JEE Chemistry 2023 Question with Solution

Given: The reaction is zero order and $t_{1/2} = 50 \, \text{min}$.

Find: The time when concentration becomes $\frac{[A]_0}{4}$.

For a zero-order reaction, the integrated rate law is

$$[A]_t = [A]_0 - kt$$

and the half-life is

$$t_{1/2} = \frac{[A]_0}{2k}$$

Using $t_{1/2} = 50 \, \text{min}$,

$$50 = \frac{[A]_0}{2k}$$

So,

$$k = \frac{[A]_0}{100}$$

Now for concentration reducing to one-fourth,

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

Substitute in the zero-order equation:

$$\frac{[A]_0}{4} = [A]_0 - kt$$ $$kt = \frac{3[A]_0}{4}$$

Substituting $k = \frac{[A]_0}{100}$,

$$\frac{[A]_0}{100} t = \frac{3[A]_0}{4}$$ $$t = \frac{3}{4} \times 100 = 75 \, \text{min}$$

Therefore, the required time is $75 \, \text{min}$.

Given: Zero-order reaction with $t_{1/2} = 50 \, \text{min}$.

Find: Time for $[A]$ to become $\frac{[A]_0}{4}$.

In a zero-order reaction, concentration decreases linearly with time. In $50 \, \text{min}$, the concentration falls from $[A]_0$ to $\frac{[A]_0}{2}$, so the amount consumed is $\frac{[A]_0}{2}$.

To reach $\frac{[A]_0}{4}$, the amount consumed must be

$$[A]_0 - \frac{[A]_0}{4} = \frac{3[A]_0}{4}$$

This is $\frac{3}{2}$ times the half-life consumption $\left(\frac{[A]_0}{2}\right)$. Therefore time is also $\frac{3}{2}$ times:

$$t = \frac{3}{2} \times 50 = 75 \, \text{min}$$

Therefore, the required time is $75 \, \text{min}$.