Question

In the brewing of beer, ethanal, which smells like green apples, is an intermediate in the formation of ethanol. Ethanal decomposes in the following first-order reaction: \(\mathrm{CH}_{3} \mathrm{CHO}(\mathrm{g}) \longrightarrow \mathrm{CH}_{4}(\mathrm{~g})+\) \(\mathrm{CO}(\mathrm{g})\). At an elevated temperature the rate constant for the decomposition is \(1.5 \times 10^{-3} \mathrm{~s}^{-1}\). What concentration of ethanal, which had an initial concentration of \(0.120 \mathrm{~mol} \cdot \mathrm{L}^{-1}\), remains \(20.0 \mathrm{~min}\) after the start of its decomposition at this temperature?

Short answer

After 20 minutes, the concentration of ethanal remaining is approximately 0.020 mol L^{-1}.

Step-by-step solution

Understand the Type of Reaction

This is a first-order reaction where ethanal decomposes into methane and carbon monoxide. A first-order reaction has a rate proportional to the concentration of one reactant.

Write the First-Order Rate Law

For a first-order reaction, the rate law is given by the equation \[\text{Rate} = k[\mathrm{CH}_{3} \mathrm{CHO}]\] where \(k\) is the rate constant and \([\mathrm{CH}_{3} \mathrm{CHO}]\) is the concentration of ethanal.

Apply the Integrated Rate Law for First-Order Reactions

The integrated rate law for a first-order reaction is \[\ln\left(\frac{[\mathrm{A}]_0}{[\mathrm{A}]}\right) = kt\] where \([\mathrm{A}]_0\) and \([\mathrm{A}]\) are the initial and remaining concentrations of the reactant, \(t\) is time, and \(k\) is the rate constant.

Calculate the Remaining Concentration of Ethanal

Here, \(k = 1.5 \times 10^{-3} \mathrm{s}^{-1}\), \(t = 20 \text{ min} \times 60 \mathrm{s/min}\), and \([\mathrm{CH}_{3} \mathrm{CHO}]_0 = 0.120 \mathrm{mol} \cdot \mathrm{L}^{-1}\). Plugging these values into the integrated rate law gives\[\ln\left(\frac{0.120}{[\mathrm{CH}_{3} \mathrm{CHO}]}\right) = (1.5 \times 10^{-3} \mathrm{s}^{-1})(20 \times 60 \mathrm{s})\]

Simplify and Solve for \([\mathrm{CH}_{3} \mathrm{CHO}]\)

First calculate the right side of the equation: \[1.5 \times 10^{-3} \mathrm{s}^{-1} \times 20 \times 60 = 1.8\] Then, simplify the equation to solve for the remaining concentration: \[\ln\left(\frac{0.120}{[\mathrm{CH}_{3} \mathrm{CHO}]}\right) = 1.8\] Using the properties of logarithms: \[\frac{0.120}{[\mathrm{CH}_{3} \mathrm{CHO}]} = e^{1.8}\] Thus, \[ [\mathrm{CH}_{3} \mathrm{CHO}] = \frac{0.120}{e^{1.8}}\].

Calculate the Numerical Value

Using a calculator to find \(e^{1.8}\) and dividing 0.120 by this value gives the concentration of ethanal remaining after 20 minutes.

Key concepts

Rate Law

Understanding the rate law is crucial in studying the speed at which chemical reactions occur. In the context of a first-order reaction, such as the decomposition of ethanal in beer brewing, the rate law expresses the relationship between the reaction rate and the concentration of the reactant involved.

The general form of the rate law for a first-order reaction can be written as:\[\begin{equation}\text{Rate} = k[\text{A}]\end{equation}\]where \(k\) is the reaction rate constant and \([\text{A}]\) is the concentration of the reactant. For our specific example, it means that the rate at which ethanal decomposes is directly proportional to its concentration at any given moment.

It's important to note that first-order rate laws are common in processes where the rate of reaction depends only on the concentration of a single reactant. In more complex reactions involving multiple reactants, the rate law could have a different form, with orders possibly being fractional or greater than first.

Integrated Rate Law

Moving beyond the immediate rate of reaction, the integrated rate law plays a pivotal role in predicting how reactant concentrations change over time. Specifically, for a first-order reaction, the integrated rate law links the initial concentration of a reactant to its concentration at a later time.

The integrated rate law for a first-order reaction is expressed as:\[\begin{equation}\ln\left(\frac{[\text{A}]_0}{[\text{A}]}\right) = kt\end{equation}\]where \([\text{A}]_0\) is the initial concentration, \([\text{A}]\) is the concentration at time \(t\), and \(k\) is the rate constant. This equation is derived by integrating the rate law over time and allows us to calculate how much reactant remains after a certain period.

In practical terms, if we know the rate constant and the initial concentration of ethanal, we can use the integrated rate law to find out how much ethanal will remain at any subsequent point in time. The natural logarithm in the equation provides an exponential relationship between concentration and time, reflecting the nature of first-order kinetics.

Reaction Rate Constant

The reaction rate constant \(k\) is a fundamental parameter in the study of kinetics, symbolizing the intrinsic speed at which a chemical reaction proceeds. It is unique for each chemical reaction and is influenced by various factors such as temperature, pressure, and the presence of a catalyst.

In the example of ethanal's decomposition, the provided rate constant is \(1.5 \times 10^{-3} \mathrm{s}^{-1}\), indicating the speed of reaction under the given conditions at an elevated temperature. Understanding the value of \(k\) is essential because it can be plugged into the rate law or the integrated rate law to find out the rate of the reaction at a particular concentration, or how that concentration changes over time, respectively.

It is also worth noting that the units of the rate constant vary depending on the order of the reaction. For first-order reactions, the rate constant has units of inverse time, such as \(\text{s}^{-1}\), because it represents the probability of a single reactant molecule reacting per unit of time.