Chemistry - Chemical Kinetics Question with Solution | TestHub

(a) is correct as explained below:

$k=A{e}^{-\frac{E_a}{RT}}$

$\Rightarrow \frac{k}{A}=e^{-\frac{E_a}{RT}}$

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

$\Rightarrow \ln \left(\frac{A}{k}\right)=\frac{E_a}{RT}$

(b) is not correct.

$\frac{d(\ln A)}{dT}-\frac{d(\ln k)}{dT}=\frac{d}{dT}\left(\frac{E_a}{RT}\right)$

$-\frac{d(\ln k)}{dT}=-\frac{E_a}{R{T}^2}$ [$\ln A$ does not change with $T$]

$\frac{d(\ln k)}{dT}=\frac{E_a}{R{T}^2}$

(c) is correct.

$k=A{e}^{-\frac{E_a}{RT}}$

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

$\Rightarrow \ln A=\ln k+\frac{E_a}{RT}$

$\Rightarrow \log A=\log k+\frac{E_a}{2.303RT}$.

Hence, (d) is not correct; it follows from the explanations of (a), (b), and (c).