Question

What is the activation energy for a reaction whose rate constant doubles when temperature changes from 30°C to 40°C?

Answer 1

This small amount of energy input necessary for all chemical reactions to occur is called the activation energy (or free energy of activation) and is abbreviated EA. Activation energy: Activation energy is the energy required for a reaction to proceed; it is lower if the reaction is catalyzed.

Answer 2

Answer:

$E_{a}=54658.45 Joules.$

Explanation:

Since we know the relation between the rate of reaction and temperature of the reaction.

Assume $K_{1}$ and $K_{2}$ are the rate constant and $T_{1}$ and $T_{2}$ are the temperature then,

$log\dfrac{K_{2} }{K_{1}} =\dfrac{E_{a}(T_{2}-T_{1}) }{2.303RT_{1}T_{2}}$...............(1)

We have

$\dfrac{K_{2} }{K_{1}} = 2$ and $T_{1}=30+273=303 k$ and $T_{2}=273+40=313k$

Using equation (1) we get,

$E_{a} = \dfrac{log2\times2.303\times 8.314\times 303\times 313}{10}$

After solving we get,

$E_{a}=54658.45 Joules.$