What is the activation energy and rate constant?

Question

A reaction has a rate constant of 1.25×10⁻⁴ s−1 at 27  ∘C and 0.230 s−1 at 79  ∘C.

Determine the activation barrier for the reaction.  

What is the value of the rate constant at 17 ∘C?

Answer 1

k1 = 1.25x10^-4s^-1

T1 = 27 + 273 = 300K

k2 = 0.230 s^-1

T2 = 79 + 273 = 352K

Ea = ?

R = 8.314 J/Kmol = 0.008314 kJ/Kmol

Arrhenius Equation: ln (k2/k1) = -Ea/R (1/T2 - 1/T1)

ln (0.230 / 1.25x10-4) = -Ea / 0.008314 (1/352 - 1/300)

7.52 = -Ea /0.008314 (0.00284 - 0.00333)

7.52 = -Ea /0.008314 (-0.00049)

Ea = 128 kJ/mol

To find the rate constant @ 17ºC, use the Arrhenius equation again where...

T2 = 17 + 273 = 290K

k2 = ?

Ea = 128 kJ/mol

And solve for k2