Question

The rate constants of some reactions double with every 10-degree rise in temperature. Assume that a reaction takes place at 295 K and 305 K. What must the activation energy be for the rate constant to double as described?

Answer 1

Answer : The activation energy for the reaction is, 51.9 kJ

Explanation :

According to the Arrhenius equation,

$K=A\times e^{\frac{-Ea}{RT}}$

or,

$\log (\frac{K_2}{K_1})=\frac{Ea}{2.303\times R}[\frac{1}{T_1}-\frac{1}{T_2}]$

where,

$K_1$ = rate constant at 295 K

$K_2$ = rate constant at 305 K = $2K_1$

Ea = activation energy for the reaction = ?

R = gas constant = 8.314 J/mole.K

$T_1$ = initial temperature = 295 K

$T_2$ = final temperature = 305 K

Now put all the given values in this formula, we get:

$\log (\frac{2K_1}{K_1})=\frac{Ea}{2.303\times 8.314J/mole.K}[\frac{1}{295K}-\frac{1}{305K}]$

$Ea=51879.96J=51.9kJ$

Therefore, the activation energy for the reaction is, 51.9 kJ