A student has studied the decomposition of a gas $A B_{3}$ at $25 ° C$ . He obtained the following data
$p (m m H g)$ $50$ $100$ $200$ $400$
Relative $t_{\frac{1}{2}} (s)$ $4$ $2$ $1$ $0.5$
The order of the reaction is

Question

A student has studied the decomposition of a gas A B 3 at 25 ° C . He obtained the following data p ( m m H g ) 50 100 200 400 Relative t 1 2 s 4 2 1 0 . 5 The order of the reaction is

TestHub · Accepted Answer

By observing the data given, by doubling the pressure, half-life is reduced to half. This result is following second order kinetics. Mathematically it can be proved as follows,

The half-life and initial pressure of $n^{th}$ order reaction can be related as follows,

$t_{\frac{1}{2}} \propto {(P_{o})}^{1 - n}$

$\frac{{(t_{\frac{1}{2}})}_{1}}{{(t_{\frac{1}{2}})}_{2}} = \frac{{(P_{0})}_{1}^{1 - n}}{{(P_{0})}_{2}^{1 - n}}$

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

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

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

$\Rightarrow n - 1 = 1$

$\Rightarrow n = 2$

Hence, the order of the reaction is second order.

Chemistry - Chemical Kinetics Question with Solution | TestHub

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Doubts & Discussion

$p (m m H g)$	$50$	$100$	$200$	$400$
Relative $t_{\frac{1}{2}} (s)$	$4$	$2$	$1$	$0.5$