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Chemical Kinetics and Nuclear Chemistry question

2022 · 28 Jul · Shift 2 · Q19
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Chemical Kinetics and Nuclear Chemistry question

2022 · 28 Jul · Shift 2 · Q19

JEE MainChemistryChemical Kinetics and Nuclear ChemistryNumerical+4 / −1
For a reaction, given below is the graph of ln⁡k\ln klnk vs 1T{1 \over T}T1​. The activation energy for the reaction is equal to ‾cal mol−1\underline{\hspace{2cm}}\mathrm{cal} \,\mathrm{mol}^{-1}​calmol−1. (nearest integer) (Given : R=2 cal K−1  mol−1\mathrm{R}=2 \,\mathrm{cal} \,\mathrm{K}^{-1} \,\mathrm{~mol}^{-1}R=2calK−1 mol−1 ) JEE Main 2022 (Online) 28th July Evening Shift Chemistry - Chemical Kinetics and Nuclear Chemistry Question 72 English
Numerical answer
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Correct answer: 8

  1. For the Arrhenius equation,

k=Ae−Ea/(RT)k=Ae^{-E_a/(RT)}k=Ae−Ea​/(RT)

Taking natural log:

ln⁡k=ln⁡A−EaR(1T)\ln k = \ln A - \frac{E_a}{R}\left(\frac{1}{T}\right)lnk=lnA−REa​​(T1​)

  1. Thus, in a graph of ln⁡k\ln klnk vs 1T\dfrac{1}{T}T1​, the slope is

m=−EaRm=-\frac{E_a}{R}m=−REa​​

  1. From the given straight-line graph, the slope is obtained as

m=−4m=-4m=−4

  1. Therefore,

−EaR=−4-\frac{E_a}{R}=-4−REa​​=−4

So,

EaR=4\frac{E_a}{R}=4REa​​=4

  1. Given R=2 cal K−1mol−1R=2\,\text{cal K}^{-1}\text{mol}^{-1}R=2cal K−1mol−1,

Ea=4×2=8 cal mol−1E_a = 4 \times 2 = 8\,\text{cal mol}^{-1}Ea​=4×2=8cal mol−1

  1. Hence, the activation energy is

8 cal mol−1\boxed{8\,\text{cal mol}^{-1}}8cal mol−1​

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