A certain first-order reaction has a rate constant of 2.75×10^{- 2} s^{- 1} at 20 ° C. What is the value of k at 60 ° C if the energy of activation equals 75.5 kJ/mol?

Question

Quizplus · Accepted Answer

The Arrhenius equation relates the rate constant (k) to the activation energy (Ea), temperature (T), and the gas constant (R).

k = A * exp(-Ea/(R*T))

where A is the pre-exponential factor.

To solve for k at 60 °C, we can use the given rate constant at 20 °C and the activation energy:

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

where k1 and T1 refer to the rate constant and temperature at 20 °C, and k2 and T2 refer to the rate constant and temperature at 60 °C.

R = 8.314 J/(mol*K)

Plugging in the values:

ln(k1/k2) = (75,500 J/mol) / (8.314 J/(mol*K)) * ((1/333 K) - (1/293 K))

ln(k1/k2) = 19.607

k1/k2 = exp(19.607)

k2 = k1 / exp(19.607)

k2 = (2.75×10^-2 s^-1) / exp(19.607)

k2 ≈ 1.14 s^-1

Therefore, the answer is choice C, k = 1.14 s^-1.

A Certain First-Order Reaction Has a Rate Constant of 2