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

Expert verified
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

01

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.
02

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 \( t \).\[ t = \frac{[A]_0 - [A]_t}{k} \]\[ t = \frac{8.9 \times 10^{16} - 4.45 \times 10^{16}}{1.92 \times 10^{13}} \]
03

Solve for Time \( t \)

Plugging in the values:\[ t = \frac{8.9 \times 10^{16} - 4.45 \times 10^{16}}{1.92 \times 10^{13}} \]\[ t = \frac{4.45 \times 10^{16}}{1.92 \times 10^{13}} \]\[ t = 2.3177 \times 10^3 \text{ seconds} \]
04

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 \( [A]_{10} \):\[ [A]_{10} = 8.9 \times 10^{16} - (1.92 \times 10^{13})(10) \]
05

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

Calculate the surface coverage after 10 seconds:\[ [A]_{10} = 8.9 \times 10^{16} - 1.92 \times 10^{14} \]\[ [A]_{10} = 8.708 \times 10^{16} \text{ molecules/cm}^2 \]
06

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:\[ \text{Fraction remaining} = \frac{[A]_{10}}{[A]_0} \]\[ \text{Fraction remaining} = \frac{8.708 \times 10^{16}}{8.9 \times 10^{16}} \]

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The rate constant \((k)\) for a reaction was measured as a function of temperature. A plot of In \(k\) versus \(1 / T\) (in \(\mathrm{K}\) ) is linear and has a slope of -7445 K. Calculate the activation energy for the reaction.

A reaction in which \(A, B,\) and \(C\) react to form products is zero order in A, one-half order in B, and second order in C. a. Write a rate law for the reaction. b. What is the overall order of the reaction? c. By what factor does the reaction rate change if [A] is doubled (and the other reactant concentrations are held constant)? d. By what factor does the reaction rate change if [B] is doubled (and the other reactant concentrations are held constant)? e. By what factor does the reaction rate change if [C] is doubled (and the other reactant concentrations are held constant)? f. By what factor does the reaction rate change if the concentrations of all three reactants are doubled?

A chemical reaction is endothermic and has an activation energy that is twice the value of the enthalpy change of the reaction. Draw a diagram depicting the energy of the reaction as it progresses. Label the position of the reactants and products and indicate the activation energy and enthalpy of reaction.

Consider a simple reaction in which reactant A forms products: \(\mathrm{A} \longrightarrow\) products What is the rate law if the reaction is zero order with respect to A? First order? Second order? For each case, explain how a doubling of the concentration of A would affect the rate of reaction.

A certain substance X decomposes. Fifty percent of X remains after 100 minutes. How much \(X\) remains after 200 minutes if the reaction order with respect to \(X\) is (a) zero order, (b) first order, (c) second order?

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free