Chemistry - Chemical Kinetics Question with Solution | TestHub

According to the Arrhenius equation:

$\ln k=\ln A-\frac{E_a}{RT}$

For a plot of $\ln k$ versus $1/T$, the y-intercept corresponds to $\ln A$. From the given graph, the line intersects the y-axis at $6$.

Therefore, $\ln A=6$.

We are given:
$E_a=6.64{\text{ kJ mol}}^{-1}=6640{\text{ J mol}}^{-1}$
$R=8.3{\text{ J K}}^{-1}{\text{ mol}}^{-1}$
$k=e^2{\text{ min}}^{-1}$, which gives $\ln k=\ln (e^2)=2$

Substituting these values into the Arrhenius equation:

$2=6-\frac{6640}{8.3\times T}$

$\frac{6640}{8.3\times T}=6-2$

$\frac{6640}{8.3\times T}=4$

$T=\frac{6640}{4\times 8.3}$

$T=\frac{6640}{33.2}$

$T=200\text{ K}$

Answer: $200$ K