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Chemical Kinetics and Nuclear Chemistry question

2020 · 5 Sep · Shift 2 · Q14
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  5. /2020 · 5 Sep · Shift 2 · Q14

Chemical Kinetics and Nuclear Chemistry question

2020 · 5 Sep · Shift 2 · Q14

JEE MainChemistryChemical Kinetics and Nuclear ChemistryMCQ+4 / −1
The rate constant (k) of a reaction is measured at differenct temperatures (T), and the data are plotted in the given figure. The activation energy of the reaction in kJ mol–1 is : (R is gas constant) JEE Main 2020 (Online) 5th September Evening Slot Chemistry - Chemical Kinetics and Nuclear Chemistry Question 122 English
  1. A
    R
  2. B
    2R
  3. C
    1R{1 \over R}R1​
  4. D
    12R{1 \over {2R}}2R1​
View written solutionFree

Correct answer: B

  1. Use the Arrhenius equation

    The Arrhenius equation is k=Ae−Ea/(RT)k = A e^{-E_a/(RT)}k=Ae−Ea​/(RT)

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

    This is of the form: y=c+mxy = c + mxy=c+mx where slope m=−EaRm = -\frac{E_a}{R}m=−REa​​

  2. Interpret the graph

    For a plot of ln⁡k\ln klnk versus 1/T1/T1/T, the slope is negative and equals −EaR-\frac{E_a}{R}−REa​​

    From the given figure, the slope is −2-2−2.

    Therefore, −EaR=−2-\frac{E_a}{R} = -2−REa​​=−2

  3. Calculate activation energy

    EaR=2\frac{E_a}{R} = 2REa​​=2 Ea=2RE_a = 2REa​=2R

  4. Match with options

    The correct option is: 2R\boxed{2R}2R​

    So, Option B is correct.

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