Question

For the reaction \[\ce{A -> B + C}\], the data given below were obtained. Prove that the reaction is of first order.

t	0	90	180
[A]	50.8	19.7	7.62

Answer 1

To prove that the reaction \[\ce{A -> B + C}\] is first order, we must show that it obeys the first-order rate law, whose integrated form is

`ln([A]_0/([A])) = kt`

or `log([A]_0/([A])) = (kt)/2.303`

This implies if the reaction is first-order, a plot of log⁡[A] versus t should be linear, and the rate constant k computed for each time interval should be constant.

By using the first-order formula `k = 2.303/t log ([A]_0/([A]))`

Between 0 and 90 sec:

`k_1 = 2.303/90 log (50.8/19.7)`

= `2.303/90 * log(2.578)`

= `2.303/90 * 0.4116`

= `0.948/90`

= 0.01053 s⁻¹

Between 0 and 180 sec:

`k_2 = 2.303/180 log (50.8/7.62)`

= `2.303/180 * log(6.666)`

= `2.303/180 * 0.8239`

= `1.897/180`

= 0.01045 s⁻¹

Since k₁ ≈ k₂ ≈ 0.0105 s⁻¹, the rate constant is independent of time, which confirms that the reaction follows first-order kinetics.

∴ The reaction is of first order.