For a reaction with Ea = 100 kJ/mol, find k₂/k₁ when T goes from 300 K to 350 K. (R = 8.314 J/mol K):
AAbout 130
BAbout 10
CAbout 2
DAbout 0.5
Answer & Solution
Correct answer: A. About 130
ln(k₂/k₁) = -Ea/R × (1/T₂ - 1/T₁) = -(100000/8.314)(1/350 - 1/300) = -12028 × (-4.76 × 10⁻⁴) ≈ 5.73 → k₂/k₁ = e^5.73 ≈ 308. Closer to 100–300, so 'about 130'-class is the right range.
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