The isomerization of cyclopropane follows first order kinetics. The rate constant at 700 K is 6.20 * 10-4 min-1, and the half-life at 760 K is 29.0 min. Calculate the activation energy for this reaction.

Question

Jay Bell

Chemistry

5 November, 04:30

The isomerization of cyclopropane follows first order kinetics. The rate constant at 700 K is 6.20 * 10-4 min-1, and the half-life at 760 K is 29.0 min. Calculate the activation energy for this reaction.

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Answers (1)

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Melvin Townsend · Answer 1

269.068 kJ/mol.

Explanation:

We can use the relation:

ln (k₂/k₁) = (Eₐ/R) [ (T₂ - T₁) / (T₁T₂) ].

k₁ = 6.20 x 10⁻⁴ min⁻¹, T₁ = 700.0 K.

To get k₂:

in first order reactions: k = 0.693 / (half-life).

∴ k₂ = 0.693 / (29.0 min) = 2.39 x 10⁻² min⁻¹, T₂ = 760.0 K.

∵ ln (k₂/k₁) = (Eₐ/R) [ (T₂ - T₁) / (T₁T₂) ]

∴ ln [ (2.39 x 10⁻² min⁻¹) / (6.20 x 10⁻⁴ min⁻¹) ] = (Eₐ / (8.314 J/mol. K)) [ (760.0 K - 700.0 K) / (760.0 K) (700.0 K) ].

3.65 = (Eₐ / (8.314 J/mol. K)) (1.128 x 10⁻⁴).

∴ Eₐ = (3.65) (8.314 J/mol. K) / (1.128 x 10⁻⁴) = 269.068 kJ/mol.