The rate constant for a reaction at 40.0°C is exactly 6 times that at 20.0°C.Calculate the Arrhenius energy of activation for the reaction.(R = 8.314 J/K.0mol)

Question

Quizplus · Accepted Answer

According to the Arrhenius equation, the ratio of rate constants (k2/k1) is equal to exp[(Ea/R) * (1/T1 - 1/T2)].

Let k1 be the rate constant at 20.0°C and k2 be the rate constant at 40.0°C.

We are given that k2 = 6k1. Therefore:

6k1/k1 = exp[(Ea/R) * (1/293 - 1/313)]

6 = exp[(Ea/8.314) * (0.0034)]

Taking the natural logarithm of both sides:

ln(6) = (Ea/8.314) * (0.0034)

Solving for Ea:

Ea = (ln 6) * (8.314/0.0034)

Ea = 68.3 kJ/mol.

Therefore, the correct answer is (C).

The Rate Constant for a Reaction at 40