Question

A certain reaction has an activation energy of 66.41 kJ/mol. At what Kelvin temperature will the reaction proceed 3.00 times faster than it did at 293 K?

Answer 1

Answer : The temperature will be, 392.462 K

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 $T_1$  = $K_1$

$K_2$ = rate constant at $T_2$ = $3K_1$

$Ea$ = activation energy for the reaction = 66.41 kJ/mole = 66410 J/mole

R = gas constant = 8.314 J/mole.K

$T_1$ = initial temperature = 293 K

$T_2$ = final temperature = ?

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

$\log (\frac{3K_1}{K_1})=\frac{66410J/mole}{2.303\times 8.314J/mole.K}[\frac{1}{293K}-\frac{1}{T_2}]$

$T_2=392.462K$

Therefore, the temperature will be, 392.462 K