Question

For a chemical reaction the specific rate constant at 600K is $1.60\times {10}^6{s}^{-1}$, calculate the rate constant for the reaction at 700K, if energy of activation for the reaction is $2.209KJ{mol}^{-1}$.

Answer 1

Arrhenius equation for calculation of rate constant at different temperatures is given by

$K=A$ $e^{-Ea/RT}$

where,

$K$= rate constant at temperature $TK$

$A$= Arrhenius constant

$Ea$= activation energy of the reaction

$R$= universal gas constant

$T$= temperature in $K$

Let, us suppose, at temperature $T,K$ and $T_{2}K$ the rate constants are $K_1$ and $K_2$ respectively,so,

$K_1=A$ $e^{-Ea/R{T}_1}$ $-\left(i\right)$

$K_2=A$ $e^{-Ea/R{T}_{2}2}$ $-\left(ii\right)$

$\left(ii\right)/\left(i\right)$

$\Rightarrow \frac{K_2}{K_1}=\frac{e^{-Ea/R{T}_2}}{e^{-Ea/R{t}_1}}$

Taking log on both sides:-

$\Rightarrow \log \left(\frac{K_2}{K_1}\right)=\log \left(e^{\frac{-Ea}{R}(\frac{1}{T_2}-\frac{1}{T_1})}\right)$

$\Rightarrow \log \frac{K_2}{K_1}=\frac{Ea}{R}(\frac{1}{T_1}-\frac{1}{T_2})$

Now, according to question,

$T_1=600K$, $K_1=1.60\times {10}^6{s}^{-1}$

$K_2=?$, $T_2=700K$, $Ea=2.209\times {10}^3$ $J/mol$

$R=8.314$ $J{K}^{-1}mo{l}^{-1}$

$\Rightarrow \log \left(\frac{K_2}{K_1}\right)=\frac{+2209}{8.314}[\frac{1}{600}-\frac{1}{700}]$

$\Rightarrow \log \left(\frac{K_2}{K_1}\right)=\frac{2209\times 100}{83140\times 6\times 7}$

$\Rightarrow \log \left(\frac{K_2}{1.6\times {10}^6}\right)=0.63$

Taking antilog on both sides:-

$\Rightarrow \frac{K_2}{1.6\times {10}^6}=$ $Antilog\left(0.63\right)$

$\Rightarrow K_2=1.87\times 1.6\times {10}^6$

$\Rightarrow K_2=3\times {10}^6{s}^{-1}$