Question

The rate constants of some reactions double with every 10 degree rise in temperature. Assume that a reaction takes place at 271 K and 281 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, 119.7 J

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 271 K

$K_2$ = rate constant at 281 K = $2K_1$

$Ea$ = activation energy for the reaction = ?

R = gas constant = 8.314 J/mole.K

$T_1$ = initial temperature = 271 K

$T_2$ = final temperature = 281 K

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

$\log (\frac{2K_1}{K_1})=\frac{Ea}{2.303\times 8.314J/mole.K}[\frac{1}{271K}-\frac{1}{281K}]$

$Ea=119.7J$

Therefore, the activation energy for the reaction is, 119.7 J