A reaction has a rate constant of 1.24×10−4s−1 at 27 ∘C and 0.231s−1 at 75 ∘C .
a) Determine the activation barrier for the reaction.
Express your answer in units of kilojoules per mole.
b) What is the value of the rate constant at 18 ∘C ?
Express your answer in units of inverse seconds.
1 Answer

a)
ln(k2/k1) = (Ea/R)(1/T1 – 1/T2)
ln(0.231 / 1.24e4) = (Ea / 8.314 J/molK)(1/(27 + 273K) – 1/(75 + 273K))
7.530 = (5.530e5)(Ea)
Ea = 1.36e5 J/mol or 136 kJ/mol
b)
ln(k2 / 1.24e4) = (1.36e5 J/mol / 8.314 J/molK)(1/(27 + 273K)) – 1/(18 + 273K))
ln(k2 / 1.24e4) = 1.686
k2 = (1.24e4 s^1)e^(1.686)
k2 = 2.30e5 s^1