Question

The following data were obtained at a certain temperature for the decomposition of ammonia in contact with tungsten:

$\begin{array}{cccc}\text{Relative}{t}_{1/2}& 3.64& 1.82& 0.91\\ \text{p(mm)}& 50& 100& 200\end{array}$
($t_{1/2}$ is the half life and $\text{p}$ is the partial pressure of ammonia at different intervals.)
Find the order of the reaction.

• A. $1$
• B. $2$
• C. $0$
• D. $3$

Answer 1

The correct option is B $2$

We know that
$t_{1/2}\propto \frac{1}{(a{)}^{n-1}}$

Here $a$ refers to the concentration of reaction mixture. For gaseous reactants the concentration is similar to the pressure of the gas.
$\therefore \frac{(t_{1/2}{)}_1}{(t_{1/2}{)}_2}={\left(\frac{a_2}{a_1}\right)}^{n-1}={\left(\frac{p_2}{p_1}\right)}^{n-1}\Rightarrow \frac{3.64}{1.82}={\left(\frac{100}{50}\right)}^{n-1}\Rightarrow 2=2^{n-1}\Rightarrow n-1=1\Rightarrow n=2$