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

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

Step by step solution

01

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.
02

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.
03

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.
04

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\[\ln\left(\frac{0.120}{[\mathrm{CH}_{3} \mathrm{CHO}]}\right) = (1.5 \times 10^{-3} \mathrm{s}^{-1})(20 \times 60 \mathrm{s})\]
05

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, \[ [\mathrm{CH}_{3} \mathrm{CHO}] = \frac{0.120}{e^{1.8}}\].
06

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.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

For the reversible, one-step reaction \(\mathrm{A}+\mathrm{B} \rightleftarrows \mathrm{C}+\mathrm{D}\), the forward rate constant is \(52.4 \mathrm{~L} \cdot \mathrm{mol}^{-1} \cdot \mathrm{h}^{-1}\) and the rate constant for the reverse reaction is \(32.1 \mathrm{~L} \cdot \mathrm{mol}^{-1} \cdot \mathrm{h}^{-1}\). The activation energy was found to be \(35.2 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}\) for the forward reaction and \(44.0 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}\) for the reverse reaction. (a) What is the equilibrium constant for the reaction? (b) Is the reaction exothermic or endothermic? (c) What will be the effect of raising the temperature on the rate constants and the equilibrium constant?

Express the units for rate constants when the concentrations are in moles per liter and time is in seconds for (a) zero-order reactions; (b) first-order reactions; (c) second-order reactions.

The pre-equilibrium and the steady-state approximations are two different approaches to deriving a rate law from a proposed mechanism. For the following mechanism, determine the rate law (a) by the pre-equilibrium approximation and (b) by the steady-state approximation. (c) Under what conditions do the two methods give the same answer? (d) What will the rate laws become at high concentrations of \(\mathrm{Br}^{-}\)? $$ \begin{aligned} &\mathrm{CH}_{3} \mathrm{OH}+\mathrm{H}^{+} \rightleftharpoons \mathrm{CH}_{3} \mathrm{OH}_{2}^{+} \text {(fast equilibrium) } \\ &\mathrm{CH}_{3} \mathrm{OH}_{2}^{+}+\mathrm{Br}^{-} \longrightarrow \mathrm{CH}_{3} \mathrm{Br}+\mathrm{H}_{2} \mathrm{O} \text { (slow) } \end{aligned} $$

Determine whether each of the following statements is true or false. If a statement is false, explain why. (a) The equilibrium constant for a reaction equals the rate constant for the forward reaction divided by the rate constant for the reverse reaction. (b) In a reaction that is a series of equilibrium steps, the overall equilibrium constant is equal to the product of all the forward rate constants divided by the product of all the reverse rate constants. (c) Increasing the concentration of a product increases the rate of the reverse reaction, and so the rate of the forward reaction must then increase, too.

Write the overall reaction for the mechanism proposed below and identify any reaction intermediates. Step \(1 \mathrm{C}_{4} \mathrm{H}_{9} \mathrm{Br} \longrightarrow \mathrm{C}_{4} \mathrm{H}_{9}{ }^{+}+\mathrm{Br}^{-}\) Step \(2 \mathrm{C}_{4} \mathrm{H}_{9}{ }^{+}+\mathrm{H}_{2} \mathrm{O} \longrightarrow \mathrm{C}_{4} \mathrm{H}_{9} \mathrm{OH}_{2}{ }^{+}\) Step \(3 \mathrm{C}_{4} \mathrm{H}_{9} \mathrm{OH}_{2}^{+}+\mathrm{H}_{2} \mathrm{O} \longrightarrow \mathrm{C}_{4} \mathrm{H}_{9} \mathrm{OH}+\mathrm{H}_{3} \mathrm{O}^{+}\)

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free