The rate of a reaction decreased by 3.555 times when the temperature was changed from 40ᵒC to 30ᵒC. The activation energy (in kJ mol –1 ) of the reaction is _______. (Round off to the nearest integer)
Take; R = 8.314 J mol –1 K –1 ln 3.555 = 1.268

Solution

Chemistry ma’am · 4 steps

The rate ratio is the rate-constant ratio, so this is two-temperature Arrhenius once you switch to kelvin.

1. Rate ratio. Concentrations are the same at both temperatures, so the rate ratio equals the rate-constant ratio. The rate is 3.555 times larger at the higher temperature, so k₂/k₁ = 3.555.

   ln k₂/k₁ = Eₐ/R(1/T₁-1/T₂)

2. Convert to kelvin. T₁ = 30 + 273 = 303 K and T₂ = 40 + 273 = 313 K. The bracket is 1/303-1/313 = 10/94839 = 1.054 × 10⁻⁴ K⁻¹.
3. Solve for Ea. Eₐ = (8.314 × 1.268)/(1.054 × 10⁻⁴) = (10.542 × 94839)/10 ≈ 9.998 × 10⁴ J mol⁻¹.
4. So. Eₐ ≈ 99.98 kJ mol⁻¹, which rounds to 100.

The common slip: 1. Putting 30 and 40 straight into 1/T₁-1/T₂ without adding 273 makes the bracket 1/120 and gives Eₐ ≈ 1.27 kJ mol⁻¹, which rounds to 1, not 100.

The idea to keep. A ratio of rates at two temperatures is a ratio of rate constants. Always put temperatures in kelvin before taking reciprocals, and convert J to kJ only at the end.

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