Problem 97 - Chemistry Structure and Properties

Question

The desorption 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\(/ \mathrm{s}\) at 150 \(\mathrm{K}\) . \begin{equation} \begin{array}{l}{\text { a. What is the haff-life of the desorption reaction? }} \\ {\text { b. If the surface is initially completely covered with } n \text { -butane at }} \\ {150 \mathrm{K}, \text { how long will it take for } 25 \% \text { of the molecules to desorb? For }} \\ {50 \% \text { to desorb? }}\\\\{\text { c. If the surface is initially completely covered, what fraction will remain }} \\ {\text { covered after } 10 \text { s? After } 20 \mathrm{s?}}\end{array} \end{equation}

Step-by-Step Solution

Answer

The half-life is \( \frac{\ln(2)}{0.128} \) seconds, which is approximately 5.42 seconds. To desorb 25% of the molecules, it will take \( \frac{\ln(4/3)}{0.128} \) seconds. After 10 s, about 0.233 of the initial amount will remain, and after 20 s, about 0.054 of the initial amount will remain.

Step 1: Determine half-life for first-order reactions

The half-life for a first-order reaction is given by the equation: \( t_{1/2} = \frac{\ln(2)}{k} \), where \( k \) is the rate constant. Plug in the given rate constant \( k = 0.128 / \mathrm{s} \) to find the half-life.

Step 2: Calculate the half-life

Substitute the given rate constant into the half-life equation: \( t_{1/2} = \frac{\ln(2)}{0.128 / \mathrm{s}} \).

Step 3: Determine the time for 25% desorption

Using the first-order reaction formula \( N = N_0 e^{-kt} \), where \( N \) is the number of molecules remaining, \( N_0 \) is the initial number, \( k \) is the rate constant, and \( t \) is time, solve for \( t \) when \( N = 0.75 N_0 \).

Step 4: Determine the time for 50% desorption

For a first-order reaction, the time for 50% desorption is the same as the half-life, which was already calculated in Step 2.

Step 5: Calculate the fraction remaining after 10 s

Substitute \( t = 10 \mathrm{s} \) into the first-order reaction formula to calculate the fraction of molecules remaining.

Step 6: Calculate the fraction remaining after 20 s

Substitute \( t = 20 \mathrm{s} \) into the first-order reaction formula to calculate the fraction of molecules remaining.

Key Concepts

Reaction Rate Constant

When studying chemical kinetics, the reaction rate constant is a crucial parameter that quantifies the speed of a reaction. In a first-order reaction, such as the desorption of n-butane from aluminum oxide, the rate of reaction is directly proportional to the concentration of a single reactant.

For a reaction described by the equation \( A \rightarrow Product \), the rate can be expressed as \( \text{rate} = k[A] \), where \( [A] \) is the concentration of reactant A and \( k \) is the reaction rate constant. It has units such as \(1/\text{s}\) or \(1/\text{min}\), depending on how time is measured. A larger value of \( k \) indicates a faster reaction. In the given exercise, the reaction rate constant \( k \) is 0.128 \(1/\text{s}\), which plays a pivotal role in determining the reaction's half-life and the time it takes for certain percentages of the reactant to desorb.

Half-life of a Reaction

The half-life of a reaction is defined as the time required for half the amount of the reactant to be converted into product. For first-order reactions, the half-life, symbolized as \( t_{1/2} \), is a constant that does not depend on the initial concentration of the reactant. The relationship between the half-life and the rate constant \( k \) for a first-order reaction is given by the equation \( t_{1/2} = \frac{\ln(2)}{k} \).

This formula is derived from the natural logarithm properties and the fact that at half-life, the concentration of the reactant is half of its initial value. The half-life provides a convenient way to compare the speed of different reactions; a shorter half-life indicates a faster reaction. By plugging the rate constant from the exercise into the established equation, we can calculate the exact half-life for the n-butane desorption process at 150 K.

Desorption Process

A desorption process is the reverse of adsorption, meaning it involves the release of a substance from a surface. Desorption can be physical or chemical, such as when a gas molecule detaches from a solid surface. The rate at which desorption occurs can be crucial for various applications, including catalysis, purification processes, and the production of semiconductor materials.

In the context of our exercise, we are focused on the desorption of n-butane from a single crystal of aluminum oxide, which follows first-order kinetics. The desorption process insights allow us to predict and calculate how long it takes for a certain percentage of the layer to desorb and to understand how much of the layer remains at any given time. Understanding the kinetics of desorption can help in controlling the thickness of layers in manufacturing processes and in the analysis of reaction mechanisms on surfaces.