Question

The rate constant for the reaction, $COCl_2\left(g\right) \rightarrow CO\left(g\right)+Cl_2\left(g\right)$ is given by $\left[k / \left(min^{-1}\right)\right]=-11067 /T \,K+31.33.$.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,
$\ln k=\ln A-\frac{E_{a}}{R T}$
$\frac{-E_{a}}{R T}=-\frac{11067}{T}$
$\Rightarrow \frac{-E_{a}}{R}=-11067$
Also, $\ln A=3133$
If the rate constant gets doubled, then
$\log \left(\frac{2 k}{k}\right)=\frac{E_{a}}{2303 R}\left[\frac{1}{T_{1}}-\frac{1}{T_{2}}\right]$
When $T_{1}=25^{\circ} C =298 \,K$
$\therefore \log \frac{2 k}{k}=\frac{11067}{2.303}\left[\frac{1}{298}-\frac{1}{T_{2}}\right]$
$0.301=4805.47\left[\frac{1}{298}-\frac{1}{T_{2}}\right]$
$6.62 \times 10^{-5}=\frac{1}{298}-\frac{1}{T_{2}}$
$ \frac{1}{T_{2}} =\frac{1}{298}-6.62 \times 10^{-5} $
$ \frac{1}{T_{2}} =3.289 \times 10^{-3} $
$ T_{2} =304.1\, K$
or $31^{\circ} C $