Question

If a reaction has a rate constant of 3.7 x 10-3 s-1 at 25oC, and an activation energy of 43.6 kJ/mol, what will be the rate constant for this reaction at 75oC?

Answer 1

Answer:

The rate constant for the reaction at 75 °C = 4.6 × 10^(-2) s^(-1)

Explanation:

By Arrhenius equitation

k = Ae^(-Ea/(RT))

Therefore

3.7 x 10-3 s-1 = A×e^(43600J/mol/(8.3145×298))

Therefore A = 160900.320

Therefore at 75 °C T = 348.15 K and k = 160900.320×e^(-43600/(8.314×348.15)

k = 4.6 × 10^(-2) s^(-1)

the rate constant for this reaction at 75 °C is k = 4.6 × 10^(-2) s^(-1)

Which shows that the rate of reaction increases as temperature increases that is the proportions of effective collisions with energy equal to or greater than the activation energy increases.