It has been found that for a chemical reaction with rise in temperature by 9 K the rate constant gets doubled. Assuming a reaction to be occurring at 300 K, the value of activation energy is found to be ____________ kJ mol⁻¹. [nearest integer]
(Given ln10 = 2.3, R = 8.3 J K⁻¹ mol⁻¹, log 2 = 0.30)

Solution

Chemistry ma’am · 4 steps

The rate constant doubles over 9 K, so take the ratio of the two Arrhenius equations and solve for Eₐ.

1. Set up the ratio. Write k = A e^(-Eₐ/RT) at T₁ = 300 K and T₂ = 309 K, then divide one by the other. The factor A cancels, so Eₐ is the only unknown.

   ln k₂/k₁ = Eₐ/R(1/T₁-1/T₂) = Eₐ/R·(T₂ - T₁)/(T₁ T₂)

2. Put in numbers. k₂/k₁ = 2, so ln 2 = ln 10 × log 2 = 2.3 × 0.30 = 0.69. The temperature factor is 9/(300 × 309) = 9/92700 K⁻¹.
3. Solve for Ea. Eₐ = (0.69 × 8.3 × 92700)/9 = 5.727 × 10300 ≈ 58988 J mol⁻¹.
4. So. Eₐ ≈ 58.99 kJ mol⁻¹, which rounds to 59.

The common slip: 26. Putting log 2 = 0.30 straight into the ln form, without multiplying by 2.3, gives about 26 instead of 59.

The idea to keep. When k changes by a known factor between two temperatures, use the two-temperature Arrhenius form. Convert log to ln with 2.303 (here 2.3), and keep R in joules until the end.

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