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Chemical Kinetics and Nuclear Chemistry question

2017 · Shift 0 · Q14
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Chemical Kinetics and Nuclear Chemistry question

2017 · Shift 0 · Q14

JEE MainChemistryChemical Kinetics and Nuclear ChemistryMCQ+4 / −1
Two reactions R1R_1R1​ and R2R_2R2​ have identical pre-exponential factors. Activation energy of R1R_1R1​ exceeds that of R2R_2R2​ by 10 kJ mol–1. If k1k_1k1​ and k2k_2k2​ are rate constants for reactions R1R_1R1​ and R2R_2R2​ respectively at 300 K, then ln(k2k_2k2​/k1k_1k1​) is equal to : (RRR = 8.314 J mol–1 K–1)
  1. A
    12
  2. B
    6
  3. C
    4
  4. D
    8
View written solutionFree

Correct answer: C

  1. Use the Arrhenius equation:

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

Since both reactions have identical pre-exponential factor AAA,

k1=Ae−Ea1/(RT),k2=Ae−Ea2/(RT)k_1 = A e^{-E_{a1}/(RT)}, \qquad k_2 = A e^{-E_{a2}/(RT)}k1​=Ae−Ea1​/(RT),k2​=Ae−Ea2​/(RT)

  1. Form the ratio:

k2k1=Ae−Ea2/(RT)Ae−Ea1/(RT)=e(Ea1−Ea2)/(RT)\frac{k_2}{k_1} = \frac{A e^{-E_{a2}/(RT)}}{A e^{-E_{a1}/(RT)}} = e^{(E_{a1}-E_{a2})/(RT)}k1​k2​​=Ae−Ea1​/(RT)Ae−Ea2​/(RT)​=e(Ea1​−Ea2​)/(RT)

  1. Take natural logarithm:

ln⁡(k2k1)=Ea1−Ea2RT\ln\left(\frac{k_2}{k_1}\right) = \frac{E_{a1}-E_{a2}}{RT}ln(k1​k2​​)=RTEa1​−Ea2​​

Given:

Ea1−Ea2=10 kJ mol−1=10000 J mol−1E_{a1} - E_{a2} = 10\,\text{kJ mol}^{-1} = 10000\,\text{J mol}^{-1}Ea1​−Ea2​=10kJ mol−1=10000J mol−1

T=300 K,R=8.314 J mol−1K−1T = 300\,\text{K}, \qquad R = 8.314\,\text{J mol}^{-1}\text{K}^{-1}T=300K,R=8.314J mol−1K−1

So,

ln⁡(k2k1)=100008.314×300\ln\left(\frac{k_2}{k_1}\right) = \frac{10000}{8.314 \times 300}ln(k1​k2​​)=8.314×30010000​

=100002494.2≈4.01= \frac{10000}{2494.2} \approx 4.01=2494.210000​≈4.01

  1. Therefore,

ln⁡(k2k1)≈4\ln\left(\frac{k_2}{k_1}\right) \approx 4ln(k1​k2​​)≈4

Hence, the correct option is C.

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