Question

Calculate the rate constant of a reaction at $293$ $K$ when the energy of activation is $103$ $kJ$ ${mol}^{-1}$ and the rate constant at $273$ $K$ is $7.87\times {10}^{-7}$ $s^{-1}$.
$(R=8.314\times {10}^{-3}kJ{mol}^{-1}{K}^{-1})$

Answer 1

The Arrhenius equation is,
${\log }_{10}\frac{k_2}{k_1}=\frac{E_a}{2.303R}\left[\frac{T_2-T_1}{T_1{T}_2}\right]$
Given: $k_1=7.87\times {10}^{-7}$ $s^{-1};$ $E_a=103$ $kJ$ ${mol}^{-1};$
$R=8.314\times {10}^{-3}$ $kJ$ ${mol}^{-1}$ $K^{-1};$
$T_1=273$ $K$ and $T_2=293$ $K$
Substituting the values in Arrhenius equation,
${\log }_{10}\frac{k_2}{7.87\times {10}^{-7}}=\frac{103\times 20}{2.303\times 8.314\times {10}^{-3}\times 293\times 273}$
$=1.345$
$k_2=1.74\times {10}^{-5}$ $s^{-1}$