Question

At a certain temperature,the half life periods for the catalytic decomposition of ammonia were found to be as;
Pressure (mm) 50 100 200
$t_{1/2}$ 3.52 1.92 1.0
What is the order of reaction?

• A. 0
• B. 1
• C. 2
• D. 3

Answer 1

Answer: (C)

2

The data given in the problem

Pressure (mm) $50$ $100$ $200$

$t_{1/2}$ $3.52$ $1.92$ $1.0$

(I) (II) (III)

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

Since, we have

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

where, $po$= initial pressure of reactant gas

So, for $n=2$; $t_{1/2}\propto \left(\frac{1}{P_o}\right)$

That means for second order reaction $t_{1/2}$ is inversely proportional to pressure of reactant gas. So, the given reaction is of second order.