JEE Chemistry 2026 Question with Solution

Given: A plot of $[AB]$ versus time is provided, and the line decreases linearly with time.

Find: The half-life $x$ in minutes.

From the graph, the $y$-axis is $[AB]$, not $\ln[AB]$ or $1/[AB]$. Therefore, the reaction is zero order.

For a zero-order reaction:

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

From the graph:

• at $t = 0$, $[AB]_0 \approx 0.60 \, \text{M}$
• at $t = 100 \, \text{s}$, $[AB] \approx 0.55 \, \text{M}$
• at $t = 200 \, \text{s}$, $[AB] \approx 0.50 \, \text{M}$

So the rate constant is obtained from the slope magnitude:

$$k = \frac{0.60 - 0.55}{100} = \frac{0.05}{100} = 5 \times 10^{-4} \, \text{M s}^{-1}$$

For a zero-order reaction, half-life is:

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

Substituting the values:

$$t_{1/2} = \frac{0.60}{2 \times 5 \times 10^{-4}} = \frac{0.60}{10^{-3}} = 600 \, \text{s}$$

Convert into minutes:

$$600 \, \text{s} = 10 \, \text{min}$$

Therefore, the value of $x$ is $10$.

The solution states an inconsistent second-order interpretation and a final value of $4$, but that does not match the provided graph or the listed correct answer. Using the graph, the correct numerical answer is $10$.

Given: The graph is a straight-line plot of $[AB]$ against time.

Find: The half-life.

A straight-line decrease of concentration with time corresponds to a zero-order reaction, because:

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

Half-life for zero order is not constant in general form and is given by:

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

Now read the graph directly. The concentration falls from $0.60 \, \text{M}$ to $0.50 \, \text{M}$ in $200 \, \text{s}$. Hence,

$$k = \frac{0.60 - 0.50}{200} = \frac{0.10}{200} = 5 \times 10^{-4} \, \text{M s}^{-1}$$

Initial concentration is:

$$[AB]_0 = 0.60 \, \text{M}$$

So,

$$t_{1/2} = \frac{0.60}{2 \times 5 \times 10^{-4}} = 600 \, \text{s}$$

Thus,

$$t_{1/2} = 10 \, \text{min}$$

Therefore, the nearest integer value is $10$.