Question

The rate of a gaseous reaction is halved when the volume of the vessel is doubled. What is the order of reaction?

Answer 1

Given:

The volume of the vessel is doubled, which means the concentration of gases is halved.

The rate of reaction is halved as well.

Let the initial concentration be C, and the rate law be:

Rate = k[C]ⁿ

After doubling the volume, 

New concentration = `C/2`

New rate = `k(C/2)^n`

= `1/2 * kC^n`

So,

`(1/2)^n = 1/2`

⇒ n = 1

∴ The order of reaction = 1

∴ It is a first-order reaction.