A student has studied the decomposition of a gas A B _ 3 at 25 ° C...

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,

$t\dfrac{1}{2} ∝ {\left({P}_{o}\right)}^{1 - n}$

$\dfrac{{\left({t}_{\dfrac{1}{2}}\right)}_{1}}{{\left(t\dfrac{1}{2}\right)}_{2}} = {\left({P}_{0}\right)}_{1}^{1 - n}{\left({P}_{0}\right)}_{2}^{1 - n}$

$⇒ \left(\dfrac{4}{2}\right) = {\left(\dfrac{50}{100}\right)}^{1 - n}$

$⇒ 2 = {\left(\dfrac{1}{2}\right)}^{1 - n}$

$⇒ 2 = {\left(2\right)}^{n - 1}$

$⇒ n - 1 = 1$

$⇒ n = 2$

Hence, the order of the reaction is second order.