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Chemical Kinetics and Nuclear Chemistry question

2022 · 28 Jul · Shift 1 · Q18
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Chemical Kinetics and Nuclear Chemistry question

2022 · 28 Jul · Shift 1 · Q18

JEE MainChemistryChemical Kinetics and Nuclear ChemistryNumerical+4 / −1
For the given first order reaction A→B\mathrm{A} \rightarrow \mathrm{B}A→B the half life of the reaction is 0.3010 min0.3010 \mathrm{~min}0.3010 min. The ratio of the initial concentration of reactant to the concentration of reactant at time 2.0 min2.0 \mathrm{~min}2.0 min will be equal to ‾\underline{\hspace{2cm}}​. (Nearest integer)
Numerical answer
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Correct answer: 100

  1. For a first-order reaction, the half-life is given by

t1/2=0.693kt_{1/2} = \frac{0.693}{k}t1/2​=k0.693​

Given:

t1/2=0.3010 mint_{1/2} = 0.3010\ \text{min}t1/2​=0.3010 min

So,

k=0.6930.3010k = \frac{0.693}{0.3010}k=0.30100.693​

k≈2.302 min−1k \approx 2.302\ \text{min}^{-1}k≈2.302 min−1

  1. For a first-order reaction, the integrated rate law is

k=2.303tlog⁡[A]0[A]tk = \frac{2.303}{t} \log \frac{[A]_0}{[A]_t}k=t2.303​log[A]t​[A]0​​

We need the ratio at t=2.0t=2.0t=2.0 min.

So,

log⁡[A]0[A]t=kt2.303\log \frac{[A]_0}{[A]_t} = \frac{kt}{2.303}log[A]t​[A]0​​=2.303kt​

Substitute values:

log⁡[A]0[A]t=(2.302)(2.0)2.303≈1.998≈2\log \frac{[A]_0}{[A]_t} = \frac{(2.302)(2.0)}{2.303} \approx 1.998 \approx 2log[A]t​[A]0​​=2.303(2.302)(2.0)​≈1.998≈2

Thus,

[A]0[A]t=102=100\frac{[A]_0}{[A]_t} = 10^2 = 100[A]t​[A]0​​=102=100

  1. Therefore, the required ratio is

100\boxed{100}100​

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