Chemical Kinetics Question 2

Question 2 - 24 January - Shift 2

A student has studied the decomposition of a gas $AB_3$ at $25^{\circ} C$. He obtained the following data.

$p(mm Hg)$ 50 100 200 400
Relative $ _{1 / 2}(s)$ 4 2 1 0.5

The order of the reaction is

(1) 0.5

(2) 2

(3) 1

(4) 0 (zero)

Show Answer

Answer: (2)

Solution:

$t _{1 / 2} \propto(P_O)^{1-n}$

$\frac{(t _{1 / 2}^{t h})_1}{(t _{1 / 2})_2}=\frac{(P_0)_1^{1-n}}{(P _{0_2})_2^{1-n}}$

$\Rightarrow(\frac{4}{2})=(\frac{50}{100})^{1-n}$

$\Rightarrow 2=(\frac{1}{2})^{1-n}$

$\Rightarrow 2=(2)^{n-1}$

$\Rightarrow n-1=1$

$\Rightarrow n=2$