For a reaction,the activation energy E_{a} = | vedclass

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(B) According to the Arrhenius equation: $\log \left( \frac{K_{2}}{K_{1}} \right) = \frac{E_{a}}{2.303 \ R} \left( \frac{1}{T_{1}} - \frac{1}{T_{2}} \

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$1.6 	imes 10^{6} \ s^{-1}$

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(B) According to the Arrhenius equation: $\log \left( \frac{K_{2}}{K_{1}} ight) = \frac{E_{a}}{2.303 \ R} \left( \frac{1}{T_{1}} - \frac{1}{T_{2}} ight)$.Given that the activation energy $E_{a} = 0$.Substituting $E_{a} = 0$ into the equation,we get: $\log \left( \frac{K_{2}}{K_{1}} ight) = \frac{0}{2.303 \ R} \left( \frac{1}{T_{1}} - \frac{1}{T_{2}} ight) = 0$.This implies $\frac{K_{2}}{K_{1}} = 10^{0} = 1$,which means $K_{2} = K_{1}$.Therefore,the rate constant at $400 \ K$ remains the same as at $200 \ K$,which is $1.6 	imes 10^{6} \ s^{-1}$.

For a reaction,the activation energy $E_{a} = 0$ and the rate constant at $200 \ K$ is $1.6 \times 10^{6} \ s^{-1}$. The rate constant at $400 \ K$ will be (given $R = 8.314 \ J \ K^{-1} \ mol^{-1}$):

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Subtract $(i)$ $\ln \, k_1 = - \frac{E_a}{R T_1} + \ln A$ and $(ii)$ $\ln \, k_2 = - \frac{E_a}{R T_2} + \ln A$ and write the resulting equation.

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