The desorption (leaving of the surface) of a single molecular layer of n-butane from a single crystal of aluminum oxide is found to be first order with a rate constant of 0.128 s−1 at 150 K. If the surface is initially completely covered, what fraction will remain covered after 10 s? After 20 s?

Verified step by step guidance

1. Identify that the desorption process follows first-order kinetics, which can be described by the equation: \( [A] = [A]_0 e^{-kt} \), where \([A]\) is the concentration at time \(t\), \([A]_0\) is the initial concentration, \(k\) is the rate constant, and \(t\) is the time.
2. Recognize that the problem asks for the fraction of the surface that remains covered, which corresponds to \( \frac{[A]}{[A]_0} \).
3. Substitute the given rate constant \(k = 0.128 \text{ s}^{-1}\) and the time \(t = 10 \text{ s}\) into the first-order kinetics equation to find the fraction remaining after 10 seconds: \( \frac{[A]}{[A]_0} = e^{-0.128 \times 10} \).
4. Repeat the substitution for \(t = 20 \text{ s}\) to find the fraction remaining after 20 seconds: \( \frac{[A]}{[A]_0} = e^{-0.128 \times 20} \).
5. Interpret the results: the calculated values from the equations will give you the fraction of the surface that remains covered after 10 seconds and 20 seconds, respectively.