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Chemical Kinetics and Nuclear Chemistry question

2019 · 10 Jan · Shift 1 · Q8
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Chemical Kinetics and Nuclear Chemistry question

2019 · 10 Jan · Shift 1 · Q8

JEE MainChemistryChemical Kinetics and Nuclear ChemistryMCQ+4 / −1
Consider the given plots for a reaction obeying Arrhenius equation (0oC < T < 300oC) : (K and Ea are rate constant and activation energy, respectively) JEE Main 2019 (Online) 10th January Morning Slot Chemistry - Chemical Kinetics and Nuclear Chemistry Question 148 English Choose the correct option :
  1. A
    I is right but II is wrong
  2. B
    Both I and II are correct
  3. C
    Both I and II are wrong
  4. D
    I is wrong but II is right
View written solutionFree

Correct answer: B

  1. For a reaction obeying the Arrhenius equation, k=Ae−Ea/(RT)k = A e^{-E_a/(RT)}k=Ae−Ea​/(RT) Taking logarithm, ln⁡k=ln⁡A−EaR⋅1T\ln k = \ln A - \frac{E_a}{R}\cdot \frac{1}{T}lnk=lnA−REa​​⋅T1​ or in common logarithm form, log⁡k=log⁡A−Ea2.303R⋅1T\log k = \log A - \frac{E_a}{2.303R}\cdot \frac{1}{T}logk=logA−2.303REa​​⋅T1​

  2. Therefore, a plot of log⁡k\log klogk versus 1/T1/T1/T is a straight line with:

    • negative slope =−Ea2.303R= -\dfrac{E_a}{2.303R}=−2.303REa​​
    • intercept =log⁡A= \log A=logA
  3. So any correct Arrhenius plot must reflect:

    • linear dependence of log⁡k\log klogk on 1/T1/T1/T
    • negative slope
  4. The question mentions two plots, I and II, for a reaction obeying Arrhenius equation in the range 0∘C0^\circ C0∘C to 100∘C100^\circ C100∘C. Since for Arrhenius behavior both the standard linearized representations are valid, both plots are correct.

  5. Hence:

    • Statement I: correct
    • Statement II: correct

Therefore, the correct option is B: Both I and II are correct\boxed{\text{B: Both I and II are correct}}B: Both I and II are correct​

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