The rate constant of a chemical reaction increased from 0.100 s?1 to 3.30 s?1 up

Question

The rate constant of a chemical reaction increased from 0.100 s?1 to 3.30 s?1 upon raising the temperature from 25.0 ?C to 53.0 ?C .

A) Calculate the value of (1/T2?1/T1) where T1 is the initial temperature and T2 is the final temperature.

Express your answer numerically.

B)Calculate the value of ln(k1k2) where k1 and k2 correspond to the rate constants at the initial and the final temperatures as defined in part A.

Express your answer numerically.

C) What is the activation energy of the reaction?

Express your answer numerically in kilojoules per mole.

Explanation / Answer

              Given K1 = 0.100 s-1   K2 = 3.30 s-1

           Since the rate constant units, k are s-1, it is a first order reaction.

           And T1 = 25oC = (25 + 273) k = 298k

           T2 = 53oC = (53 + 273) k = 326k

A) (1/ T2 – 1/ T1) = (1/ 326k – 1/ 298k) = -2.88 x 10-4 k-1

B) ln(K1/ K2) = ln(0.100/ 3.30) = -3.5      And ln(K1 x K2) = ln(0.100 x 3.30) = -1.11 s-2

C) Arrhenius equation, k = A*exp(-Ea/R*T)  

We don't have A, but we have the k for two reactions for two different temperatures. Since A is not temperature dependent, ratio the two (k1/k2) and A will cancel out.

k1 = Ao exp(-Ea/RT1)
k2 = Ao exp(-Ea/RT2)

We know the ratio of (k1/k2), A cancels out, and all else is known except activation energy.

Dividing

(k1/k2) = (A/A)*[exp(-Ea/RT1) / exp(-Ea/RT2)]

Simplifying

(k1/k2) = exp[(Ea/R) x (1/T2 - 1/T1)]

  Simplifying [R = 8.314 J/mol*K]

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

-3.5 = (Ea/8.314 J/mol*K) x (-2.88 x 10-4 k-1)

Therefore

Ea = 101038.19 J/mol

Ea = 101.04 kJ/mol

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