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Ch.14 - Chemical Kinetics
Chapter 14, Problem 98

What fraction of the film is left after 10 s, assuming the same initial coverage as in part a, given that 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 * 10^13 molecules/cm^2 * s at 120 K?

Verified step by step guidance
1
Identify that the problem involves zero-order kinetics, where the rate of reaction is constant and does not depend on the concentration of the reactant.
Use the zero-order rate equation: \[ [A] = [A]_0 - kt \] where \([A]\) is the concentration at time \(t\), \([A]_0\) is the initial concentration, \(k\) is the rate constant, and \(t\) is the time.
Determine the initial concentration \([A]_0\) in terms of molecules/cm². Since the problem states the initial coverage is the same as in part a, use that initial concentration value.
Substitute the given rate constant \(k = 1.92 \times 10^{13}\) molecules/cm²·s and the time \(t = 10\) s into the zero-order rate equation.
Calculate the concentration \([A]\) at \(t = 10\) s and then find the fraction of the film remaining by dividing \([A]\) by \([A]_0\).

Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Zero-Order Kinetics

Zero-order kinetics refers to a reaction rate that is independent of the concentration of the reactant. In this case, the rate of evaporation of the n-pentane film remains constant over time, meaning that the amount of film evaporated per unit time does not change as the film thickness decreases. This concept is crucial for calculating the remaining fraction of the film after a specific time.
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Zero-Order Reactions

Rate Constant

The rate constant is a proportionality factor in the rate equation that quantifies the speed of a chemical reaction. For zero-order reactions, the rate constant (in this case, 1.92 * 10^13 molecules/cm^2 * s) indicates how many molecules evaporate per square centimeter per second, providing a basis for determining how much of the film will remain after a given time period.
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Rate Constant Units

Fraction Remaining

The fraction remaining of a substance after a certain time can be calculated using the initial amount and the amount that has evaporated. In this scenario, knowing the rate of evaporation and the time elapsed allows for the calculation of how much of the 120-nm film of n-pentane is left after 10 seconds, which is essential for answering the question.
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Mole Fraction Formula
Related Practice
Open 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 s−1 at 150 K. If the surface is initially completely covered, what fraction will remain covered after 10 s? After 20 s?
Textbook 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/s at 150 K. a. What is the half-life of the desorption reaction?-

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Textbook 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/s at 150 K. b. If the surface is initially completely covered with n-butane at 150 K, how long will it take for 25% of the molecules to desorb (leave the surface)? For 50% to desorb?

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Textbook 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⨉1013 molecules/cm2•s at 120 K. a. If the initial surface coverage is 8.9⨉1016 molecules/cm2, how long will it take for one-half of the film to evaporate?

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Textbook Question

The kinetics of this reaction were studied as a function of temperature. (The reaction is first order in each reactant and second order overall.)

C2H5Br(aq) + OH- (aq) → C2H5OH(l) + Br- (aq)

Temperature (°C) k (L,mol •s)

25 8.81⨉10-5

35 0.000285

45 0.000854

55 0.00239

65 0.00633

a. Determine the activation energy and frequency factor for the reaction.

Textbook Question

The kinetics of this reaction were studied as a function of temperature. (The reaction is first order in each reactant and second order overall.)

C2H5Br(aq) + OH- (aq) → C2H5OH(l) + Br- (aq)

Temperature (°C) k (L,mol •s)

25 8.81⨉10-5

35 0.000285

45 0.000854

55 0.00239

65 0.00633

b. Determine the rate constant at 15 °C.

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