Question

The half-life period and initial concentration for a reaction are as follows. What is the order of reaction?

Initial conc. [mol L−1]	350	540	158
t1/2 (sec)	425	275	941

Answer 1

For a reaction of order n,

`t_(1//2)/t_(1//2)^' = (([A]_0^')/([A]_0))^(n - 1)`

Putting t_1/2 = 425 s, `t_(1//2)^'` = 275 s,

[A]₀ = 350 mol L⁻¹ and

`[A]_0^'` = 540 mol L⁻¹, we get

`425/275 = (540/350)^(n - 1)`

or, `log_10  425/275 = (n - 1) log_10  (540/350)`

or, 0.1890 = (n − 1) × 0.1883,

or, n − 1 = `0.1890/0.1883` = 1.004

or, n = 1 + 1.004

= 2.004 

= 2

Hence, the given reaction is of second order.