Problem 97 - Chemistry A Molecular Approach

Question

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 \(0.128 / \mathrm{s}\) at \(150 \mathrm{~K}\). a. What is the half-life of the desorption reaction? b. If the surface is initially completely covered with \(n\) -butane at \(150 \mathrm{~K},\) how long will it take for \(25 \%\) of the molecules to desorb (leave the surface)? For \(50 \%\) to desorb? c. If the surface is initially completely covered, what fraction will remain covered after 10 s? After 20 s?

Step-by-Step Solution

Answer

a. Half-life of the desorption reaction is \( \frac{\ln(2)}{0.128 / \mathrm{s}} \). b. For 25% to desorb, solve \( \ln(0.75) = -0.128 \times t \); for 50%, use the half-life. c. The remaining fraction after 10s can be found by \( \ln(r) = -0.128 \times 10 \) and after 20s by \( \ln(r) = -0.128 \times 20 \) respectively.

Step 1: Calculating the Half-Life of the Desorption Reaction

To calculate the half-life \( t_{1/2} \) for a first-order reaction, we use the formula \( t_{1/2} = \frac{\ln(2)}{k} \), where \( k \) is the rate constant. Here, we have \( k = 0.128 / \mathrm{s} \) and \( \ln(2) \) is the natural logarithm of 2. Substituting the values, find \( t_{1/2} \) for the reaction.

Step 2: Calculating Time for 25% Desorption

For 25% desorption, we need to find the time \( t \) when the fraction remaining \( r \) is 75%. Since the reaction is first order, we use the formula \( \ln(r) = -kt \). Solve for \( t \) when \( r = 0.75 \) and \( k = 0.128 / \mathrm{s} \) to find the time for 25% to desorb.

Step 3: Calculating Time for 50% Desorption

50% desorption is the same as the half-life of the reaction. Thus, the time taken for 50% desorption is the half-life, which has been calculated in Step 1.

Step 4: Calculating Fraction Remaining after 10 seconds

To find the fraction remaining after 10 seconds, we use the formula \( \ln(r) = -kt \) where \( t = 10 \mathrm{s} \) and solve for \( r \) to determine what fraction of \( n \) -butane remains covered after 10 seconds.

Step 5: Calculating Fraction Remaining after 20 seconds

Similarly, use the formula \( \ln(r) = -kt \) where \( t = 20 \mathrm{s} \) and solve for \( r \) to determine what fraction of \( n \) -butane remains covered after 20 seconds.

Key Concepts

Half-Life of Reaction

The concept of half-life is a crucial aspect when studying reaction kinetics, particularly in first-order reactions. Half-life, or t_1/2, is the amount of time required for half of the reactant to be consumed or transform into another phase. This is a constant and particularly important because it does not depend on the initial concentration of the reactant.

In the exercise provided, the half-life can be determined using the formula \( t_{1/2} = \frac{\ln(2)}{k} \) where \( k \) is the rate constant. Given that the rate constant (k) is \(0.128 / \mathrm{s}\), we plug in this value to find the half-life of butane desorption from the aluminum oxide surface. This calculation is straightforward but provides valuable insight into how long it will take for half of the surface to be free of butane. Understanding half-life helps in predicting the duration of the reaction as well as in designing systems that require precise timing in the desorption process, like in catalysis or surface treatments.

Rate Constant

The rate constant, symbolized as \( k \), is a key component in the kinetic analysis of a chemical reaction. In first-order reactions, it signifies the proportionality between the reaction rate and the concentration of the reactant. Simply put, the rate constant tells us the speed at which a reaction proceeds under certain conditions of temperature and pressure.

In the provided example, the rate constant is given as \(0.128 / \mathrm{s}\) at a temperature of \(150 \mathrm{~K}\). This numerical value reflects how rapidly butane desorbs from the aluminum oxide surface. The higher the rate constant, the faster the desorption process. Importantly, rate constants are influenced by external factors such as temperature, and this dependency is often explained by the Arrhenius equation. An understanding of how to compute and interpret the rate constant allows students to not just solve homework problems, but also to predict the outcomes and design of real-world chemical processes.

Chemical Desorption Process

Chemical desorption refers to the process by which molecules detach from the surface they are adsorbed on. It is a critical process in many industrial and research applications, including catalysis, material science, and environmental studies.

The exercise under discussion involves the desorption of \( n \) -butane from an aluminum oxide surface. This first-order reaction implies that the rate of desorption is directly proportional to the concentration of \( n \) -butane adsorbed on the surface. Desorption kinetics provide insights on how surfaces can be cleaned or regenerated and how reactive intermediates are formed and consumed in catalytic reactions. This knowledge is especially valuable in industries where surface properties like cleanliness and reactiveness are crucial for product quality and efficiency.