Question

A student has studied the decomposition of a gas $A{B}_3$ at $25^{\circ }C$. He obtained the following data.

$p(mmHg)$	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)

$t_{1/2}\propto {\left(P_O\right)}^{1-n}$
$\frac{{\left(t_{1/2}\right)}_1}{{\left(t_{1/2}\right)}_2}=\frac{{\left(P_0\right)}_1^{1-n}}{{\left(P_{0_2}\right)}_2^{1-n}}$
$\Rightarrow \left(\frac{4}{2}\right)-{\left(\frac{50}{100}\right)}^{1-n}$
$\Rightarrow 2={\left(\frac{1}{2}\right)}^{1-n}$
$\Rightarrow 2=(2{)}^{n-1}$
$\Rightarrow n-1=1$
$n=2$