Question

The following data were obtained at a certain temperature for the decomposition of ammonia;

$\begin{array}{cccc}p\left(mm\right)& 50& 100& 200\\ t1/2& 3.64& 1.82& 0.91\end{array}$
The order of the reaction is :

• A. 0
• B. 1
• C. 2
• D. 3

Answer 1

The correct option is C 2

As given in problem

$p\left(mm\right)$ $50$ $100$ $200$

$t_{1/2}$ $3.64$ $1.02$ $0.91$

$\left(I\right)$ $\left(II\right)$ $\left(III\right)$

Comparing $\left(I\right)$ and $\left(II\right)$ data, we can see that as the pressure is doubled, half life of the reaction $t_{1/2}$ is halved. Similar observation can be made by comparing $\left(II\right)$ and $\left(III\right)$.

Since we have,

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

where, $po$= initial pressure of the reactant

So, for $n=2$ ; $t_{1/2}\propto \frac{1}{po}$

That means for second order reaction $t_{1/2}$ is inversely proportional to pressure of the reactant gas. The similar observation can be made in the data set given so the reaction is of second order.