Question

At 1100 k n-nonane thermally cracks 20 times as rapidly as at 1000 k. Find the activation energy for this decomposition\

Answer 1

Answer : The activation energy for the reaction is  27402 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 $1000K$ = x

$K_2$ = rate constant at $1100K$ = 20x

$Ea$ = activation energy for the reaction = ?

R = gas constant = 8.314 J/mole.K

$T_1$ = initial temperature = 1000 K

$T_2$ = final temperature = 1100 K

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

$\log (\frac{20x}{x})=\frac{Ea}{2.303\times 8.314J/mole.K}[\frac{1}{1000K}-\frac{1}{1100K}]$

$Ea=27402J/mole$

Therefore, the activation energy for the reaction is 27402 Joules.