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(B) According to the Arrhenius equation,the rate constant $k$ at temperature $T$ is given by $k = A e^{-E_a/RT}$.For two different temperatures $T_1$

Vedclass · Accepted Answer

$\log \frac{k_1}{k_2} = \frac{E_a}{2.303 R} \left( \frac{T_1 - T_2}{T_1 T_2} \right)$

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(B) According to the Arrhenius equation,the rate constant $k$ at temperature $T$ is given by $k = A e^{-E_a/RT}$.For two different temperatures $T_1$ and $T_2$,we have:$\ln k_1 = \ln A - \frac{E_a}{RT_1}$ and $\ln k_2 = \ln A - \frac{E_a}{RT_2}$.Subtracting the two equations:$\ln k_2 - \ln k_1 = \frac{E_a}{RT_1} - \frac{E_a}{RT_2} = \frac{E_a}{R} \left( \frac{1}{T_1} - \frac{1}{T_2} \right) = \frac{E_a}{R} \left( \frac{T_2 - T_1}{T_1 T_2} \right)$.Converting to base $10$ logarithm:$\log \frac{k_2}{k_1} = \frac{E_a}{2.303 R} \left( \frac{T_2 - T_1}{T_1 T_2} \right)$.Multiplying by $-1$ on both sides:$\log \frac{k_1}{k_2} = \frac{E_a}{2.303 R} \left( \frac{T_1 - T_2}{T_1 T_2} \right)$.

For a first-order reaction,if the rate constant is $k_1$ at temperature $T_1$ and $k_2$ at temperature $T_2$,which of the following relations is correct? ($E_a$ = activation energy)

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