Question

The rate constant for the reaction, $COC{l}_2\left(g\right)\to CO\left(g\right)+C{l}_2\left(g\right)$ is given by $\left[k/\left(mi{n}^{-1}\right)\right]=-11067/TK+\mathrm{31.33.}$.The temperatureat which the rate of this reaction will be doubled from that at $25^{\circ }C$ is

• A. $75^{\circ }C$
• B. $100^{\circ }C$
• C. $31^{\circ }C$
• D. $50^{\circ }C$

Answer 1

Answer: (C)

Given,
$\ln k=-\frac{11067}{T}k+31.33$
Comparing with Arrhenius equation,
$\ln k=\ln A-\frac{E_a}{RT}$
$\frac{-E_a}{RT}=-\frac{11067}{T}$
$\Rightarrow \frac{-E_a}{R}=-11067$
Also, $\ln A=3133$
If the rate constant gets doubled, then
$\log \left(\frac{2k}{k}\right)=\frac{E_a}{2303R}\left[\frac{1}{T_1}-\frac{1}{T_2}\right]$
When $T_1=25^{\circ }C=298K$
$\therefore \log \frac{2k}{k}=\frac{11067}{2.303}\left[\frac{1}{298}-\frac{1}{T_2}\right]$
$0.301=4805.47\left[\frac{1}{298}-\frac{1}{T_2}\right]$
$6.62\times 10^{-5}=\frac{1}{298}-\frac{1}{T_2}$
$\frac{1}{T_2}=\frac{1}{298}-6.62\times 10^{-5}$
$\frac{1}{T_2}=3.289\times 10^{-3}$
$T_2=304.1K$
or $31^{\circ }C$