Answer to Question #108839 in Physical Chemistry for Mitul

Question

The rate constant of a certain first order reaction is doubled when the temperature is increased from
27 °C to 37 °C. Calculate the energy of activation and Arrhenius factor for the reaction.

Answer 1

As $K=Ae^{-\frac{E_a}{RT}}$

So, at $T=27°C=300K$

$K=Ae^{-\frac{E_a}{300R}}$

At $T=37°C=310K$

$2K=Ae^{-\frac{E_a}{310R}}$

Dividing the two equations, we get

$2=e^{\frac{E_a}{300R}-\frac{E_a}{310R}}$

$log2=\frac{E_a}{300R}-\frac{E_a}{310R}$

$E_a=53582.8986J$

Now, $K=Ae^{-\frac{E_a}{300R}}=4.6779\times10^{-10}A$

Hence, Arrhenius factor $A=0.2137\times10^{10}K$