Question

A reaction has a rate constant of 1.27x10-4/s at 29 ℃ and .227/s at 77 ℃. Determine the activation barrier for the reaction. What is the value of the rate constant at 18 ℃?

Answer 1

We can solve this problem by Arrhenius Equation

where,
k = Rate constant
A = Frequency factor or Preexponential factor
E_a = Activation energy
T = Kelvin temparature
R = The gas constant
e = Mathematical quantity, e

The frequency factor, A, in the equation is approximately constant for such a small temperature change.

Given
k₁ = 1.27x10^-4/s, T1 = 302 Kelvin
k₂ = 0.227/s, T2 = 350 Kelvin

ln (k) = ln (A) - E_A/RT
Therefore,
ln (1.27x10^-4) = ln (A) - E_A/(8.31x302) ---------------------------- (1)
ln ( 0.227   ) = ln (A) - E_A/(8.31x350) ---------------------------- (2)

(1) - (2) gives us
ln (1.27x10^-4) - ln (0.227) = E_A{1/(8.31x350) - 1/(8.31x302)}
-7.488 = E_A (-0.0000546469)

Therefore, Activation Energy barrier, E_A = 137 Kilo joules per mole ------------------- (3)

ln (A) = ln(0.227) + 137000/(8.31x350) From (2)
ln (A) = -1.01177208289
Therefore A = 0.36357412488

For rate constant at 18^o C i.e.,

k = Ae-^E_A/RT

k = (0.36357412488)e^{(-137000/8.31*291)}

Therefore k_18^oC = 1.07 x 10^-25 / s