Question

A reaction has a rate constant of 1.21×10⁻⁴ s−1 at 26  ∘C and 0.229 s−1 at 75  ∘C .

You may want to reference (Page) section 13.5 while completing this problem

Part A

Determine the activation barrier for the reaction.

(Express your answer in units of kilojoules per mole.)

Part B

What is the value of the rate constant at 18  ∘C ?

(Express your answer in units of inverse seconds.)

Answer 1

Temperature T1 = 26 + 273 = 299 K

Rate constant k1 = 1.21 x 10^-4 s-1

Temperature T2 = 75 + 273 = 348 K

Rate constant k2 = 0.229 s-1

Part a

From Arrhenius equation

ln(k2/k1) = (E/R) (1/T1 - 1/T2)

ln(0.229/1.21 x 10^-4) = (E/8.314 J/mol·K) (1/299 - 1/348)

7.5456 = E x 5.664 x 10^-5

E = 133216 J/mol x 1kJ/1000 J

Activation energy E = 133.216 kJ/mol

Part b

Temperature T = 18 +273 = 391 K

Let rate constant = k

From Arrhenius equation

ln(k/k1) = (E/R) (1/T1 - 1/T)

ln(k/1.21 x 10^-4) = (133216/8.314) (1/299 - 1/291)

ln(k/1.21 x 10^-4) = - 1.4732

k/(1.21 x 10^-4) = exp (- 1.4732) = 0.2292

k = 2.77 x 10^-5 s-1