Using the Arrhenius Equation to find the temperature

Question

A certain reaction has an activation energy of 34.31 kJ/mol.  At what Kelvin temperature will the reaction proceed 5.50 times faster than it did at 311k?

Answer 1

This question is related to the kinetic energies of reactants at two different temperatures, comparing them with the minimum energy needed for the reaction, E_a. At the higher temperature a much greater fraction of the molecules have a much greater fraction of the molecules have kinetic energy greater than E_a, which leads to a greater rate of reaction. To solve this problem, we use this formula (part of Arrhenius equation) f =e^-Ea/RT

E_a = 34.31 kJ = 34310 J T₁ = ? T₂= 311K

f₁ at T₁ / f₂ at 311K = 5.50 - E_a / RT = - 34310 J / (8.314 J/ mol-K x 311K) = -13.27

f₂ = e^-13.27 = 1.73 x 10^-6 f₁ at T₁ = 5.50 x 1.73 x 10^-6 = 9.49 x 10^-6 f₁ =e^-Ea/RT = 9.49 x 10^-6

- E_a / RT₁ = ln 9.49 x 10^-6 = -11.6 T₁ = - 34310 J / (8.314 J/ mol-K x 11.6) = 356 K