Arrhenius Equation And Activation Energy
The rate constant of a reaction at 500 K is 2.0 × 10⁻³ s⁻¹ and at 700 K is 4.0 × 10⁻² s⁻¹. The activation energy of the reaction is approximately: (R = 8.314 J mol⁻¹ K⁻¹)
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About This Question
- Subject
- chemistry
- Chapter
- chemical kinetics
- Topic
- arrhenius equation and activation energy
- Difficulty
- Hard
- Year
- 2025
Solution
Correct Answer:
43.6 kJ mol⁻¹
Using the Arrhenius equation in the two-temperature form: log(k₂/k₁) = Eₐ/(2.303R) × (1/T₁ − 1/T₂). Here k₁ = 2.0 × 10⁻³ s⁻¹ at T₁ = 500 K and k₂ = 4.0 × 10⁻² s⁻¹ at T₂ = 700 K. log(4.0 × 10⁻² / 2.0 × 10⁻³) = log(20) = 1.301. (1/T₁ − 1/T₂) = (1/500 − 1/700) = (7 − 5)/3500 = 2/3500 = 5.714 × 10⁻⁴ K⁻¹. Eₐ = 1.301 × 2.303 × 8.314 / 5.714 × 10⁻⁴ = 1.301 × 19.147 / 5.714 × 10⁻⁴ ≈ 43,600 J mol⁻¹ ≈ 43.6 kJ mol⁻¹.
This hard difficulty chemistry question is from the chapter chemical kinetics, covering the topic of arrhenius equation and activation energy. It appeared in the 2025 exam.
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