Question

At a certain temperature, the half life periods of a reaction at different initial pressure were as follows:
$\begin{array}{cccc}p\left(kPa\right)& 100& 66.67& 33.33\\ t\left(min\right)& 105& 235& 950\end{array}$
Find out the order of the reaction.

Answer 1

$P\left(KPa\right)$ $100$ $66.67$ $33.33$

$t\left(min\right)$ $105$ $235$ $950$

As we will consider partial pressure as concentration

of reactant (initial) in for case gascase reaction.

As you can see partial pressure is not directly

proportional to the change in half life hence

it is not zero order reaction Let is check

for second order reaction

$k_1\frac{1}{\frac{11}{2}p}$

$k_1\frac{1}{105}\times 60\times 100\times {10}^3=1.58\times {10}^{-9}P{a}^{-1}se{c}^{-1}$

$k_2=\frac{1}{66.67}\times 235\times 60\times {10}^3=1.06\times {10}^{-9}P{a}^{-1}Se{c}^{-1}$

$K_3=\frac{1}{33.33}\times 750\times 60\times {10}^3=0.526\times {10}^{-1}P{a}^{-1}se{c}^{-1}$

As $k_1\approx k_2\approx k_3$

Hence it is a second order reaction.