Question

The desorption of a single molecular layer of n-butane from a single crystal of aluminum oxide was found to be first order with a rate constant of 0.128/s at 150 K.

A. What is the half-life of the desorption reaction?

Answer 1

The half-life of the first-order desorption reaction of n-butane from aluminum oxide at 150 K, with a rate constant of 0.128 s⁻¹, is approximately 5.41 seconds.

Step-by-step explanation:

The question revolves around calculating the half-life of the desorption reaction of n-butane from aluminum oxide, which occurs with a first-order rate constant at a given temperature. In first-order reactions, the half-life does not depend on the initial concentration of the reactant. To find the half-life (t1/2) for a first-order reaction, the formula t1/2 = ln(2)/k is used, where k is the rate constant.

Given that the rate constant (k) for the desorption of n-butane is 0.128 s⁻¹, the half-life can be calculated as follows:

t1/2 = ln(2) / 0.128 s⁻¹

ln(2), which is the natural logarithm of 2, is approximately 0.693.

Therefore, the half-life (t1/2) is:

t1/2 = 0.693 / 0.128 s⁻¹ = 5.4140625 seconds

Hence, the half-life of the desorption reaction of n-butane from aluminum oxide at 150 K is approximately 5.41 seconds.