A rightarrow B (first reaction)
C rightarrow | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) For the first reaction $A \xrightarrow{k_1} B$:
Using the Arrhenius equation: $\ln\left(\frac{k_{1, 500}}{k_{1, 300}}\right) = \frac{E_{a1}}{R} \

Vedclass · Accepted Answer

$5$

Vedclass · Answer

(C) For the first reaction $A \xrightarrow{k_1} B$:
Using the Arrhenius equation: $\ln\left(\frac{k_{1, 500}}{k_{1, 300}}ight) = \frac{E_{a1}}{R} \left(\frac{1}{300} - \frac{1}{500}ight)$.
Given $k_{1, 500} = 2 k_{1, 300}$,so $\ln(2) = \frac{E_{a1}}{R} \left(\frac{2}{1500}ight) \implies E_{a1} = \frac{1500 R \ln 2}{2} = 750 R \ln 2$.
For the second reaction $C \xrightarrow{k_2} D$,$E_{a2} = \frac{E_{a1}}{2} = 375 R \ln 2$.
At $500 \ K$,for the first reaction,$t_{1/2} = 2 \ hours$,so $k_{1, 500} = \frac{\ln 2}{t_{1/2}} = \frac{\ln 2}{2} \approx 0.3466 \ hour^{-1}$.
Given $k_{2, 500} = 2 k_{1, 500} = 2 	imes \frac{\ln 2}{2} = \ln 2 \approx 0.6931 \ hour^{-1}$.
Using the Arrhenius equation for the second reaction:
$\ln\left(\frac{k_{2, 500}}{k_{2, 300}}ight) = \frac{E_{a2}}{R} \left(\frac{1}{300} - \frac{1}{500}ight) = \frac{375 R \ln 2}{R} \left(\frac{2}{1500}ight) = 375 \ln 2 	imes \frac{1}{750} = \frac{\ln 2}{2} = \ln(\sqrt{2})$.
Therefore,$\frac{k_{2, 500}}{k_{2, 300}} = \sqrt{2} \implies k_{2, 300} = \frac{k_{2, 500}}{\sqrt{2}} = \frac{\ln 2}{\sqrt{2}} \approx \frac{0.6931}{1.414} \approx 0.4902 \ hour^{-1}$.
$k_{2, 300} = 4.9 	imes 10^{-1} \ hour^{-1}$.

Similar Questions

The rate constant,the activation energy and the Arrhenius parameter of a chemical reaction at $25\,^oC$ are $3.0 \times 10^{-4}\,s^{-1}$,$104.4\,kJ\,mol^{-1}$ and $6.0 \times 10^{14}\,s^{-1}$ respectively. The value of the rate constant as $T \to \infty$ is

$A$ large increase in the rate of reaction for a small increase in temperature is due to:

Activation energy of a reaction is

At $T \ K$,the following equation is obtained for a first order reaction: $\log \frac{k}{A} = -\frac{x}{T}$. The activation energy for this reaction is equal to $(R = \text{gas constant})$

Vedclass Test Series

Exam Paper Generator

Online Exam Module