Catalyst A reduces the activation energy for a reaction by 10 kJ mol⁻¹ at 300 K. The ratio of rate constants, (ᵏT, Catalysed)/(ᵏT, Uncatalysed) is eˣ. The value of x is ___________. [nearest integer]
[Assume that the pre-exponential factor is same in both the cases. Given R = 8.31 J K⁻¹ mol⁻¹]

Solution

Chemistry ma’am · 4 steps

Same A, same T: divide the two Arrhenius equations and only the drop in Eₐ is left in the exponent.

1. Divide out A. Write k = A e^(-Eₐ/RT) for the catalysed and the uncatalysed reaction and divide. A is the same in both, so it cancels.

   k_(cat)/k_(uncat) = e^((E_(a,uncat) - E_(a,cat))/RT) = e^(Δ Eₐ/RT)

2. Read off x. So x = (Δ Eₐ)/RT. The catalyst lowers Eₐ by Δ Eₐ = 10 kJ mol⁻¹ = 10000 J mol⁻¹.
3. Put in numbers. RT = 8.31 × 300 = 2493 J mol⁻¹, so x = 10000/2493 = 4.01.
4. So. x ≈ 4.01, which rounds to 4. The catalysed reaction is about e⁴ ≈ 55 times faster.

The common slip: 0. Putting Δ Eₐ in as 10 kJ with R in joules gives x = 0.004, which rounds to 0. Convert to 10000 J first.

The idea to keep. When only Eₐ changes, the ratio of rate constants is e^(Δ Eₐ/RT). Lowering Eₐ makes the exponent positive, and Δ Eₐ must be in joules to match R.

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