Question

The half-lives against initial pressure are given below. Calculate the order of the reaction.
$\begin{array}{cccc}p\left(mm\right)& 750& 500& 250\\ t\left(min\right)& 105& 235& 950\end{array}$
Find out the order of the reaction.

Answer 1

For a general $n^{th}$ order reaction we know that

$t_{1/2}\propto \frac{1}{a_0^{n-1}}$

where $a_0$ is initial concentration,

& $n$ is order

Also we know $a_0\propto pressure\left(initial\right)$

Hence $t_{1/2}\propto \frac{1}{P_0^{n-1}}\left(1\right)$

where $P_0$ is initial ressure

For two different cases, using quation $\left(1\right)$ we can write,

$\frac{t_{1/21}}{t_{1/22}}=\frac{P_{02}^{n-1}}{P_{01}^{n-1}}$

taking log on both sides

$log\left(\frac{t_{1/21}}{t_{1/22}}\right)=(n-1)log\left(\frac{P_{02}}{P_{01}}\right)$

$\Rightarrow n-1=\frac{log\left(\frac{t_{1/21}}{t_{1/22}}\right)}{log\left(\frac{P_{O2}}{P_{01}}\right)}$

putting the first two given data we get

$n=1+\frac{log\left(\frac{105}{235}\right)}{log\left(\frac{500}{750}\right)}$

$\Rightarrow n\approx 2.98\approx 3$

Hence order of reaction is $3$.