The first order rate constant k is related to | vedclass

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Question

(A) The Arrhenius equation is given by $\log \, k = \log \, A - \frac{E_a}{2.303 RT}$.Comparing this with the given equation $\log \, k = 15.0 - \frac

Vedclass · Accepted Answer

$A = 10^{15}$ and $E_a = 1.9 	imes 10^{4} \, kJ$

Vedclass · Answer

(A) The Arrhenius equation is given by $\log \, k = \log \, A - \frac{E_a}{2.303 RT}$.Comparing this with the given equation $\log \, k = 15.0 - \frac{10^{6}}{T}$:$1$. $\log \, A = 15.0$,which implies $A = 10^{15}$.$2$. $\frac{E_a}{2.303 R} = 10^{6}$.Taking $R = 8.314 \, J \, K^{-1} \, mol^{-1}$,we get $E_a = 10^{6} 	imes 2.303 	imes 8.314 \approx 1.915 	imes 10^{7} \, J \, mol^{-1} = 1.915 	imes 10^{4} \, kJ \, mol^{-1}$.Thus,the correct pair is $A = 10^{15}$ and $E_a = 1.9 	imes 10^{4} \, kJ$.

The first order rate constant $k$ is related to temperature $T$ as $\log \, k = 15.0 - (10^{6} / T)$. Which of the following pairs of values for the Arrhenius factor $A$ and activation energy $E_a$ is correct?

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