Question

The time for half change for a gaseous reaction was measured for various initial pressure and the following data were obtained:
$\begin{array}{cccc}p\left(kPa\right)& 26.6& 40& 53.3\\ t\left(min\right)& 150& 99.8& 75.3\end{array}$
Find out the order of the reaction.

Answer 1

$p\left(kPa\right)$ $26.6$ $40$ $53.3$

$t\left(min\right)$ $150$ $99.8$ $75.3$

We will consider partial pressure as initial concentration

in case of gaseous reaction

As you can see partial pressure is not during proportional;

to the change in half life hence it is not zero

order reaction.

Let's check for second order reaction

$K=\frac{1}{t_{\frac{1}{2}}p}$

$K_1=\frac{1}{150}\times 60\times 26.6\times {10}^3=4.17\times {10}^{-9}P{a}^{-1}se{c}^{-1}$

$K_2=\frac{1}{40}\times 99.8\times 60\times {10}^3=4.17\times {10}^{-9}$ $P{a}^{-1}se{c}^{-1}$

$K_3=\frac{1}{53.3}\times {10}^3\times 75.3\times 60=4.15\times {10}^{-9}P{a}^{-1}se{c}^{-1}$

As $K_1\sim K_2\sim K_3$

Hence it is a second order reaction