Chemistry Q54 – JEE-MAIN 2025

Home Chemistry Chemical KineticsQuestion #287

In a first order decomposition reaction, the time taken for the decomposition of reactant to one fourth and one eighth of its initial concentration are

t_{1}

and

t_{2}

(s), respectively. The ratio

t_{1} / t_{2}

will be:

Clue 1 of 3: First-Order Kinetics

Understand the integrated rate law for a first-order reaction.

🥷

Ninja Strategy•Logical Ratio Check

Recognize that decomposing to one-eighth of the initial concentration ( $t_{2}$ ) will take longer than decomposing to one-fourth ( $t_{1}$ ), so the ratio $t_{1} / t_{2}$ must be less than 1.

🧠 Full Solution Path

Step 1: Write the integrated rate law for a first-order reaction✦ Active

For a first-order reaction, the integrated rate law relating time $t$ , rate constant $k$ , initial concentration $[A]_{0}$ , and concentration at time $t$ , $[A]_{t}$ , is:

t = \frac{2.303}{k} \log \frac{[A]_{0}}{[A]_{t}}

Step 2: Express

t_{1}

and

t_{2}

using the rate law○ Expand

For $t_{1}$ , the concentration becomes one-fourth of the initial, so $[A]_{t} = \frac{1}{4} [A]_{0}$ . Substituting this into the rate law:

t_{1} = \frac{2.303}{k} \log \frac{[A]_{0}}{\frac{1}{4} [A]_{0}} = \frac{2.303}{k} \log 4 = \frac{2.303}{k} (2 \log 2)

For $t_{2}$ , the concentration becomes one-eighth of the initial, so $[A]_{t} = \frac{1}{8} [A]_{0}$ . Substituting this into the rate law:

t_{2} = \frac{2.303}{k} \log \frac{[A]_{0}}{\frac{1}{8} [A]_{0}} = \frac{2.303}{k} \log 8 = \frac{2.303}{k} (3 \log 2)

Step 3: Calculate the ratio

t_{1} / t_{2}

○ Expand

Now, we find the ratio $t_{1} / t_{2}$ :

\frac{t_{1}}{t_{2}} = \frac{\frac{2.303}{k} (2 \log 2)}{\frac{2.303}{k} (3 \log 2)} = \frac{2}{3}

Thus, the ratio $t_{1} / t_{2}$ is $\frac{2}{3}$ .

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Chemistry Question 54 – JEE-MAIN 2025

Clue 1 of 3: First-Order Kinetics

🧠 Full Solution Path