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Chemical Kinetics and Nuclear Chemistry question

2021 · 26 Feb · Shift 2 · Q18
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Chemical Kinetics and Nuclear Chemistry question

2021 · 26 Feb · Shift 2 · Q18

JEE MainChemistryChemical Kinetics and Nuclear ChemistryNumerical+4 / −1
If the activation energy of a reaction is 80.9 kJ mol −-− 1, the fraction of molecules at 700 K, having enough energy to react to form products is e −-− x. The value of x is ‾\underline{\hspace{2cm}}​. (Rounded off to the nearest integer) [Use R = 8.31 J K −-− 1 mol −-− 1]
Numerical answer
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Correct answer: 14

  1. Use the Boltzmann/Arrhenius fraction expression

    The fraction of molecules having energy greater than or equal to the activation energy is:

    fraction=e−Ea/(RT)\text{fraction} = e^{-E_a/(RT)}fraction=e−Ea​/(RT)

    Given that this fraction is written as e−xe^{-x}e−x, we must have

    x=EaRTx = \frac{E_a}{RT}x=RTEa​​

  2. Convert the activation energy into consistent units

    Ea=80.9 kJ mol−1=80.9×103 J mol−1=80900 J mol−1E_a = 80.9\,\text{kJ mol}^{-1} = 80.9 \times 10^3\,\text{J mol}^{-1} = 80900\,\text{J mol}^{-1}Ea​=80.9kJ mol−1=80.9×103J mol−1=80900J mol−1

  3. Substitute the given values

    R=8.31 J K−1mol−1,T=700 KR = 8.31\,\text{J K}^{-1}\text{mol}^{-1}, \qquad T = 700\,\text{K}R=8.31J K−1mol−1,T=700K

    So,

    x=809008.31×700x = \frac{80900}{8.31 \times 700}x=8.31×70080900​

  4. Calculate the denominator

    8.31×700=58178.31 \times 700 = 58178.31×700=5817

    Therefore,

    x=809005817≈13.91x = \frac{80900}{5817} \approx 13.91x=581780900​≈13.91

  5. Round to the nearest integer

    x≈14x \approx 14x≈14

  6. Final answer

    14\boxed{14}14​

The fraction of molecules is therefore approximately:

e−14e^{-14}e−14

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