JEE Chemistry 2025 Question with Solution

Given: The reaction is $R \rightarrow P$. The half-life data are $[A_0] = 0.100$ with $t_{1/2} = 200 \, \text{min}$ and $[A_0] = 0.025$ with $t_{1/2} = 100 \, \text{min}$. Find: Which statements are true.

For a reaction of order $n$,

$$t_{1/2} \propto [A_0]^{1-n}$$

Hence,

$$\frac{t_{1/2,1}}{t_{1/2,2}} = \left(\frac{[A_1]}{[A_2]}\right)^{1-n}$$

Substituting the given values,

$$\frac{200}{100} = \left(\frac{0.100}{0.025}\right)^{1-n}$$ $$2 = 4^{1-n}$$

Taking logarithm,

$$\log 2 = (1-n)\log 4$$

Given $\log 2 = 0.30$, so $\log 4 = 0.60$. Therefore,

$$0.30 = (1-n)(0.60)$$ $$1-n = 0.5$$ $$n = 0.5 = \frac{1}{2}$$

So statement A is true.

For a half-order reaction,

$$t_{1/2} \propto \frac{1}{\sqrt{[A_0]}}$$

Using $[A_0] = 0.1 \, \text{M}$ and $t_{1/2} = 200 \, \text{min}$, for $[A_0] = 1 \, \text{M}$,

$$t_{1/2} = 200 \times \sqrt{\frac{0.1}{1}} = 200 \times \frac{1}{\sqrt{10}} = \frac{200}{\sqrt{10}} \, \text{min}$$

So statement B is true.

The order of a reaction does not change with concentration for the same reaction mechanism, so statement C is false.

For $[A_0] = 1.6 \, \text{M}$,

$$t_{1/2} = 200 \times \sqrt{\frac{1.6}{0.1}}$$

Since $\frac{1.6}{0.1} = 16$,

$$t_{1/2} = 200 \times 4 = 800 \, \text{min}$$

So statement D is true.

Therefore, the correct statements are A, B and D only.

Given: Two half-life values at different initial concentrations. Find: The true statements.

The key relation used in the solution is

$$t_{1/2} \propto [A_0]^{1-n}$$

Compare the two data points:

$$[A_0]: 0.100 \to 0.025$$ $$t_{1/2}: 200 \to 100$$

Thus,

$$\frac{200}{100} = 2, \qquad \frac{0.100}{0.025} = 4$$

So,

$$2 = 4^{1-n}$$

Write $4 = 2^2$. Then,

$$2 = (2^2)^{1-n} = 2^{2(1-n)}$$

Equating powers of $2$,

$$1 = 2(1-n)$$ $$1-n = \frac{1}{2}$$ $$n = \frac{1}{2}$$

Hence statement A is correct.

Now use

$$t_{1/2} \propto \frac{1}{\sqrt{[A_0]}}$$

For statement B,

$$t_{1/2}(1 \, \text{M}) = 200 \times \sqrt{\frac{0.1}{1}} = \frac{200}{\sqrt{10}} \, \text{min}$$

So B is correct.

For statement C, changing concentration does not change the order of reaction. Therefore C is incorrect.

For statement D, following the working shown in the solution,

$$t_{1/2} = 200 \times \sqrt{\frac{1.6}{0.1}} = 200 \times 4 = 800 \, \text{min}$$

So D is treated as correct according to the provided solution working.

Therefore, the correct answer is A,B,D.