Verified Solution • 2025–26 Session moderate

QJEEM20Ch082b: JEE Main 2020 Chemistry - Class 12 JEEM20Ch082b Solution: For the following reactions A ⟶ 700 K Product it was found that …

For the following reactions
A⟶700KProduct
A⟶catalyst500KProductA\mathrel{\mathop{\kern0pt\longrightarrow}\limits_{catalyst}^{500K}} {\mathop{\rm Product}\nolimits}
it was found that Ea is decreased by 30 kJ/mol in the presence of catalyst.
If the rate remains unchanged, the activation energy for catalysed reaction is (Assume pre exponential factor is same):

Multiple Choice Options

  • A 198 kJ/mol
  • B 135 kJ/mol
  • C 105 kJ/mol
  • D 75 kJ/mol

Step-by-Step Verified Solution

K1 = A e−EaR×700{e^{ - {{{E_a}} \over {R \times 700}}}}

K2 = A e−(Ea−30)R×500{e^{ - {{\left( {{E_a} - 30} \right)} \over {R \times 500}}}}

∵\because Rate is same

∴\therefore Rate constant will also be same

K1 = K2

A e−EaR×700{e^{ - {{{E_a}} \over {R \times 700}}}} = A e−(Ea−30)R×500{e^{ - {{\left( {{E_a} - 30} \right)} \over {R \times 500}}}}

⇒\Rightarrow (Ea−30)R×500{{{\left( {{E_a} - 30} \right)} \over {R \times 500}}} = EaR×700{{{{E_a}} \over {R \times 700}}}

⇒\Rightarrow 5Ea = 7Ea – 210

⇒\Rightarrow 210 = 2Ea

⇒\Rightarrow Ea = 105 kJ/mole

∴\therefore Activation energy in the presence of catalyst = 105 – 30 = 75 kJ/mol

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