Question

For a reaction, given below is the graph of ln K vs `1/T`. The activation energy for the reaction is equal to ______  Cal mol⁻¹ (Nearest integer).
(Given R = 2 Cal K⁻¹ mol⁻¹).

Answer 1

For a reaction, given below is the graph of ln K vs `1/T`. The activation energy for the reaction is equal to 8 Cal mol⁻¹.

Explanation:

We are given a graph of ln K vs `1/T`.

From the Arrhenius equation:

`ln K = ln A - E_a/R * 1/T`

It is of the form y = mx + c

Where slope m = `- E_a/R`

We need to find activation energy (E_a) in Cal mol⁻¹, given R = 2 Cal K⁻¹mol⁻¹.

From the graph:
At `1/T` = 0, ln⁡ K = 2.
At `1/T` = 51, ln⁡ K = 0.

slope = `(0 - 20)/(5 - 0)`

= `(-20)/5`

= −4

i.e., `- E_a/R = -4`

But, E_a ​= 4 R

E_a​ = 4 × 2 = 8 Cal mol⁻¹