Question

The evaporation of a 120 -nm film of \(n\) -pentane from a single crystal of aluminum oxide is zero order with a rate constant of \(1.92 \times 10^{13} \mathrm{molecules} / \mathrm{cm}^{2} \cdot \mathrm{s}\) at \(120 \mathrm{~K}\) a. If the initial surface coverage is \(8.9 \times 10^{16}\) molecules \(/ \mathrm{cm}^{2}\), how long will it take for one-half of the film to evaporate? b. What fraction of the film is left after 10 s? Assume the same initial coverage as in part a.

Short answer

a. It will take approximately 2318 seconds (or about 38.6 minutes) for half of the film to evaporate.b. After 10 seconds, approximately 0.978 fraction of the film is left.

Step-by-step solution

Understand Zero Order Kinetics

Zero order kinetics imply that the rate of reaction is constant and does not depend on the concentration of the reactants. The general formula for zero order kinetics is \[ [A]_t = [A]_0 - kt \]where \( [A]_t \) = concentration at time \( t \), \( [A]_0 \) = initial concentration, and \( k \) = zero order rate constant.

Calculate Time for Half the Film to Evaporate

For half the film to evaporate, the final coverage should be half of the initial coverage, which is \( 4.45 \times 10^{16} \) molecules/cm^2 (half of \( 8.9 \times 10^{16} \) molecules/cm^2). Using the zero order kinetics formula:\[ [A]_t = [A]_0 - kt \]we set \( [A]_t = 4.45 \times 10^{16} \) molecules/cm^2 and solve for \( t \).\[ t = \frac{[A]_0 - [A]_t}{k} \]\[ t = \frac{8.9 \times 10^{16} - 4.45 \times 10^{16}}{1.92 \times 10^{13}} \]

Solve for Time \( t \)

Plugging in the values:\[ t = \frac{8.9 \times 10^{16} - 4.45 \times 10^{16}}{1.92 \times 10^{13}} \]\[ t = \frac{4.45 \times 10^{16}}{1.92 \times 10^{13}} \]\[ t = 2.3177 \times 10^3 \text{ seconds} \]

Calculate the Fraction of Film Left After 10 s

To find the fraction of the film left after 10 s, use the initial surface coverage and the elapsed time with the rate constant. Again, use the zero order reaction rate formula:\[ [A]_{10} = [A]_0 - kt \]where \( t = 10 \) s. Solve for \( [A]_{10} \):\[ [A]_{10} = 8.9 \times 10^{16} - (1.92 \times 10^{13})(10) \]

Solve for \( [A]_{10} \)

Calculate the surface coverage after 10 seconds:\[ [A]_{10} = 8.9 \times 10^{16} - 1.92 \times 10^{14} \]\[ [A]_{10} = 8.708 \times 10^{16} \text{ molecules/cm}^2 \]

Calculate the Fraction Remaining

The fraction of the film remaining after 10 s is the ratio of the surface coverage after 10 s to the initial surface coverage:\[ \text{Fraction remaining} = \frac{[A]_{10}}{[A]_0} \]\[ \text{Fraction remaining} = \frac{8.708 \times 10^{16}}{8.9 \times 10^{16}} \]

Key concepts

Chemical Kinetics

Chemical kinetics is the study of rates at which chemical processes occur and the factors that influence them. It includes investigations of how different conditions affect the speed of chemical reactions and the creation of models that can predict the behavior of chemical systems over time. Central to chemical kinetics is the idea that reactions proceed at a rate determined by the reactants' concentration, temperature, presence of catalysts, and other factors.

Understanding kinetics helps explain various natural phenomena and practical applications, such as food preservation, combustion engines, and the stability of medications. The rate of a reaction and how it varies with different conditions makes it possible to optimize processes, ensuring maximum desired yield while minimizing unwanted by-products.

Reaction Rate Constant

The reaction rate constant, denoted by the symbol 'k' in chemical equations, is a fundamental quantity in kinetics that characterizes the speed of a reaction. It is unique for each chemical reaction and depends on factors such as temperature and the physical state of the reactants. For zero order reactions, the rate constant has units of concentration divided by time, indicating the rate at which the reactants are consumed per unit time, regardless of their concentration.

Thus, the value of 'k' gives insight into how fast a reaction proceeds under specified conditions. In zero order kinetics, where the rate constant is given, it tells us that the reaction rate is fixed and does not change with varying reactant concentrations. Knowing 'k' allows chemists to calculate how long it will take for reactants to be used up or products to be formed.

Surface Coverage

Surface coverage refers to the proportion of a surface that is occupied by a particular species, such as atoms, molecules, or particles. In chemical reactions, especially those involving a solid surface, such as catalysts or adsorbents, the surface coverage is an essential factor influencing the reaction rate. High surface coverage can lead to faster reaction rates due to more reactant molecules being in close proximity to the reactive sites on the surface.

In the context of the evaporation of a film from a crystal surface, as in the given exercise, the initial and final surface coverage provides a measure of how much reactant remains on the surface at a given time. By tracking the surface coverage over time, we can determine the progress of the reaction and the remaining quantity of material. This concept is instrumental in fields such as catalysis, material science, and when dealing with reactions that involve phase changes.