A second order reaction has rate constant 1.28 x 107 L/mol/s at 59°C, and 3.55x 106 at -30°C.?

Question

What is the activation energy in J/mol of this reaction?

Answer 1

Use Arrhenius equation:

k = A·exp{- Ea/(R·T)}

You've got  the rate constants at tow different temeperatures:

k₁ = A·exp{- Ea/(R·T₁)}

k₂ = A·exp{- Ea/(R·T₂)}

Take the ratio of the rate constants to eliminate unknwon collision factor A. Then solve for Activation energy Ea:

k₁/k₂ = [ A·exp{- Ea/(R·T₁)} ] / [ A·exp{- Ea/(R·T₂)} ]

<=>

k₁/k₂ = exp{- Ea/(R·T₁)} / exp{- Ea/(R·T₂)}

<=>

k₁/k₂ = exp{ (Ea/R)·(1/T₂ - 1/T₁) }

<=>

ln{k₁/k₂} = (Ea/R)·(1/T₂ - 1/T₁)

=>

Ea = R · ln{k₁/k₂} / (1/T₂ - 1/T₁)

= 8.314472J/molK · ln{1.28×10⁷L/mols / 3.55×10⁶L/mols} / (1/243K - 1/332K)

= 9666J/mol