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Chemical Kinetics and Nuclear Chemistry question

2021 · 31 Aug · Shift 2 · Q18
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Chemical Kinetics and Nuclear Chemistry question

2021 · 31 Aug · Shift 2 · Q18

JEE MainChemistryChemical Kinetics and Nuclear ChemistryNumerical+4 / −1
For the reaction A →\to→ B, the rate constant k(in s −-− 1) is given by log⁡10k=20.35−(2.47×103)T{\log _{10}}k = 20.35 - {{(2.47 \times {{10}^3})} \over T}log10​k=20.35−T(2.47×103)​ The energy of activation in kJ mol −-− 1 is ‾\underline{\hspace{2cm}}​. (Nearest integer) [Given : R = 8.314 J K −-− 1 mol −-− 1]
Numerical answer
View written solutionFree

Correct answer: 47

  1. Use the Arrhenius equation in base 10 form

The Arrhenius equation is k=Ae−Ea/(RT)k = A e^{-E_a/(RT)}k=Ae−Ea​/(RT) Taking common logarithm, log⁡10k=log⁡10A−Ea2.303RT\log_{10} k = \log_{10} A - \frac{E_a}{2.303RT}log10​k=log10​A−2.303RTEa​​

  1. Compare with the given equation

Given: log⁡10k=20.35−2.47×103T\log_{10} k = 20.35 - \frac{2.47\times 10^3}{T}log10​k=20.35−T2.47×103​

Comparing with log⁡10k=log⁡10A−Ea2.303RT\log_{10} k = \log_{10} A - \frac{E_a}{2.303RT}log10​k=log10​A−2.303RTEa​​ we get Ea2.303R=2.47×103\frac{E_a}{2.303R} = 2.47\times 10^32.303REa​​=2.47×103

So, Ea=2.303R(2.47×103)E_a = 2.303R(2.47\times 10^3)Ea​=2.303R(2.47×103)

  1. Substitute the value of RRR

Ea=2.303×8.314×2.47×103 J mol−1E_a = 2.303 \times 8.314 \times 2.47\times 10^3 \text{ J mol}^{-1}Ea​=2.303×8.314×2.47×103 J mol−1

First calculate: 2.303×8.314≈19.1472.303 \times 8.314 \approx 19.1472.303×8.314≈19.147

Then, Ea≈19.147×2470E_a \approx 19.147 \times 2470Ea​≈19.147×2470 Ea≈47293 J mol−1E_a \approx 47293 \text{ J mol}^{-1}Ea​≈47293 J mol−1

  1. Convert to kJ mol−1^{-1}−1

Ea≈47.293 kJ mol−1E_a \approx 47.293 \text{ kJ mol}^{-1}Ea​≈47.293 kJ mol−1

  1. Nearest integer

47\boxed{47}47​

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