Question

the rate of chemical reaction doubles with every 10 rise in temperature. if this generalisation holds good for a reaction in temperature range 295K and 305K. what would be the value of activation energy for this reaction?

Answer 1

As we know
$\text{ln(}\frac{{\text{k}}_1}{{\text{k}}_2})=(\frac{\text{1}}{{\text{T}}_2}-\frac{\text{1}}{{\text{T}}_1})\frac{{\text{E}}_a}{\text{R}}$

$\text{ln(}\frac{1}{2})=(\frac{1}{305}-\frac{1}{295})\frac{{\text{E}}_a}{8.314}$

-0.6931$\times$8.314 = -.00011 E_a
E_a = 52385.76 J
E_a = 52.386 kJ