Q25 / 75 · JEE Main 2026 (05 April Shift 2) · Chemistry

For a reaction $A \rightarrow P$ at $T\text{ K}$, the half life $(t_{1/2})$ is plotted as a function of initial concentration $[A]_0$ of $A$ as given below.
Figure
The value of $x$ in the given figure is _______ s (Nearest integer)

Correct answer: 90

Solution

From the given graph, the plot of half-life $t_{1/2}$ versus initial concentration $[A]_0$ is a straight line passing through the origin.

This indicates that $t_{1/2}$ is directly proportional to $[A]_0$, which means $t_{1/2} \propto [A]_0$.

Therefore, the ratio $\dfrac{t_{1/2}}{[A]_0}$ is constant.

Using the coordinates from the graph, we get:

$\dfrac{240}{4 \times 10^{-3}} = \dfrac{x}{1.5 \times 10^{-3}}$

Solving for $x$:

$x = \dfrac{240 \times 1.5 \times 10^{-3}}{4 \times 10^{-3}}$

$x = 60 \times 1.5 = 90$

The value of $x$ is $90$.

Answer: $90$

Hence, the correct answer is 90.

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