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 \(\mathrm{K}\) . Concentration versus time data were collected for this reaction, and a plot of [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 \(1.25 \times 10^{-2} \mathrm{M}\) \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\) to decompose?

Short answer

The rate law for the given reaction is \(r = k[C_{2}H_{5}OH]\), and the integrated rate law is \(\ln{\dfrac{[C_{2}H_{5}OH]_t}{[C_{2}H_{5}OH]_0}} = -kt\). The rate constant (k) is \(4.00 \times 10^{-5} \mathrm{L /mol \cdot s}\). The half-life of the reaction is \(1.73 \times 10^{4}s\). The time required for the complete decomposition of ethanol is \(9.21 \times 10^4 s\).

Step-by-step solution

Find the Rate Law and the Integrated Rate Law

The given reaction: \[C_{2}H_{5}OH(g) \longrightarrow C_{2}H_{4}(g)+H_{2}O(g)\] The concentration vs. time graph is a straight line with a slope, which means the reaction is a first-order reaction. Thus, the rate law is: Rate Law: \[r = k[C_{2}H_{5}OH]\] Now, let's find the integrated rate law. For a first-order reaction, the integrated rate law is: Integrated Rate Law: \[\ln{\dfrac{[C_{2}H_{5}OH]_t}{[C_{2}H_{5}OH]_0}} = -kt\]

Determine the Value of the Rate Constant

We are given that the slope of the concentration vs. time graph is \(-4.00 \times 10^{-5} \mathrm{mol} / \mathrm{L} \cdot \mathrm{s}\) . Since the reaction is first-order, the rate constant (k) is equal to the negative slope. Rate Constant: \[k = -(-4.00 \times 10^{-5} \mathrm{mol} / \mathrm{L} \cdot \mathrm{s}) = 4.00 \times 10^{-5} \mathrm{L /mol \cdot s}\]

Calculate the Half-Life

To calculate the half-life of a first-order reaction, we use the formula: Half-Life: \[t_{1/2} = \dfrac{\ln 2}{k}\] Using the rate constant obtained previously, we have: \[t_{1/2} = \dfrac{\ln 2}{4.00 \times 10^{-5} \mathrm{L/mol\cdot s}}\] Calculating the half-life: \[t_{1/2} = 1.73 \times 10^{4}s\]

Determine the Time Required for Complete Decomposition

Given the initial concentration of ethanol \([C_{2}H_{5}OH]_0 = 1.25 \times 10^{-2} M\) and the integrated rate law earlier, we want to find the time when the concentration \([C_{2}H_{5}OH]_t = 0\) . Plugging in the values into the integrated rate law, we have: \[\ln{\dfrac{0}{1.25 \times 10^{-2} M}} = -kt\] However, the above equation has a zero in the numerator, and the natural logarithm of zero is undefined. In this context, complete decomposition implies that the concentration of ethanol is close enough to zero that it is practically negligible. So we can compute the time required for, say, 99.99% of \(C_{2}H_{5}OH\) to decompose: \[\ln{\dfrac{(1 - 0.9999) \times 1.25 \times 10^{-2} M}{1.25 \times 10^{-2} M}} = -kt\] \[t = \dfrac{\ln{0.0001}}{-4.00 \times 10^{-5} \mathrm{L/mol\cdot s}}\] Calculating the time: \[t = 9.21 \times 10^4 s\] Thus, the time required for the concentration of ethanol to reach a negligible level is \(9.21 \times 10^4 s\).

Key concepts

First-Order Reaction

In chemistry, reactions are classified based on their order, which reflects how the rate depends on the concentration of reactants. A first-order reaction is one where the reaction rate is directly proportional to the concentration of one reactant. This means if the concentration of the reactant doubles, the rate of the reaction doubles as well.

First-order reactions often involve processes where a single reactant breaks down or forms products over time. A characteristic feature of first-order reactions is that, when the concentration vs. time data is plotted, it creates a straight line, which indicates a constant rate of decomposition. Understanding this concept helps in predicting how long a reaction will take under given conditions.

Rate Law

The rate law of a reaction is an expression that shows the relationship between the concentration of reactants and the rate of the reaction. For first-order reactions, the rate law can be written as:

• \[ r = k[A] \]

Here, \( r \) is the rate of the reaction, \( k \) is the rate constant, and \([A]\) is the concentration of the reactant.

• The rate constant \( k \) is a crucial part of the rate law, as it represents how fast the reaction proceeds.
• For first-order reactions, the rate is directly proportional to the concentration of a single reactant.

Understanding the rate law helps chemists to control and manipulate reactions in industrial and laboratory settings, ensuring that reactions occur at desired rates.

Integrated Rate Law

The integrated rate law is especially useful for calculating how concentrations change over time for a particular reaction order. For first-order reactions, the integrated rate law is expressed as:

• \[ \ln{\left(\dfrac{[A]_t}{[A]_0}\right)} = -kt \]

In this equation,

• \([A]_t\) is the concentration of the reactant at time \( t \).
• \([A]_0\) is the initial concentration of the reactant.
• The term \( -kt \) helps in determining the rate of decrease in the concentration over time.

By rearranging this formula, you can predict either the concentration at a certain time or the time required to reach a particular concentration of reactant. It is particularly useful because it provides a clear mathematical relationship that allows for the prediction of reaction progress over time.

Rate Constant

The rate constant, denoted as \( k \), is a pivotal factor in reaction kinetics, providing a quantifiable measure of reaction speed. For first-order reactions, the rate constant is directly derived from the slope of the concentration versus time curve.

• In our example, the slope was given as \(-4.00 \times 10^{-5} \text{ mol/L}\cdot\text{s}\), resulting in a rate constant of \(4.00 \times 10^{-5} \text{ L/mol}\cdot\text{s}\).
• This positive value of \( k \) indicates how rapidly the ethanol decomposes on the alumina surface.

The rate constant incorporates the effects of temperature and the inherent characteristics of the reactants and catalysts involved in the reaction. It is a crucial parameter that allows chemists to manipulate and optimize reactions for various applications.

Half-Life Calculation

The half-life of a reaction is the time it takes for the concentration of a reactant to fall to half its initial value. For first-order reactions, the half-life is independent of the initial concentration and is given by the formula:

• \[ t_{1/2} = \dfrac{\ln 2}{k} \]

This indicates that once you know the rate constant \( k \), you can easily determine the half-life of the reaction.

• In the ethanol decomposition example, the half-life was calculated as \(1.73 \times 10^{4} \text{ s}\).
• It provides a convenient measure of reaction speed over time, as each half-life reduces the concentration significantly, iterating through a simple logarithmic scale.

Understanding half-life calculations is especially vital in fields such as pharmacokinetics, radioactive decay studies, and many more, where determining the time scale of processes is critical.