Question

A student has studied the decomposition of a gas AB₃​ at 25^∘C. He obtained the following data.

p(mmHg)	50	100	200	400
\(relative\space  t_{1/2}​(s)\)	4	2	1	0.5

The order of the reaction is……….

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

Hint:

For reactions with variable half-lives, use the relation: \[t_{1/2} \propto (P_0)^{1-n}\]to determine the order of the reaction.

Answer 1

The Correct Option is D

For a reaction, the half-life \(t_{1/2}\) is related to the initial pressure (\(P_0\)) by:
\[t_{1/2} \propto (P_0)^{1-n},\]where \(n\) is the order of the reaction.
Taking the ratio of half-lives at different pressures:
\[\frac{t_{1/2}(P_1)}{t_{1/2}(P_2)} = \left( \frac{P_2}{P_1} \right)^{n-1}.\]
Using the given data:
\[\frac{t_{1/2}(50)}{t_{1/2}(100)} = \left( \frac{100}{50} \right)^{n-1}.\]
Substitute \(t_{1/2}(50) = 4 \, \text{s}\) and \(t_{1/2}(100) = 2 \, \text{s}\):
\[\frac{4}{2} = (2)^{n-1}.\]
Simplify:
\[2 = 2^{n-1}.\]
Taking the logarithm:
\[n - 1 = 1 \implies n = 2.\]
Thus, the order of the reaction is \(n = 2\).