Question

The first order rate constant for the decomposition of ethyl iodide by the reaction:

$C_2{H}_{5}I\left(g\right)\to C_2{H}_4\left(g\right)+HI\left(g\right)$

at $600K$ is $1.60\times {10}^{-5}{s}^{-1}$. Its energy of activation is $209kJmo{l}^{-1}$. The rate constant of the reaction at $700K$ will be:

• A. $2.88\times {10}^{-3}{s}^{-1}$
• B. $2.96\times {10}^{-3}{s}^{-1}$
• C. $2.76\times {10}^{-3}{s}^{-1}$
• D. $Noneofthese$

Answer 1

2The correct option is D. None of these
Given;
$k=1.60\times {10}^{-5}{s}^{-1}$
$E_a=209KJmo{l}^{-1}=209\times {10}^{3}Jmo{l}^{-1}$
$T_1=600K$
$T_2=700K$
$R=8.314Jmo{l}^{-1}{K}^{-1}$
We know that
$log\frac{K_2}{K_1}=\frac{E_a}{2.303R}\times \frac{T_2-T_1}{T_1{T}_2}$
$log\frac{K_2}{K_1}=\frac{209\times {10}^3}{2.303\times 8.314}\times \frac{700-600}{600\times 700}$

$log\frac{K_2}{K_1}=2.5989$
$\frac{K_2}{K_1}=antilog2.5989=397.1$
$K_2=397.1\times K_1=397.1\times 1.6\times {10}^{-5}=6.35\times {10}^{-3}{s}^{-1}$