JEE Main 2020 · 9 January · morning · Single correct · +4 / −1

For the following reactions
A→(700K) Product
A→ _(catalyst)^(500K) Product
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):

  1. (a) 198 kJ/mol
  2. (b) 135 kJ/mol
  3. (c) 105 kJ/mol
  4. (d) 75 kJ/mol
Answer and solution

Answer (d) 75 kJ/mol

Solution

Chemistry ma’am · 4 steps

Same rate and same A means the exponent Eₐ/RT must be equal at both temperatures.

  1. Equal rates. From k = A e^(-Eₐ/RT), if k and A are the same for both paths, then Eₐ/T must also be the same. R cancels too.
  2. Set it up. Let the uncatalysed activation energy be Eₐ at 700 K. The catalysed one is Eₐ - 30 at 500 K.

    Eₐ/700 = (Eₐ - 30)/500

  3. Solve. Cross-multiplying gives 500Eₐ = 700Eₐ - 21000, so 200Eₐ = 21000 and Eₐ = 105 kJ mol⁻¹ for the uncatalysed reaction.
  4. So. The catalysed reaction has 105 - 30 = 75 kJ mol⁻¹, which is option (d).

The idea to keep. When two Arrhenius rate constants are equal and A is the same, Eₐ/T is equal for both. Then a one-line ratio gives the answer.

Answer from the NTA final answer key. Solution by PrepWiser.

Study Chemical Kinetics

Asked about 150 times in JEE Main since 2020. This question is from 2020.