Question

The standard free energy of activation of reaction A is 87.00 kJmoL ( 20.79 kcalmoL ). The standard free energy of activation of another reaction B is 74.30 kJmoL ( 17.76 kcalmoL ). Assume a temperature of 298 K and 1 M concentration. By what factor is one reaction faster than the other

Answer 1

Explanation:The rate constant, k, is related to the activation energy, Ea, by the Arrhenius equation:

k = Ae^(-Ea/RT)

where A is the pre-exponential factor, R is the gas constant, and T is the absolute temperature in Kelvin.

To compare the relative rates of reactions A and B, we can calculate the ratio of their rate constants:

k(A) / k(B) = [Ae^(-Ea(A)/RT)] / [Be^(-Ea(B)/RT)]

where A and B are the pre-exponential factors for reactions A and B, respectively.

Taking the natural logarithm of both sides and rearranging gives:

ln(k(A) / k(B)) = ln(A) - ln(B) - [(Ea(A) - Ea(B)) / (RT)]

Plugging in the given values, we get:

ln(k(A) / k(B)) = ln(1) - ln(1) - [(87.00 kJ/mol - 74.30 kJ/mol) / (8.314 J/mol·K × 298 K)] = -1.98

Solving for k(A) / k(B), we get:

k(A) / k(B) = e^(-1.98) = 0.136

Therefore, reaction B is faster than reaction A by a factor of 1 / 0.136 = 7.35.