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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.

Respuesta :

superman1987

Answer:

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.

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