JEE Chemistry 2024 Question with Solution

Given: A first-order reaction and the required extent of completion is $99.9\%$.

Find: Time required in terms of the half-life $t_{1/2}$.

For a first-order reaction,

$$N = N_0 e^{-kt}$$

where $N_0$ is the initial concentration and $N$ is the concentration at time $t$.

For half-life,

$$N = \frac{N_0}{2}$$

So,

$$\frac{N_0}{2} = N_0 e^{-k t_{1/2}}$$

Thus,

$$t_{1/2} = \frac{\ln 2}{k}$$

For $99.9\%$ completion, only $0.1\%$ reactant remains, so

$$N = 0.001 N_0$$

Hence,

$$0.001 N_0 = N_0 e^{-k t}$$

Therefore,

$$-kt = \ln(0.001)$$

Using $\ln(0.001) = -6.90775$,

$$t = \frac{6.90775}{k}$$

Now compare with half-life:

$$\frac{t}{t_{1/2}} = \frac{6.90775/k}{0.693/k} \approx 9.9658 \approx 10$$

Therefore, the required time is $10$ times the half-life, so the answer is $10$.

Given: A first-order reaction with $99.9\%$ completion.

Find: The multiple of half-life required.

Use the first-order relation

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

For $99.9\%$ completion,

$$\frac{[A]}{[A]_0} = 0.001$$

So,

$$t = \frac{2.303}{k} \log \frac{1}{0.001}$$

Since

$$\log(10^3) = 3$$

we get

$$t = \frac{2.303}{k} \times 3$$

Also, for a first-order reaction,

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

Thus,

$$t = \frac{6.909}{k} \approx 10 \times t_{1/2}$$

This works because $99.9\%$ completion means only $10^{-3}$ of the reactant remains. Therefore, the answer is $10$.