Question

A) The activation energy of a certain reaction is 48.0 kJ/mol . At 27 ∘C , the rate constant is 0.0120s−1. At what temperature in degrees Celsius would this reaction go twice as fast?

B) Given that the initial rate constant is 0.0120s−1 at an initial temperature of 27 ∘C , what would the rate constant be at a temperature of 170. ∘C for the same reaction described in Part A?

Answer 1

A) We need to use the Arrhenius equation; the integrated form is given as

ln k₂/k₁ = E_a/R*(1/T₁ – 1/T₂)

where k₁ is the rate of the reaction at absolute temperature T₁ and k₂ is the rate of the reaction at absolute temperature T₂.

Given E_a = 48.0 kJ/mol = (48.0 kJ/mol)*(1000 J/1 kJ) = 48000 J/mol, T₁ = 27°C ≡ (27 + 273) K = 300 K, k₁ = 0.0120 s^-1 and k₂ = 2*0.0120 s^-1 = 0.0240 s^-1, plug in values and obtain

ln (0.0240 s^-1)/(0.0120 s^-1) = (48000 J/mol)/(8.314 J/mol.K)*(1/300 – 1/T₂) K^-1

====> ln (2) = (5773.394275)*(0.003333 – 1/T₂)

====> 0.69315 = (5773.394275)*(0.003333 – 1/T₂)

====> 0.003333 – 1/T₂ = 0.69315/5773.394275 = 0.000120

====> 1/T₂ = 0.003213

====> T₂ = 1/0.003213 = 311.2356 ≈ 311.23

The desired temperature is 311.23 K ≡ (311.23 – 273)°C = 38.23°C (ans).

B) We shall use the Arrhenius equation again.

Given E_a = 48.0 kJ/mol = (48.0 kJ/mol)*(1000 J/1 kJ) = 48000 J/mol, T₁ = 27°C ≡ (27 + 273) K = 300 K, T₂ = 170°C ≡ (170 + 273) K = 443 K and k₁ = 0.0120 s^-1, plug in values and obtain

ln k₂/k₁ = E_a/R*(1/T₁ – 1/T₂)

====> ln k₂/(0.0120 s^-1) = (48000 J/mol)/(8.314 J/mol.K)*(1/300 – 1.443) K^-1

====> ln k₂/(0.0120 s^-1) = (5773.394275)*(0.001076) = 6.2122

====> k₂/(0.0120 s^-1) = exp^(6.2122) = 498.7974

====> k₂ = 498.7974*(0.0120 s^-1) = 5.98556 s^-1 ≈ 5.9856 s^-1 (ans).