Chemistry Q56 – NEET-UG 2025

Home Chemistry Chemical KineticsQuestion #272

If the half-life (

t_{1 / 2}

) for a first order reaction is 1 minute, then the time required for 99.9% completion of the reaction is closest to :

Clue 1 of 3: First-Order Kinetics

Understand the integrated rate law and half-life concept for first-order reactions.

🧠 Full Solution Path

Step 1: Calculate the rate constant (

k

)✦ Active

For a first-order reaction, the half-life ( $t_{1 / 2}$ ) is related to the rate constant ( $k$ ) by the formula:

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

Given $t_{1 / 2} = 1$ minute, we can calculate $k$ :

k = \frac{0.693}{t_{1 / 2}} = \frac{0.693}{1 min} = 0.693 {min}^{- 1}

Step 2: Determine the ratio of initial to final concentration○ Expand

For 99.9% completion of the reaction, the percentage of reactant remaining is $100 % - 99.9 % = 0.1 %$ . If the initial concentration is $[A]_{0}$ , then the concentration at time $t$ , $[A]_{t}$ , is $0.1 %$ of $[A]_{0}$ :

[A]_{t} = 0.001 [A]_{0}

Therefore, the ratio $\frac{[A]_{0}}{[A]_{t}}$ is:

\frac{[A]_{0}}{[A]_{t}} = \frac{[A]_{0}}{0.001 [A]_{0}} = \frac{1}{0.001} = 1000

💡 Teacher's Secret Hint

Remember that 'completion' refers to the amount of reactant consumed, so you need to find the amount remaining.

Step 3: Calculate the time (

t

) for 99.9% completion○ Expand

The integrated rate law for a first-order reaction is:

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

Substitute the values of $k$ and $\frac{[A]_{0}}{[A]_{t}}$ :

t = \frac{2.303}{0.693 {min}^{- 1}} \log (1000)

Since $\log (1000) = 3$ :

t = \frac{2.303}{0.693} \times 3 min

t = \frac{6.909}{0.693} min

t \approx 9.969 min

This value is closest to 10 minutes.

💡 Teacher's Secret Hint

Alternatively, recognize that $2^{10} \approx 1000$ . Since $0.1 % = (1 / 2)^{n}$ , $1 / 1000 = (1 / 2)^{n}$ , so $2^{n} = 1000$ , which means $n \approx 10$ half-lives. Thus, $t = 10 \times t_{1 / 2} = 10 \times 1 min = 10 min$ .

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Chemistry Question 56 – NEET-UG 2025

Clue 1 of 3: First-Order Kinetics

🧠 Full Solution Path