The number of molecules with energy greater than the threshold energy for a reaction increases five fold by a rise of temperature from 27ᵒC to 42ᵒC. Its energy of activation in J/mol is _____. (Round off to the nearest integer)
(Take ln 5 = 1.6094; R = 8.314 J mol –1 K –1 )

Solution

Chemistry ma’am · 4 steps

The fraction of molecules above the threshold is e^(-Eₐ/RT), so a five-fold rise gives Eₐ directly.

1. Fraction above threshold. The fraction of molecules with energy at least Eₐ is e^(-Eₐ/RT). Write it at both temperatures and divide; the total number of molecules cancels, and their ratio is 5.

   ln 5 = Eₐ/R(1/T₁-1/T₂)

2. Convert to kelvin. T₁ = 27 + 273 = 300 K and T₂ = 42 + 273 = 315 K. The bracket is 1/300-1/315 = 15/94500 = 1/6300 K⁻¹.
3. Solve for Ea. Eₐ = R ln 5 × 6300 = 8.314 × 1.6094 × 6300 = 13.3806 × 6300 = 84297.5 J mol⁻¹.
4. So. The activation energy is 84297 J mol⁻¹, so the answer is 84297.

The common slip: 1012. Putting 27 and 42 straight into 1/T₁-1/T₂ without adding 273 gives about 1012. The Arrhenius equation needs absolute temperature.

The idea to keep. The number of molecules crossing the threshold scales exactly like k, as e^(-Eₐ/RT). Treat their ratio like k₂/k₁, and always put temperatures in kelvin.

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