Question

A reaction has a rate constant of 1.26×10⁻⁴ s−1 at 26  ∘C and 0.229 s−1 at 77  ∘C .

A) Determine the activation barrier for the reaction. ( kJ/mol )

B) What is the value of the rate constant at 16  ∘C ? (s^-1)

Answer 1

Given that

K₁ = 1.26*10^-4 s^-1

temperature (T₁) = 26^oC = 299.15K

k₂ = 0.229 s^-1

Temperature (T₂) = 77^oC = 77+ 273.15 = 350.15 K

Gas constant (R) = 8.314 J/mol.K

As we know

ln (k₂/k₁) = (E_a /R)(1/T₁ - 1/T₂)

ln ( 0.229 s^-1 / 1.26*10^-4 s^-1) = (E_a / 8.314 J/mol.K) (1/299.15K - 1/350.15K)

E_a = Activation energy = 128157.89 J = 128.2 kJ /mol

(B)Again if T₂ = 16^oC = 16+273.15 = 289.15 K and k₂ = ?

ln (k₂/k₁) = (E_a /R)(1/T₁ - 1/T₂)

ln ( k₂ / 1.26*10^-4 s^-1) = (128157.89 J / 8.314 J/mol.K) (1/299.15K - 1/289.15K)

k₂ = 2.12 *10^-5 s^-1