

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(C) Given,$\ln k = -\frac{11067}{T} + 31.33$. Comparing with the Arrhenius equation,$\ln k = \ln A - \frac{E_a}{RT}$,we get $\frac{E_a}{R} = 11067 \,

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$31$

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(C) Given,$\ln k = -\frac{11067}{T} + 31.33$. Comparing with the Arrhenius equation,$\ln k = \ln A - \frac{E_a}{RT}$,we get $\frac{E_a}{R} = 11067 \, K$. For the rate constant to double,we use the relation $\log \left( \frac{k_2}{k_1} ight) = \frac{E_a}{2.303 R} \left[ \frac{1}{T_1} - \frac{1}{T_2} ight]$. Here,$T_1 = 25^{\circ} C = 298 \, K$ and $\frac{k_2}{k_1} = 2$. Substituting the values: $\log 2 = \frac{11067}{2.303} \left[ \frac{1}{298} - \frac{1}{T_2} ight]$. $0.3010 = 4805.47 \left[ 0.0033557 - \frac{1}{T_2} ight]$. $6.2637 	imes 10^{-5} = 0.0033557 - \frac{1}{T_2}$. $\frac{1}{T_2} = 0.003293$. $T_2 \approx 303.66 \, K \approx 31^{\circ} C$.

The rate constant for the reaction,$COCl_{2(g)} \longrightarrow CO_{(g)} + Cl_{2(g)}$ is given by $\ln[k / (min^{-1})] = -11067 / T(K) + 31.33$. The temperature at which the rate of this reaction will be doubled from that at $25^{\circ} C$ is $..... \, ^{\circ} C$.

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