At 30^\circ C , the half life for the decomposition of AB_2 is 200 ~s and is independent of the initial concentration of AB_2 . The time required for 80 \% of the AB_2 to decompose is Given: \log 2=0.30\quad \log 3=0.48

Chemical Kinetics and Nuclear Chemistry PYQ — JEE Main 2022 · 26 Jul · Shift 2 · Q4 with Solution

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Correct answer: C

The half-life is given to be independent of the initial concentration. This is the characteristic of a first-order reaction.

For a first-order reaction,

t_{1/2} = \frac{0.693}{k}

Given:

t_{1/2} = 200\,\text{s}

So,

k = \frac{0.693}{200}

If $80\%$ of $\mathrm{AB_2}$ decomposes, then $20\%$ remains.

So,

\frac{[A]_t}{[A]_0} = 0.20

For a first-order reaction,

t = \frac{2.303}{k} \log \frac{[A]_0}{[A]_t}

Thus,

t = \frac{2.303}{k} \log \frac{1}{0.2} = \frac{2.303}{k} \log 5

Now,

\log 5 = \log \frac{10}{2} = \log 10 - \log 2 = 1 - 0.30 = 0.70

Hence,

t = \frac{2.303}{k} \times 0.70

Substitute $k = \frac{0.693}{200}$ :

t = \frac{2.303 \times 0.70 \times 200}{0.693}

Since

2.303 \times 0.70 = 1.6121

so,

t = \frac{1.6121 \times 200}{0.693} = \frac{322.42}{0.693} \approx 465\,\text{s}

This is closest to:

\boxed{467\,\text{s}}

Therefore, the correct option is C.