Problem 41 - Chemistry An Atoms First Approach

Question

The decomposition of ethanol \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\right)\) on an alumina \(\left(\mathrm{Al}_{2} \mathrm{O}_{3}\right)\) surface$$\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(g) \longrightarrow \mathrm{C}_{2} \mathrm{H}_{4}(g)+\mathrm{H}_{2} \mathrm{O}(g)$$was studied at 600 K. Concentration versus time data were collected for this reaction, and a plot of \([\mathrm{A}]\) versus time resulted in a straight line with a slope of \(-4.00 \times 10^{-5} \mathrm{mol} / \mathrm{L} \cdot \mathrm{s}\) a. Determine the rate law, the integrated rate law, and the value of the rate constant for this reaction. b. If the initial concentration of \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\) was \(1.25 \times 10^{-2}\) \(M,\) calculate the half-life for this reaction. c. How much time is required for all the \(1.25 \times 10^{-2} M\) \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\) to decompose?

Step-by-Step Solution

Answer

The rate law for the decomposition of ethanol is Rate = k[\(C_{2}H_{5}OH]\), with k = \(4.00 \times 10^{-5} \frac{\mathrm{mol}}{\mathrm{L} \cdot \mathrm{s}}\). The integrated rate law is \(\ln [\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}] = \ln [\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}]_{0} - kt\). The half-life of the reaction is 17,325 s. To decompose completely, it requires 69,396 s.

Step 1: Write an expression for the rate of reaction

Since the dependence of the concentration of ethanol on time is given as a straight line, this means that the rate law is of the form: Rate = k[\(C_{2}H_{5}OH]^{n}\), where k is the rate constant and n is the order of the reaction.

Step 2: Determine the order of reaction

Since the plot of [A] versus time results in a straight line, the reaction is first order. So, n=1 and the rate law is: Rate = k[\(C_{2}H_{5}OH]\).

Step 3: Write the integrated rate law

As the reaction is a first-order reaction, the integrated rate law is: \(\ln [\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}] = \ln [\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}]_{0} - kt\), where [\(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\)] and [\(C_{2}H_{5}OH]_{0}\) are the concentrations of ethanol at time t and initial time respectively.

Step 4: Calculate the rate constant

The slope of the plot of [\(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\)] vs time is -k. Therefore, k = \(4.00 \times 10^{-5} \frac{\mathrm{mol}}{\mathrm{L} \cdot \mathrm{s}}\). b) Find the half-life of the reaction

Step 1: Write the expression for half-life of a first-order reaction

The half-life (t₁/₂) of a first-order reaction is given by: t₁/₂ = \(\frac{0.693}{k}\).

Step 2: Calculate the half-life

Using the rate constant obtained in part a, we can find the half-life. t₁/₂ = \(\frac{0.693}{4.00 \times 10^{-5} \frac{\mathrm{mol}}{\mathrm{L} \cdot \mathrm{s}}}\) = 17325 s. c) Find the time required for all the \(1.25 \times 10^{-2} \mathrm{M}\) ethanol to decompose

Step 1: Use the integrated rate law

We will use the integrated rate law obtained in part a. \(\ln [\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}] = \ln [\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}]_{0} - kt\). Here, [\(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\)] = 0 (as all ethanol has decomposed), and [\(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\)]₀ = \(1.25 \times 10^{-2}\) M.

Step 2: Calculate the time required for complete decomposition

Now, we can solve for time t: \(\ln (0) = \ln (1.25 \times 10^{-2}\mathrm{M}) - 4.00 \times 10^{-5} \frac{\mathrm{mol}}{\mathrm{L} \cdot \mathrm{s}} \cdot t\). Divide both sides by -4.00 \(\times 10^{-5} \frac{\mathrm{mol}}{\mathrm{L} \cdot \mathrm{s}}\): t = \(\frac{\ln (1.25 \times 10^{-2}\mathrm{M})}{4.00 \times 10^{-5} \frac{\mathrm{mol}}{\mathrm{L} \cdot \mathrm{s}}}\) = 69396 s.

Key Concepts

Reaction Rate

Reaction rate is a crucial concept in chemical kinetics, representing how fast or slow a reaction proceeds. It quantifies the change in concentration of reactants or products over time. Reaction rates can provide insights into how a reaction progresses and the conditions affecting this progress.
In this exercise, the decomposition of ethanol on an alumina surface was examined, and it was determined by tracking concentration versus time data. This data showed a straight line, indicating a zero-order or linear reaction rate in which the concentration of reactant decreases uniformly over time.
This linearity in the plot suggests the simplistic nature of the reaction path, primarily predictable under a fixed set of conditions. Understanding how to determine and interpret reaction rates can significantly enhance the predictive power related to reaction conditions and outcomes.

First-Order Reactions

First-order reactions are characterized by a direct proportionality between the rate and the concentration of a single reactant. For the oxidation of ethanol, it was determined to be a first-order reaction due to the linear decrease when plotted on a concentration versus time graph.
The rate law for a first-order reaction is expressed as:

• Rate = k[\(C_{2}H_{5}OH\)]
• where \( k \) is the rate constant.

This indicates that the reaction proceeds at a rate directly influenced by the concentration of ethanol. Such reactions typically exhibit exponential decay as the reaction progresses unless external conditions are altered.
This concept is vital because it informs how adjustments in concentration will alter reaction speed, aiding the optimization of reaction conditions.

Rate Constant

The rate constant \( (k) \) is a specific rate expression constant for a given reaction at a particular temperature. It provides vital information about the reaction rate without requiring knowledge of the concentrations involved.
For the reaction in focus, the rate constant was deduced as \(4.00 imes 10^{-5} \, \text{mol/L} \cdot \text{s}\) from the slope of the straight-line graph of concentration against time. The negative sign indicates a decrease in concentration over time; however, the value itself provides a comparative measure of speed across similar reactions.
Understanding \( k \) is essential because it defines each reaction's pace under given conditions and aids in computational simulations where predictions about reaction behavior are necessary.

Half-Life

The concept of half-life is particularly useful in chemical kinetics as a predictable measure of the time required for half of a reactant to undergo reaction. For a first-order reaction, half-life remains constant and does not depend on the initial concentration.
The half-life equation for a first-order reaction is:

• \( t_{1/2} = \frac{0.693}{k} \)

In the exercise, the half-life was calculated to be approximately 17325 seconds using the given rate constant. This consistency allows chemists to quickly ascertain how long a reaction will take to reach a certain point under consistent conditions.
Overall, half-life is a crucial tool for predicting the duration required for significant changes in reactant concentration, impacting areas like pharmaceutical stability and chemical process design.