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Chemical Kinetics and Nuclear Chemistry question

2012 · Shift 0 · Q18
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Chemical Kinetics and Nuclear Chemistry question

2012 · Shift 0 · Q18

JEE MainChemistryChemical Kinetics and Nuclear ChemistryMCQ+4 / −1
For a first order reaction, (A) →\to→ products, the concentration of A changes from 0.1 M to 0.025 M in 40 minutes. The rate of reaction when the concentration of A is 0.01 M is :
  1. A
    1.73 x 10–5 M/ min
  2. B
    3.47 x 10–4 M/min
  3. C
    3.47 x 10–5 M/min
  4. D
    1.73 x 10–4 M/min
View written solutionFree

Correct answer: B

  1. Use the first-order integrated rate law

For a first-order reaction,

rac{2.303}{t}\,\log\left(\frac{[A]_0}{[A]_t}\right)=k

Given:

  • Initial concentration: [A]0=0.1 M[A]_0 = 0.1\,\text{M}[A]0​=0.1M
  • Final concentration after 404040 min: [A]t=0.025 M[A]_t = 0.025\,\text{M}[A]t​=0.025M
  • Time: t=40 mint = 40\,\text{min}t=40min

So,

k=2.30340log⁡(0.10.025)k = \frac{2.303}{40}\log\left(\frac{0.1}{0.025}\right)k=402.303​log(0.0250.1​)

Now,

0.10.025=4\frac{0.1}{0.025} = 40.0250.1​=4

and

log⁡4=0.6021\log 4 = 0.6021log4=0.6021

Hence,

k=2.30340×0.6021k = \frac{2.303}{40}\times 0.6021k=402.303​×0.6021 k≈1.38640=0.03465 min−1k \approx \frac{1.386}{40} = 0.03465\,\text{min}^{-1}k≈401.386​=0.03465min−1
  1. Rate at [A]=0.01 M[A] = 0.01\,\text{M}[A]=0.01M

For a first-order reaction,

rate=k[A]\text{rate} = k[A]rate=k[A]

Thus,

rate=0.03465×0.01\text{rate} = 0.03465 \times 0.01rate=0.03465×0.01 rate=3.465×10−4 M/min\text{rate} = 3.465 \times 10^{-4}\,\text{M/min}rate=3.465×10−4M/min
  1. Match with the options
3.465×10−4≈3.47×10−4 M/min3.465 \times 10^{-4} \approx 3.47 \times 10^{-4}\,\text{M/min}3.465×10−4≈3.47×10−4M/min

So the correct option is B.

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