JEE Chemistry 2025 Question with Solution

Given: Two statements about first order reaction graphs are shown.

Find: Which statement is correct.

From the solution, the stated correct option is B.

For Statement (I), the half-life of a first order reaction is independent of initial concentration:

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

So a graph of $t_{1/2}$ versus $[R_0]$ as a constant horizontal line is valid for a first order reaction.

For Statement (II), the graph shown is of $\log \left( \frac{[R]}{[R_0]} \right)$ versus time. Using the first order integrated rate law,

$$\log [R] = \log [R_0] - \frac{k}{2.303} \, t$$

which gives

$$\log \left( \frac{[R]}{[R_0]} \right) = -\frac{k}{2.303} \, t$$

Hence the slope should be $-\frac{k}{2.303}$, not positive $\frac{k}{2.303}$ as shown in the figure.

Therefore, Statement I is true but Statement II is false.

The correct option is B.

Given: Two graphical statements for a first order reaction.

Find: Determine the truth value of each statement.

1. Analyze Statement I

For a first order reaction, half-life is constant and does not depend on initial concentration:

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

Since $t_{1/2}$ does not contain $[R_0]$, the horizontal graph of $t_{1/2}$ versus $[R_0]$ is valid.

2. Analyze Statement II

The first order integrated rate law is:

$$\log [R] = \log [R_0] - \frac{k}{2.303} \, t$$

Subtracting $\log [R_0]$ from both sides,

$$\log \left( \frac{[R]}{[R_0]} \right) = -\frac{k}{2.303} \, t$$

Thus, the graph of $\log \left( \frac{[R]}{[R_0]} \right)$ versus time is a straight line with negative slope:

$$-\frac{k}{2.303}$$

The shown graph has a positive slope $\frac{k}{2.303}$, so it is not valid.

3. Conclude

Therefore, Statement I is true and Statement II is false.

The correct option is B.