JEE Main 2025 — 28 January, Morning Shift — Question 32

$\quad \mathrm{t}_{1 / 2} \propto \frac{1}{\mathrm{~A}_{0}^{\mathrm{n}-1}}$

$\frac{\left(\mathrm{t}_{1 / 2}\right)_{1}}{\left(\mathrm{t}_{1 / 2}\right)_{2}}=\frac{\left(\mathrm{A}_{0}\right)_{2}^{n-1}}{\left(\mathrm{~A}_{0}\right)_{1}^{\mathrm{n}-1}}$

$\frac{200}{100}=\left(\frac{0.025}{0.100}\right)^{n-1}$ $2=\left(\frac{1}{4}\right)^{\mathrm{n}-1}$

$\mathrm{n}-1=-\frac{1}{2}$

$\mathrm{n}=\frac{1}{2}$ (order) $\Rightarrow \mathrm{t}_{1 / 2} \propto \sqrt{\mathrm{~A}_{0}}$

$\frac{200}{t_{1 / 2}}=\frac{(0.1)^{1 / 2}}{(1)^{1 / 2}}$ when $\mathrm{A}_{0}=1 \mathrm{M}$

$\mathrm{t}_{1 / 2}=200 \sqrt{10} \mathrm{~min}$

Ist order kinetics have $\mathrm{t}_{1 / 2}$ independent of their concentration.

So upon changing the concentration $t_{1 / 2}$ should not change for first order reaction.

$\frac{200}{t_{1 / 2}}=\frac{(0.1)^{1 / 2}}{(1.6)^{1 / 2}}$ when $\mathrm{A}_{0}=1.6 \mathrm{M}$

$\mathrm{t}_{1 / 2}=800 \mathrm{~min}$ \end{itemize}