Question

What is the activation energy for a reaction if its rate doubles when the temperature is raised from 20 degree celsius to 35 degree celsius?

Answer 1

The activation energy for the reaction is 47.87 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 $20^oC$ = k

$K_2$ = rate constant at $35^oC$ = 2k

$Ea$ = activation energy for the reaction = ?

R = gas constant = 8.314 J/mole.K

$T_1$ = initial temperature = $20^oC=273+20=293K$

$T_2$ = final temperature = $35^oC=273+35=308K$

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

$\log (\frac{2k}{k})=\frac{Ea}{2.303\times 8.314J/mole.K}[\frac{1}{293K}-\frac{1}{308K}]$

$Ea=47867.85J/mole=47.87kJ$

Therefore, the activation energy for the reaction is 47.87kJ

Learn More

Arrhenius equation

https://brainly.com/question/13043575

Effect of temperature on rate constant

https://brainly.com/question/11528407