A reaction is found to have a rate constant of 3.36 × 10⁴ L mol^-1s^-1 at 344 K and a rate constant of 7.69 L mol^-1s^-1 at 219 K. Determine the activation energy for this reaction.

Question

Quizplus · Accepted Answer

The Arrhenius equation can be used to link the rate constant with temperature: k = A exp(-Ea/RT) Rearranging the equation we get: ln(k) = ln(A) - Ea/RT Taking logarithms of both rate constants and subtracting those, we will obtain: ln(k2/k1) = ln(7.69/3.36) = 0.849 The activation energy is given by the slope of the line obtained when plotting ln(k) vs. 1/T. Using the two temperature values, we get: ln(k1) = ln(3.36 × 10⁴ L mol^-1s^-1) = 1.206 ln(k2) = ln(7.69 L mol^-1s^-1) = 2.039 Slope = -Ea/R = (-2.039 + 1.206)/(1/344 - 1/219) = 42.0 kJ mol^-1. Therefore, the correct answer is (B).

A Reaction Is Found to Have a Rate Constant of 3.36