JEE Chemistry 2023 Question with Solution

Given: First-order reaction $A \rightarrow B$ with half-life $t_{1/2} = 30 \, \text{min}$.

Find: Time required for $75\%$ completion.

For a first-order reaction,

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

For $75\%$ completion, the reactant left is $25\%$, so

$$[A] = \frac{1}{4}[A]_0$$

Substituting,

$$t = \frac{2.303}{k} \log \frac{[A]_0}{\frac{1}{4}[A]_0} = \frac{2.303}{k} \log 4$$

Now use the half-life relation for a first-order reaction,

$$k = \frac{0.693}{t_{1/2}} = \frac{0.693}{30}$$

Substituting $k$ and $\log 4 = 0.602$,

$$t = \frac{2.303 \times 0.602}{0.693/30} = 60 \, \text{min}$$

Therefore, the time taken is $60 \, \text{min}$.

Given: The half-life is $30 \, \text{min}$.

Find: Time for $75\%$ completion.

In a first-order reaction, after one half-life, $50\%$ of the reactant remains. After two half-lives, $25\%$ remains, which means $75\%$ completion.

So the required time is

$$t = 2 \times 30 = 60 \, \text{min}$$

Therefore, the required time is $60 \, \text{min}$.