A student has studied the decomposition of a gas $A B_{3} at 25^{\circ} C$ . 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 reactions is

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0.5

answer is B.

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Detailed Solution

$\begin{array}{l} \frac{{(t_{1 / 2})}_{1}}{{(t_{1 / 2})}_{2}} = {(\frac{a_{2}}{a_{1}})}^{n - 1} \\ \frac{4}{2} = {(\frac{100}{50})}^{n - 1} \end{array}$
$2 = 2^{n - 1} \Rightarrow 2^{n - 1} = 2^{1} \Rightarrow n - 1 = 1 \Rightarrow n = 2$

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