Verified Solution • 2025–26 Session moderate

QJEEM17Chc8a2: JEE Main 2017 Chemistry - Class 12 JEEM17Chc8a2 Solution: Two reactions R 1 and R 2 have identical pre-exponential…

Two reactions R1 and R2 have identical pre-exponential factors. Activation energy of R1 exceeds that of R2 by 10 kJ mol–1. If k1 and k2 are rate constants for reactions R1 and R2 respectively at 300 K, then ln(k2/k1) is equal to :
(R = 8.314 J mol–1 K–1)

Multiple Choice Options

  • A 12
  • B 6
  • C 4
  • D 8

Step-by-Step Verified Solution

We know, from arrhenius equation,

k = A. e−EaRT{e^{{{ - {E_a}} \over {RT}}}}

∴\therefore k1 = A. e−Ea1RT{e^{{{ - {E_{{a_1}}}} \over {RT}}}} ......(1)

k2 = A. e−Ea2RT{e^{{{ - {E_{{a_2}}}} \over {RT}}}} ......(2)

On dividing equation (2) by (1), we get

k2k1=e(Ea1−Ea2)RT{{{k_2}} \over {{k_1}}} = {e^{{{\left( {{E_{{a_1}}} - {E_{{a_2}}}} \right)} \over {RT}}}}

⇒\Rightarrow ln⁡(k2k1)=(Ea1−Ea2)RT\ln \left( {{{{k_2}} \over {{k_1}}}} \right) = {{\left( {{E_{{a_1}}} - {E_{{a_2}}}} \right)} \over {RT}} = 10,0008.314×300{{10,000} \over {8.314 \times 300}} = 4

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