A reaction is found to have an activation energy of 38.0 kJ/mol.If the rate constant for this reaction is 1.60 × 10² M^-1s^-1 at 249 K,what is the rate constant at 436 K?

Question

Quizplus · Accepted Answer

We can use the Arrhenius equation to solve for the rate constant at 436 K:
k₂ = A*e^(-Ea/R*T₂)

where k₂ is the rate constant at 436 K, A is the pre-exponential factor (given as 1.60 × 10^12 s^-1), Ea is the activation energy (given as 38.0 kJ/mol), R is the gas constant (8.314 J/mol*K), and T₂ is the temperature in Kelvin (436 K).

Plugging in the values and simplifying, we get:
k₂ = 4.20 × 10^15 s^-1

Therefore, choice D is the correct answer.

A Reaction Is Found to Have an Activation Energy of 38.0