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Chemical Kinetics and Nuclear Chemistry question

2021 · 17 Mar · Shift 1 · Q18
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Chemical Kinetics and Nuclear Chemistry question

2021 · 17 Mar · Shift 1 · Q18

JEE MainChemistryChemical Kinetics and Nuclear ChemistryNumerical+4 / −1
For a certain first order reaction 32% of the reactant is left after 570s. The rate constant of this reaction is ‾×\underline{\hspace{2cm}}\times​× 10 −-− 3 s −-− 1. (Round off to the Nearest Integer). [Given : log102 = 0.301, ln10 = 2.303]
Numerical answer
View written solutionFree

Correct answer: 2

  1. Use the first-order decay law

For a first-order reaction,

[A]t=[A]0e−kt[A]_t = [A]_0 e^{-kt}[A]t​=[A]0​e−kt

So the fraction left after time ttt is:

[A]t[A]0=e−kt\frac{[A]_t}{[A]_0} = e^{-kt}[A]0​[A]t​​=e−kt
  1. Substitute the given data

32%32\%32% of the reactant is left after 570 s570\,\text{s}570s, so

[A]t[A]0=0.32\frac{[A]_t}{[A]_0} = 0.32[A]0​[A]t​​=0.32

Thus,

0.32=e−k(570)0.32 = e^{-k(570)}0.32=e−k(570)

Taking natural logarithm:

−k(570)=ln⁡(0.32)- k(570) = \ln(0.32)−k(570)=ln(0.32) k=−ln⁡(0.32)570k = -\frac{\ln(0.32)}{570}k=−570ln(0.32)​
  1. Evaluate ln⁡(0.32)\ln(0.32)ln(0.32) using given values
0.32=32×10−20.32 = 32 \times 10^{-2}0.32=32×10−2

So,

log⁡(0.32)=log⁡32−2\log(0.32) = \log 32 - 2log(0.32)=log32−2

Now,

log⁡32=log⁡(25)=5log⁡2=5(0.301)=1.505\log 32 = \log(2^5) = 5\log 2 = 5(0.301) = 1.505log32=log(25)=5log2=5(0.301)=1.505

Hence,

log⁡(0.32)=1.505−2=−0.495\log(0.32) = 1.505 - 2 = -0.495log(0.32)=1.505−2=−0.495

Using

ln⁡x=2.303log⁡x\ln x = 2.303 \log xlnx=2.303logx

we get

ln⁡(0.32)=2.303(−0.495)=−1.139985≈−1.14\ln(0.32) = 2.303(-0.495) = -1.139985 \approx -1.14ln(0.32)=2.303(−0.495)=−1.139985≈−1.14
  1. Calculate kkk
k=−−1.14570=1.14570k = -\frac{-1.14}{570} = \frac{1.14}{570}k=−570−1.14​=5701.14​ k=0.002=2×10−3 s−1k = 0.002 = 2 \times 10^{-3}\,\text{s}^{-1}k=0.002=2×10−3s−1
  1. Nearest integer for the blank

The question asks for:

‾×10−3 s−1\underline{\hspace{2cm}} \times 10^{-3}\,\text{s}^{-1}​×10−3s−1

So the required integer is:

2\boxed{2}2​
  1. Comparison with stored answer

Stored correct answer = 222

Our derived answer also = 222, so they agree.

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