Question

At a certain temperature, the half-life period for the catalytic decomposition of ammonia was found as follows:

Pressure (Pascals):	6667	13333	26666
Half-life period in hours:	3.52	1.92	1.0

Calculate the order of reaction.

Answer 1

$\frac{{\left(t_{1/2}\right)}_1}{{\left(t_{1/2}\right)}_2}={\left(\frac{a_2}{a_1}\right)}^{n-1}$ where, $n$ is order of reaction
From the given data,
$\frac{3.52}{1.92}={\left(\frac{13333}{6667}\right)}^{n-1}$ ($a\propto$ initial pressure)
$={\left(2\right)}^{n-1}$
$\log \frac{3.52}{1.92}=(n-1)\log 2=0.3010\times (n-1)$
$0.2632=0.3010\times (n-1)$
$n=1.87\approx 2$
Similar calculations are made between first and third observations, $n$ comes equal to $1.908$ $(\approx 2)$.
Thus, the reaction is of second order.