Question

How many times faster will the reaction occur at 47 ∘C than at 0 ∘C?

Express your answer using two significant figures.

The activation energy of a certain reaction is 79.4 kJ/mol .

Answer 1

We need to use the integrated rate law for the Arrhenius equation as

ln k₂/k₁ = E_a/R*(1/T₁ – 1/T₂) where k₁ is the rate constant for a reaction at T₁ = 0°C ≡ (0 + 273) K = 273 K and k₂ is the rate constant of the reaction at T₂ = 47°C ≡ (47 + 273) K = 320 K.

Plug in values and get

ln k₂/k₁ = (79.4 kJ/mol)/(8.314 J/mol.K)*[(1/273 K) – (1/320 K)]

====> ln k₂/k₁ = (79.4 kJ/mol)*(1000 J/1 kJ)/(8.314 J/mol.K)*(0.003663 – 0.003125) K^-1

====> ln k₂/k₁ = (9550.15636 K)*(0.000538 K^-1) = 5.13798

====> k₂/k₁ = exp^(5.13798) = 170.37197 ≈ 170 (if we consider only the nearest whole number)

====> k₂ = 170*k₁

Therefore, the reaction is approximately 170 times faster at 47°C than at 0°C (ans).