The pressure of a gas decomposing at the surface of a solid catalyst has been measured at different times and the results are given below:

 

t/s	0	100	200	300
p/pa	4.00 x 103	3.50 x 103	3.00 x 103	2.5 x103

Determine the order of reaction, its rate constant and half-life period?
It can be seen that rate of reaction between different time interval is

0–100 s, rate = – [3.50 – 400] x 10³ Pa/100 s = 5 Pa / s
100-200 s, rate = – [3.00 – 3.50] x 10³ Pa/100 s = 5 Pa / s
200-300 s, rate = – [2.50 – 3.00] x 10³ Pa / 100 s = 5 Pa / s

We notice that the rate remains constant, therefore, reaction is of zero order.
k = rate = 5 Pa / s

t_{1 / 2} = initial concentration or pressure/2K

= 4.00 x 10³ Pa / 2 x 5 Pa s^–1 = 400 s.

Half-life of a reaction: The half-life of a reaction represented as t_1/2 is the time required for the reactant concentration to drop to one half of its initial value. Consider the zeroth order reaction.
R → Products

Integrate Zeroth order rate equation.

[R] = – kt + [R]₀

where at time t = t_1/2, the fraction of [R] that remains [R] / [R]₀, therefore above equation can be written

[R]₀/2 = – k t_1/2 + [ R]₀
k_1/2 = [R]₀ /2
                      ${\mathrm{t}}_{1/2} = {\left[\mathrm{R}\right]}_0/2\mathrm{k}$

for the first order integrated rate equation

$\mathrm{In}{\left[\mathrm{R}\right]}_{\mathrm{t}}/{\left[\mathrm{R}\right]}_0 = -\mathrm{kt}$


at time ${\mathrm{t}}_{1/2},$  ${\left[\mathrm{R}\right]}_{\mathrm{t}}/{\left[\mathrm{R}\right]}_0 = 1/2$

$$\begin{gathered} \mathrm{In} \frac{1}{2} = -\mathrm{kt} 1/2 \\ {\mathrm{kt}}_{1/2} = {\mathrm{In}}^2 \\ {\mathrm{t}}_{1/2} = 2.303 \log 2/\mathrm{k} = 0.693 \mathrm{k} \end{gathered}$$

The half life depends on reactant concentration in different order of reactions as follows.
For zero order raction t_1/2 ∝ [R]₀. For first order reaction t_1/2 is independent of R₀, for second order reaction t_1/2 ∝ 1/[R]₀.
For nth order reaction t_1/2 ∝ 1/[R]₀^n–1.