Question

The value of rate constant for the reaction, \[\ce{A -> Products}\], is 5.6 × 10⁻³ mol⁻¹ L s⁻¹. How will the rate change when the concentration of A is halved?

Answer 1

The given reaction is

\[\ce{A -> Products}\]

and rate constant is

k = 5.6 × 10⁻³ mol⁻¹ L s⁻¹

The unit of k is mol⁻¹ L s⁻¹

This corresponds to a second-order reaction.

so the rate law is 

Rate = k[A]²

Let original concentration be [A] = a, so

Original Rate = ka²

If [A] is halved → new concentration = a/2

Then

New rate = `k(a/2)^2`

= `k * a^2/4`

= `1/4 xx "Original rate"`

∴ The rate becomes one-fourth when the concentration of A is halved.