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Chemical Kinetics and Nuclear Chemistry question

2024 · 27 Jan · Shift 2 · Q24
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Chemical Kinetics and Nuclear Chemistry question

2024 · 27 Jan · Shift 2 · Q24

JEE MainChemistryChemical Kinetics and Nuclear ChemistryNumerical+4 / −1
Time required for completion of 99.9%99.9 \%99.9% of a First order reaction is ‾\underline{\hspace{2cm}}​ times of half life (t1/2)\left(t_{1 / 2}\right)(t1/2​) of the reaction.
Numerical answer
View written solutionFree

Correct answer: 10

  1. For a first-order reaction, the integrated rate law is
kt=ln⁡[A]0[A]tkt = \ln\frac{[A]_0}{[A]_t}kt=ln[A]t​[A]0​​
  1. If 99.9%99.9\%99.9% of the reaction is completed, then 0.1%0.1\%0.1% of reactant remains.

So,

[A]t[A]0=0.001=10−3\frac{[A]_t}{[A]_0} = 0.001 = 10^{-3}[A]0​[A]t​​=0.001=10−3

Thus,

kt=ln⁡10.001=ln⁡(1000)=3ln⁡10kt = \ln\frac{1}{0.001} = \ln(1000) = 3\ln 10kt=ln0.0011​=ln(1000)=3ln10

Hence,

t=3ln⁡10kt = \frac{3\ln 10}{k}t=k3ln10​
  1. For a first-order reaction, half-life is
t1/2=ln⁡2kt_{1/2} = \frac{\ln 2}{k}t1/2​=kln2​
  1. Required ratio:
tt1/2=3ln⁡10kln⁡2k=3ln⁡10ln⁡2\frac{t}{t_{1/2}} = \frac{\frac{3\ln 10}{k}}{\frac{\ln 2}{k}} = \frac{3\ln 10}{\ln 2}t1/2​t​=kln2​k3ln10​​=ln23ln10​

Now,

ln⁡10≈2.303,ln⁡2≈0.693\ln 10 \approx 2.303, \quad \ln 2 \approx 0.693ln10≈2.303,ln2≈0.693

So,

tt1/2=3(2.303)0.693≈6.9090.693≈9.97≈10\frac{t}{t_{1/2}} = \frac{3(2.303)}{0.693} \approx \frac{6.909}{0.693} \approx 9.97 \approx 10t1/2​t​=0.6933(2.303)​≈0.6936.909​≈9.97≈10
  1. Therefore, the time required is
10\boxed{10}10​

times the half-life.

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