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/cm 2 •s at 120 K. a. If the initial surface coverage is 8.9⨉1016 molecules/cm 2 , how long will it take for one-half of the film to evaporate?

hey everyone in this example we have the decomposition of ammonia over hot tungsten as a zero order reaction, we're told the rate constant by this value here And we need to or we're also told the initial amount of Ammonia as .00475 moller. So we need to calculate the time when the concentration of ammonia is half of its original amount. So we should recall that the half life of a reaction for a zeroth order reaction according to the prompt is going to equal the concentration of our original sample initially Divided by two times the rate constant K. So plugging in what we know from the prompt, we would have that our half life is equal to our initial initial concentration of our sample of ammonia. From the prompt in the numerator would be 0. Molar. And we want to recall that polarity can be interpreted as moles per leader so we can go ahead and write this in as units of moles per leader. Then in our denominator we're going to plug in two times our rate constant, which from the prompt is given as 1.31 times 10 to the negative six power moles per leaders time seconds. And so getting rid of our units. We can cancel out leaders as well as moles and we're left with seconds for our half life and we're going to get a value equal to 1812.98. And we're left with units of seconds. However, we can go ahead and write this in scientific notation so that we get our half life equal to 1.81 times 10 to the third power seconds. And so for our final answer, this is the value that we get for the time when the concentration of ammonia is half of its original amount. So this would be the half life of our ammonia. So I hope that everything I explained was clear. If you have any questions, please leave them down below and I will see everyone in the next practice video.

Ch.14 - Chemical Kinetics

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Ch.14 - Chemical Kinetics

Problem 98a

Chapter 14, Problem 98a

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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Identify that the reaction is zero order, which means 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, and \(k\) is the rate constant.

Substitute the given values into the equation: \([A]_0 = 8.9 \times 10^{16}\) molecules/cm\(^2\), \(k = 1.92 \times 10^{13}\) molecules/cm\(^2\)•s.

Calculate the concentration when half of the film has evaporated: \([A] = \frac{1}{2} \times 8.9 \times 10^{16}\) molecules/cm\(^2\).

Rearrange the zero-order rate equation to solve for \(t\): \[ t = \frac{[A]_0 - [A]}{k} \] and substitute the known values to find the time.

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Key Concepts

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

Zero-Order Kinetics

In zero-order kinetics, the rate of a reaction is constant and independent of the concentration of the reactants. This means that the rate of evaporation of the n-pentane film remains the same regardless of how much of the film is left. The rate constant, in this case, is given as 1.92×10^13 molecules/cm²•s, which allows us to calculate the time required for a specific change in concentration.

Half-Life in Zero-Order Reactions

The half-life of a zero-order reaction is determined by the initial concentration and the rate constant. It can be calculated using the formula t½ = [A]₀ / (2k), where [A]₀ is the initial concentration and k is the rate constant. This concept is crucial for determining how long it will take for half of the n-pentane film to evaporate, given the initial surface coverage.

Surface Coverage

Surface coverage refers to the number of molecules present on a surface per unit area, expressed in molecules/cm². In this scenario, the initial surface coverage of the n-pentane film is 8.9×10^16 molecules/cm². Understanding surface coverage is essential for calculating the time it takes for the film to evaporate, as it directly influences the rate of the zero-order 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?

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.

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

c. If a reaction mixture is 0.155 M in C₂H₅Brand 0.250 M in OH^-, what is the initial rate of the reaction at 75 °C?

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?

Verified video answer for a similar problem:

Key Concepts

Zero-Order Kinetics

Half-Life in Zero-Order Reactions

Surface Coverage

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?