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Chemical Kinetics and Nuclear Chemistry question

2021 · 24 Feb · Shift 1 · Q22
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Chemical Kinetics and Nuclear Chemistry question

2021 · 24 Feb · Shift 1 · Q22

JEE MainChemistryChemical Kinetics and Nuclear ChemistryNumerical+4 / −1
Gaseous cyclobutene isomerises to butadiene in a first order process which has a 'k' value of 3.3 ×\times× 10-4 s-1 at 153°C. The time in minutes it takes for the isomerization to proceed 40% to completion at this temperature is ‾\underline{\hspace{2cm}}​. (Rounded off to the nearest integer)
Numerical answer
View written solutionFree

Correct answer: 26

  1. Given data
  • First-order rate constant:
    k=3.3×10−4 s−1k = 3.3 \times 10^{-4}\ \text{s}^{-1}k=3.3×10−4 s−1
  • Reaction has proceeded 40% to completion.

This means 40% reacted, so 60% remains.

Hence, [A]t[A]0=0.60\frac{[A]_t}{[A]_0} = 0.60[A]0​[A]t​​=0.60

  1. Use the first-order integrated rate law

For a first-order reaction, k=1tln⁡([A]0[A]t)k = \frac{1}{t} \ln\left(\frac{[A]_0}{[A]_t}\right)k=t1​ln([A]t​[A]0​​)

So, t=1kln⁡([A]0[A]t)t = \frac{1}{k} \ln\left(\frac{[A]_0}{[A]_t}\right)t=k1​ln([A]t​[A]0​​)

Substitute values: t=13.3×10−4ln⁡(10.60)t = \frac{1}{3.3\times10^{-4}} \ln\left(\frac{1}{0.60}\right)t=3.3×10−41​ln(0.601​)

  1. Calculate the logarithmic term

ln⁡(10.60)=ln⁡(1.6667)≈0.5108\ln\left(\frac{1}{0.60}\right) = \ln(1.6667) \approx 0.5108ln(0.601​)=ln(1.6667)≈0.5108

Thus, t=0.51083.3×10−4t = \frac{0.5108}{3.3\times10^{-4}}t=3.3×10−40.5108​

t≈1547.9 st \approx 1547.9\ \text{s}t≈1547.9 s

  1. Convert into minutes

t=1547.960≈25.8 mint = \frac{1547.9}{60} \approx 25.8\ \text{min}t=601547.9​≈25.8 min

  1. Round to nearest integer

t≈26 mint \approx 26\ \text{min}t≈26 min

Final Answer

The time required is: 26\boxed{26}26​

The derived answer matches the stored correct answer.

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