Question

The rate constant of a reaction is 0.01439 min⁻¹ at 25°C and its activation energy is 70,000 J mol⁻¹. What is the value of rate constant at 40°C? (Given; R = 8.314 J K⁻¹ mol⁻¹)

Answer 1

Given:

k₁​ = 0.01439 min⁻¹ at T₁ = 25°C = 298 K

T₂ = 40°C = 313 K

E_a = 70,000 J/mol

R = 8.314 J K⁻¹ mol⁻¹

We will use the Arrhenius equation in two-point form:

`ln(k_2/k_1) = E_a/R ((T_2 - T_1)/(T_1T_2))`

`ln(k_2/0.01439) = 70000/8.314 ((313 - 298)/(298 xx 313))`

`ln(k_2/0.01439) = 8415.5 ((15)/(93374))`

`ln(k_2/0.01439) = 8415.5 xx 1.606 xx 10^-4`

⇒ `k_2/0.01439 = e^1.351`

⇒ k₂ = 0.01439 × 3.861

≈ 0.0556 min⁻¹

∴ k₂ = 0.0556 min⁻¹ at 40°C