JEE Chemistry 2024 Question with Solution

Given: The reaction is first-order. The rates are $0.04 \, \text{mol} \, \text{L}^{-1} \, \text{s}^{-1}$ at $t_1 = 10$ min and $0.03 \, \text{mol} \, \text{L}^{-1} \, \text{s}^{-1}$ at $t_2 = 20$ min.

Find: The half-life $t_{1/2}$ of the reaction.

For a first-order reaction, rate is proportional to concentration, so we can use the integrated first-order form with rates:

$$k = \frac{2.303}{t_2 - t_1} \log_{10}\left(\frac{\text{Rate}_1}{\text{Rate}_2}\right)$$

Substituting the given values:

$$k = \frac{2.303}{20 - 10} \log_{10}\left(\frac{0.04}{0.03}\right)$$ $$k = \frac{2.303}{10} \log_{10}\left(\frac{4}{3}\right)$$

Using $\log(2) = 0.3010$ and $\log(3) = 0.4771$:

$$k = \frac{2.303}{10} \left(\log_{10}(4) - \log_{10}(3)\right)$$ $$k = \frac{2.303}{10} \left(0.6020 - 0.4771\right)$$ $$k = \frac{2.303}{10} \left(0.1249\right)$$ $$k \approx 0.02877 \, \text{min}^{-1}$$

Now use the first-order half-life relation:

$$t_{1/2} = \frac{0.693}{k}$$ $$t_{1/2} = \frac{0.693}{0.02877}$$ $$t_{1/2} \approx 24.1 \, \text{min}$$

Therefore, the half-life of the reaction is $24$ minutes.

Given: The rates at $10$ min and $20$ min are $0.04$ and $0.03$ respectively.

Find: The half-life $t_{1/2}$.

For a first-order reaction:

$$\text{Rate} = k[A]_0 e^{-kt}$$

At $t = 10$ min and $t = 20$ min:

$$0.04 = k[A]_0 e^{-k(10)}$$ $$0.03 = k[A]_0 e^{-k(20)}$$

Dividing the second equation by the first:

$$\frac{0.03}{0.04} = e^{-k(20-10)}$$ $$\frac{4}{3} = e^{10k}$$

Taking logarithm gives the direct half-life form:

$$t_{1/2} = \frac{10\log_{10}(2)}{\log_{10}(4/3)}$$

Now substitute values:

$$t_{1/2} = \frac{10 \times 0.3010}{0.6020 - 0.4771}$$ $$t_{1/2} = \frac{3.010}{0.1249}$$ $$t_{1/2} \approx 24.08 \, \text{min}$$

Therefore, the half-life is $24$ minutes.