Question

On introducing a catalyst at 400 K, the rate of a first order reaction is increased by 1.7 times of the one without a catalyst. If the activation energy in the presence of a catalyst is $4kJmo{l}^{-1};$ then, what is the slope of the plot of ln(k) against $\frac{1}{T}$ in the absence of the catalyst? (k = rate constant in $se{c}^{-1}$ and T is temperature in Kelvin(K); Given : $ln1.7=0.5$ and $R=\frac{25}{3}J{K}^{-1}mo{l}^{-1})$

• A. +680 K
• B. -680 K
• C. +800 K
• D. -800 K

Answer 1

The correct option is B -680 K

$\mathrm{According}\mathrm{to}\mathrm{Arrhenius}\mathrm{equation},k={\mathrm{Ae}}^{-\frac{E_a}{\mathrm{RT}}}\Rightarrow \mathrm{lnk}=\mathrm{lnA}-\frac{E_a}{\mathrm{RT}}\mathrm{So},\mathrm{the}\mathrm{required}\mathrm{slope}\mathrm{will}\mathrm{be}-\frac{E_a}{R}.\mathrm{According}\mathrm{to}\mathrm{question},k_{\mathrm{catalyst}}={\mathrm{Ae}}^{(-\frac{4000\times 3}{25\times 400})}.....\left(1\right)k_{\mathrm{without}\mathrm{catalyst}}={\mathrm{Ae}}^{(-\frac{E_a}{R\times 400})}.....\left(2\right)\mathrm{Dividing}\mathrm{equation}\left(1\right)\mathrm{by}\mathrm{equation}\left(2\right),\frac{k_{\mathrm{catalyst}}}{k_{\mathrm{without}\mathrm{catalyst}}}=e^{(-\frac{30}{25}+\frac{E_a}{R\times 400})}\Rightarrow 1.7=e^{(-\frac{30}{25}+\frac{E_a}{R\times 400})}\Rightarrow \ln (1.7)=-1.2+\left(\frac{E_a}{R\times 400}\right)\Rightarrow \frac{E_a}{R}=680K\mathrm{Since}\mathrm{the}\mathrm{slope}\mathrm{is}-\frac{E_a}{R},\mathrm{so}\mathrm{the}\mathrm{answer}\mathrm{will}\mathrm{be}-680K$