Question

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 $t_{1/2} (s)$	4	2	1	0.5

The order of the reaction is

• A. 1
• B. $0.5$
• C. 0 (zero)
• D. 2

Answer 1

Answer: (D)

$t _{1 / 2} \propto\left( P _{ O }\right)^{1- n }$
$\frac{\left( t _{1 / 2}\right)_1}{\left( t _{1 / 2}\right)_2}=\frac{\left( P _0\right)_1^{1- n }}{\left( P _{0_2}\right)_2^{1- n }}$
$\Rightarrow\left(\frac{4}{2}\right)-\left(\frac{50}{100}\right)^{1- n }$
$\Rightarrow 2=\left(\frac{1}{2}\right)^{1- n }$
$\Rightarrow 2=(2)^{ n -1}$
$\Rightarrow n -1=1$
$n =2$