Temperature Dependence Of Rate
The rate of a reaction doubles when the temperature is raised from 27°C to 37°C. The activation energy of the reaction is approximately: (R = 8.314 J mol⁻¹ K⁻¹)
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Solution
53.6 kJ mol⁻¹
T₁ = 27 + 273 = 300 K, T₂ = 37 + 273 = 310 K, k₂/k₁ = 2. Using log(k₂/k₁) = Eₐ/(2.303R) × (T₂ − T₁)/(T₁T₂): log(2) = Eₐ/(2.303 × 8.314) × (310 − 300)/(300 × 310). 0.3010 = Eₐ/19.147 × 10/93000. Eₐ = 0.3010 × 19.147 × 93000/10 = 0.3010 × 1,780,671/10 ≈ 53,600 J mol⁻¹ ≈ 53.6 kJ mol⁻¹.
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About This Question
- Subject
- chemistry
- Chapter
- chemical kinetics
- Topic
- temperature dependence of rate
- Difficulty
- Hard
- Year
- 2025
Solution
Correct Answer:
53.6 kJ mol⁻¹
T₁ = 27 + 273 = 300 K, T₂ = 37 + 273 = 310 K, k₂/k₁ = 2. Using log(k₂/k₁) = Eₐ/(2.303R) × (T₂ − T₁)/(T₁T₂): log(2) = Eₐ/(2.303 × 8.314) × (310 − 300)/(300 × 310). 0.3010 = Eₐ/19.147 × 10/93000. Eₐ = 0.3010 × 19.147 × 93000/10 = 0.3010 × 1,780,671/10 ≈ 53,600 J mol⁻¹ ≈ 53.6 kJ mol⁻¹.
This hard difficulty chemistry question is from the chapter chemical kinetics, covering the topic of temperature dependence of rate. It appeared in the 2025 exam.
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