The rate constant of a reaction increases by five times on increase in temperature from 27°C to 52°C. The value of activation energy in kJ mol⁻¹ is _________. (Rounded off to the nearest integer)
[R = 8.314 J K⁻¹mol⁻¹]

Solution

Chemistry ma’am · 4 steps

Turn both temperatures into kelvin first, then divide the two Arrhenius equations so A cancels.

1. Convert to kelvin. T₁ = 27 + 273 = 300 K and T₂ = 52 + 273 = 325 K. The rate constant becomes five times larger, so k₂/k₁ = 5.

   ln k₂/k₁ = Eₐ/R(1/T₁-1/T₂)

2. Put in numbers. The left side is ln 5 = 1.609. The bracket is 1/300-1/325 = 25/97500 = 2.564 × 10⁻⁴ K⁻¹.
3. Solve for Ea. Eₐ = (8.314 × 1.609)/(2.564 × 10⁻⁴) = 5.22 × 10⁴ J mol⁻¹.
4. So. Eₐ ≈ 52.2 kJ mol⁻¹, which rounds to 52.

The common slip: 23. Using log 5 = 0.699 without the factor 2.303 gives about 23, not 52.

The idea to keep. Arrhenius needs kelvin, never degrees Celsius. If you use log₁₀ instead of ln, bring in the factor 2.303 on the right side.

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