For a first-order reaction A rightarrow P,the | vedclass

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(D) The Arrhenius equation is given by $\log \, k = \log \, A - \frac{E_a}{2.303 \, RT}$.Comparing this with the given equation $\log \, k = - \frac{2

Vedclass · Accepted Answer

$1.0 	imes 10^6 \, s^{-1}$ and $38.3 \, kJ \, mol^{-1}$

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(D) The Arrhenius equation is given by $\log \, k = \log \, A - \frac{E_a}{2.303 \, RT}$.Comparing this with the given equation $\log \, k = - \frac{2000}{T} + 0.6$:$\log \, A = 0.6 \implies A = 10^{0.6} \approx 3.98 \, s^{-1}$ (Note: The provided option $10^6$ suggests the equation was $\log \, k = - \frac{2000}{T} + 6$).Assuming the equation is $\log \, k = - \frac{2000}{T} + 6$:$\log \, A = 6 \implies A = 10^6 \, s^{-1}$.$\frac{E_a}{2.303 \, R} = 2000$.$E_a = 2000 	imes 2.303 	imes 8.314 \, J \, mol^{-1} \approx 38297 \, J \, mol^{-1} = 38.3 \, kJ \, mol^{-1}$.

For a first-order reaction $A \rightarrow P$,the rate constant $k$ depends on temperature $(T)$ according to the equation $\log \, k = - \frac{2000}{T} + 0.6$. The pre-exponential factor $A$ and the activation energy $E_a$ are,respectively:

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