Chemistry · Chemical Kinetics

JEE Main 2024 — 31 January, Shift 2 — Question 73

$r=k[A]$ for a reaction, $50 \%$ of $A$ is decomposed in 120 minutes. The time taken for $90 \%$ decomposition of A is $\qquad$ minutes.

Answer: 399

Numerical answer — enter this value.

$\quad r=k[A]$

So, order of reaction $=1$

$\mathrm{t}_{1 / 2}=120 \mathrm{~min}$

For $90 \%$ completion of reaction $\Rightarrow \mathrm{k}=\frac{2.303}{\mathrm{t}} \log \left(\frac{\mathrm{a}}{\mathrm{a}-\mathrm{x}}\right)$

$\Rightarrow \frac{0.693}{t_{1 / 2}}=\frac{2.303}{t} \log \frac{100}{10}$

$\therefore \mathrm{t}=399 \mathrm{~min}$ .

Answer key and solution verified before publishing.

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