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Chemical Kinetics and Nuclear Chemistry question

2022 · 26 Jun · Shift 2 · Q22
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Chemical Kinetics and Nuclear Chemistry question

2022 · 26 Jun · Shift 2 · Q22

JEE MainChemistryChemical Kinetics and Nuclear ChemistryNumerical+4 / −1
Catalyst A reduces the activation energy for a reaction by 10 kJ mol −-− 1 at 300 K. The ratio of rate constants, kT, CatalysedkT, Uncatalysed{{{}^kT,\,Catalysed} \over {{}^kT,\,Uncatalysed}}kT,UncatalysedkT,Catalysed​ is ex. The value of x is ‾\underline{\hspace{2cm}}​. [nearest integer] [Assume that the pre-exponential factor is same in both the cases. Given R = 8.31 J K −-− 1 mol −-− 1]
Numerical answer
View written solutionFree

Correct answer: 4

  1. Use Arrhenius equation

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

    Given that the pre-exponential factor AAA is same for catalysed and uncatalysed reactions, the ratio of rate constants is: kcatalysedkuncatalysed=e−Ea,c/RT÷e−Ea,u/RT\frac{k_{\text{catalysed}}}{k_{\text{uncatalysed}}} = e^{-E_{a,c}/RT} \div e^{-E_{a,u}/RT}kuncatalysed​kcatalysed​​=e−Ea,c​/RT÷e−Ea,u​/RT =e(Ea,u−Ea,c)/(RT)= e^{(E_{a,u}-E_{a,c})/(RT)}=e(Ea,u​−Ea,c​)/(RT)

  2. Activation energy decrease

    Catalyst reduces activation energy by 10 kJ mol−110\,\text{kJ mol}^{-1}10kJ mol−1. So, Ea,u−Ea,c=10 kJ mol−1=10000 J mol−1E_{a,u}-E_{a,c} = 10\,\text{kJ mol}^{-1} = 10000\,\text{J mol}^{-1}Ea,u​−Ea,c​=10kJ mol−1=10000J mol−1

    Therefore, kcatalysedkuncatalysed=e10000/(8.31×300)\frac{k_{\text{catalysed}}}{k_{\text{uncatalysed}}} = e^{10000/(8.31\times 300)}kuncatalysed​kcatalysed​​=e10000/(8.31×300)

  3. Calculate exponent

    8.31×300=24938.31\times 300 = 24938.31×300=2493 100002493≈4.01\frac{10000}{2493} \approx 4.01249310000​≈4.01

    Hence, kcatalysedkuncatalysed=e4.01\frac{k_{\text{catalysed}}}{k_{\text{uncatalysed}}} = e^{4.01}kuncatalysed​kcatalysed​​=e4.01

    Comparing with the given form exe^xex, we get: x≈4.01x \approx 4.01x≈4.01

  4. Nearest integer

    x=4x = 4x=4

Final Answer: 4\boxed{4}4​

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