Question

The standard free energy of activation of one reaction A is 95.00 kJ mol–1 (22.71 kcal mol–1). The standard free energy of activation of another reaction B is 74.20 kJ mol–1 (17.73 kcal mol–1). Assume a temperature of 298 K and 1 M concentration.
By what factor is one reaction faster than the other?
Which reaction is faster?
(a) Reaction A is faster.
(b) Reaction B is faster.
(c) Cannot be determined.

Answer 1

Answer:

The answer to the questions are as follows

Reaction B is 4426.28 times faster than reaction A

(b) Reaction B is faster.

Explanation:

To solve the question we are meant to compare both reactions to see which one is faster

The values of the given activation energies are as follows

For A

Ea = 95.00 kJ mol–1 (22.71 kcal mol–1) and

for  B

Ea = 74.20 kJ mol–1 (17.73 kcal mol–1)

T is the same for both reactions and is equal to 298 k

Concentration of both reaction = 1M

The Arrhenius Law is given by

k = $Ae^{\frac{-E_{a} }{RT} }$

Where

k = rate constant

Ea = activation energy

R = universal gas constant

T = temperature  (Kelvin )

A = Arrhenius factor

Therefore

For reaction A, the rate constant k₁ is given by k₁ = $Ae^{\frac{-95000}{(8.314)(298)} }$

And for B the rate constant k₂ is given by k₂ = $Ae^{\frac{-74200 }{(8.314)(298)} }$

k₁ = A×2.225×10⁻¹⁷

k₂ = A×9.850×10⁻¹⁴

As seen from the above Reaction B is faster than reaction A by (A×9.850×10⁻¹⁴)/(A×2.225×10⁻¹⁷) or 4426.28 times