The decomposition of a hydrocarbon has a rate constant of 2.5 × 10⁴ s⁻¹ at 27 °C. At what temperature would the rate constant be 7.5 × 10⁴ s⁻¹ if the energy of activation is 19.147 × 10³ J mol⁻¹?

k₁ = 2.5 × 10⁴ s⁻¹
k₂ = 7.5 × 10⁴ s⁻¹
Eₐ = 19.147 × 10³ J/mol
T₁ = 27 + 273 = 300 K

Using the Arrhenius equation in logarithmic form:

log(k₂/k₁) = Eₐ / (2.303R) * (T₂ - T₁) / (T₂T₁)

log(7.5 × 10⁴ / 2.5 × 10⁴) = (19.147 × 10³) / (2.303 × 8.314) * (T₂ - 300) / (T₂ × 300)

0.477 = (19.147 × 10³) / (19.147) * (T₂ - 300) / (300T₂)

0.477 = 1000 * (T₂ - 300) / (300T₂)

0.1431T₂ = T₂ - 300

300 = T₂ - 0.1431T₂

300 = 0.8569T₂

T₂ = 300 / 0.8569

T₂ ≈ 350 K