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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