JEE MAIN - Chemistry (2023 - 24th January Evening Shift - No. 9)

A student has studied the decomposition of a gas AB$$_3$$ 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 reaction is

2
0.5
1
0 (zero)

Explanation

$$ \begin{aligned} & \mathrm{t}_{1 / 2} \propto\left(\mathrm{P}_{\mathrm{o}}\right)^{1-\mathrm{n}} \\\\ & \frac{\left(\mathrm{t}_{1 / 2}\right)_1}{\left(\mathrm{t}_{1 / 2}\right)_2}=\frac{\left(\mathrm{P}_o\right)_1^{1-\mathrm{n}}}{\left(\mathrm{P}_{o_2}\right)_2^{1-\mathrm{n}}} \\\\ & \Rightarrow\left(\frac{4}{2}\right)=\left(\frac{50}{100}\right)^{1-\mathrm{n}} \\\\ & \Rightarrow 2=\left(\frac{1}{2}\right)^{1-\mathrm{n}} \\\\ & \Rightarrow 2=(2)^{\mathrm{n}-1} \\\\ & \Rightarrow \mathrm{n}-1=1 \\\\ & \Rightarrow \mathrm{n}=2 \end{aligned} $$

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