A → B
The rate constants of the above reaction at 200 K and 300 K are 0.03 min⁻¹ and 0.05 min⁻¹ respectively. The activation energy for the reaction is ___________ J (Nearest integer)
(Given : ln10=2.3
R=8.3 J K⁻¹ mol⁻¹
log 5=0.70
log 3=0.48
log 2=0.30)

Solution

Chemistry ma’am · 5 steps

Divide the two Arrhenius equations so A cancels, then turn the given logs into a natural log.

1. Divide out A. Write k = A e^(-Eₐ/RT) at both temperatures and divide one by the other. The factor A cancels, so Eₐ is the only unknown left.

   ln k₂/k₁ = Eₐ/R(1/T₁-1/T₂)

2. Left side. k₂/k₁ = 0.05/0.03 = 5/3. So log 5/3 = 0.70 - 0.48 = 0.22, and ln 5/3 = 2.3 × 0.22 = 0.506.
3. Temperature bracket. 1/200 - 1/300 = (3 - 2)/600 = 1/600 K⁻¹.
4. Solve for Ea. Eₐ = 0.506 × 8.3 × 600 = 0.506 × 4980 = 2519.88 J mol⁻¹.
5. So. The activation energy is about 2520 J.

The common slip: 1096. Using log 5/3 = 0.22 directly in place of ln 5/3 drops the factor 2.3 and gives 1096, not 2520.

The idea to keep. Two rate constants at two temperatures give Eₐ through their ratio. If the data are in log base 10, multiply by 2.3 to get the natural log before using R.

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