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Chemical Kinetics and Nuclear Chemistry question

2022 · 29 Jun · Shift 2 · Q18
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Chemical Kinetics and Nuclear Chemistry question

2022 · 29 Jun · Shift 2 · Q18

JEE MainChemistryChemical Kinetics and Nuclear ChemistryNumerical+4 / −1
The equation k = (6.5 ×\times× 1012s −-− 1)e −-− 26000K/T is followed for the decomposition of compound A. The activation energy for the reaction is ‾\underline{\hspace{2cm}}​ kJ mol −-− 1. [nearest integer] (Given : R = 8.314 J K −-− 1 mol −-− 1)
Numerical answer
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Correct answer: 216

  1. Use the Arrhenius equation

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

    Given: k=(6.5×1012 s−1)e−26000 K/Tk=(6.5\times 10^{12}\,\text{s}^{-1})e^{-26000\,\text{K}/T}k=(6.5×1012s−1)e−26000K/T

  2. Compare the exponent with the standard form

    Standard exponent: −EaRT-\frac{E_a}{RT}−RTEa​​

    Given exponent: −26000T-\frac{26000}{T}−T26000​

    Hence, EaR=26000\frac{E_a}{R}=26000REa​​=26000

    So, Ea=26000×RE_a=26000\times REa​=26000×R

  3. Substitute the value of RRR

    Ea=26000×8.314  J mol−1E_a=26000\times 8.314\;\text{J mol}^{-1}Ea​=26000×8.314J mol−1

    Ea=216164  J mol−1E_a=216164\;\text{J mol}^{-1}Ea​=216164J mol−1

  4. Convert to kJ mol−1^{-1}−1

    Ea=2161641000=216.164  kJ mol−1E_a=\frac{216164}{1000}=216.164\;\text{kJ mol}^{-1}Ea​=1000216164​=216.164kJ mol−1

  5. Nearest integer

    Ea≈216  kJ mol−1E_a\approx 216\;\text{kJ mol}^{-1}Ea​≈216kJ mol−1

Final Answer: 216216216

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