Question

A certain reaction with an activation energy of 185 kJ/mol was run at 525 K and again at 545 K . What is the ratio of f at the higher temperature to f at the lower temperature?

Answer 1

Answer: The ratio of f at the higher temperature to f at the lower temperature is 4.736

Explanation:

According to the Arrhenius equation,

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

or,

$\log (\frac{f_2}{f_1})=\frac{Ea}{2.303\times R}[\frac{1}{T_1}-\frac{1}{T_2}]$

where,

$f_1$ = rate constant at 525K

$K_2$ = rate constant at 545K

$Ea$ = activation energy for the reaction = 185kJ/mol= 185000J/mol   (1kJ=1000J)

R = gas constant = 8.314 J/mole.K

$T_1$ = initial temperature = 525 K

$T_2$ = final temperature = 545 K

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

$\log (\frac{f_2}{f_1})=\frac{185000J/mol}{2.303\times 8.314J/mole.K}[\frac{1}{525K}-\frac{1}{545K}]$

$\log (\frac{f_2}{f_1})=0.6754$

$(\frac{f_2}{f_1})=4.736$

Therefore, the ratio of f at the higher temperature to f at the lower temperature is 4.736