Write the reaction quotients \(Q_{c}\) for (a) \(\mathrm{Cu}(\mathrm{s})+\mathrm{Cl}_{2}(\mathrm{~g}) \rightarrow \mathrm{CuCl}_{2}(\mathrm{~s})\) (b) \(\mathrm{NH}_{4} \mathrm{NO}_{2}\) (s) \(\rightarrow \mathrm{N}_{2} \mathrm{O}\) (g) \(+2 \mathrm{H}_{2} \mathrm{O}\) (g) (c) \(\mathrm{MgCO}_{3}(\mathrm{~s}) \rightarrow \mathrm{MgO}\) (s) \(+\mathrm{CO}_{2}\) (g)

Short Answer

Expert verified
For (a) \(Q_c = 1\) as there are no gases or aqueous species in the product; for (b) \(Q_c = [\mathrm{N}_{2}\mathrm{O}][\mathrm{H}_{2}\mathrm{O}]^2\); and for (c) \(Q_c = [\mathrm{CO}_{2}]\).

Step by step solution

01

Understanding Reaction Quotients

The reaction quotient, Qc, is the ratio of the concentrations of products raised to their stoichiometric coefficients to the concentrations of reactants raised to their stoichiometric coefficients, for a reaction at any point in time. It is important to note that solids and pure liquids do not appear in the expression for Qc as their concentrations do not change.
02

Writing Qc for Reaction (a)

For the reaction \(\mathrm{Cu}(\mathrm{s})+\mathrm{Cl}_{2}(\mathrm{~g}) \rightarrow \mathrm{CuCl}_{2}(\mathrm{~s})\), there are no gaseous or aqueous species on the product side of the reaction. Since both copper and copper(II) chloride are solid, they do not appear in the reaction quotient. The reaction quotient would involve only \(\mathrm{Cl}_2\), but since it is a reactant, Qc equals 1.
03

Writing Qc for Reaction (b)

For the reaction \(\mathrm{NH}_{4}\mathrm{NO}_{2}\) (s) \(\rightarrow \mathrm{N}_{2}\mathrm{O}\) (g) \(+ 2\mathrm{H}_{2}\mathrm{O}\) (g)\, the solid reactant does not appear in the reaction quotient. Qc is the concentration of \(\mathrm{N}_2\mathrm{O}\) times the concentration of \(\mathrm{H}_2\mathrm{O}\) squared.
04

Writing Qc for Reaction (c)

For the reaction \(\mathrm{MgCO}_{3}(\mathrm{~s}) \rightarrow \mathrm{MgO}\) (s) \(+\mathrm{CO}_{2}\) (g)\, both \(\mathrm{MgCO}_3\) and \(\mathrm{MgO}\) are solids and therefore do not appear in Qc. The reaction quotient will be equal to the concentration of \(\mathrm{CO}_2\) in the gaseous phase.

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.

Chemical Equilibrium
Chemical equilibrium is a state of a chemical reaction at which the concentrations of reactants and products remain constant over time, indicating a balance between the forward and reverse reactions. This does not mean the reactions have ceased, but rather that their rates are equal, effectively making the observable properties of the system constant. At equilibrium, the reaction quotient \(Q_c\) is equal to the equilibrium constant \(K_c\), an important figure that can be used to predict the direction of the reaction. It is crucial to assess how the system might respond to changes in concentration, pressure, or temperature, applying Le Chatelier’s principle.

For instance, if an equilibrium system is exposed to a change, say an increase in concentration of a product, the system will adjust to counteract that change, often by increasing the rate of the reverse reaction. This adjustment works to re-establish equilibrium conditions, hence maintaining the balance of reactant and product concentrations.
Stoichiometric Coefficients
Stoichiometric coefficients in a balanced chemical reaction indicate the relative quantities of each substance involved. They play a pivotal role in determining the proportions of reactants consumed and products formed during a reaction. In the context of the reaction quotient, \(Q_c\), stoichiometric coefficients determine the exponents to which the concentrations of substances are raised in the quotient's calculation.

For example, in the provided solution for reaction (b), the balanced equation \(\mathrm{NH}_{4}\mathrm{NO}_{2}\) (s) \(\rightarrow\) \(\mathrm{N}_{2}\mathrm{O}\) (g) \(+2 \mathrm{H}_{2}\mathrm{O}\) (g) shows a stoichiometric coefficient of 1 for \(\mathrm{N}_{2}\mathrm{O}\) and 2 for \(\mathrm{H}_{2}\mathrm{O}\), giving us a \(Q_c\) that considers the concentration of \(\mathrm{N}_{2}\mathrm{O}\) to the first power and \(\mathrm{H}_{2}\mathrm{O}\) to the second power. Understanding these coefficients is essential for properly writing out the reaction quotient expression and predicting the extent of a reaction.
Concentration of Reactants and Products
The concentration of reactants and products is a measure of the amount of a substance in a given volume of solution. It's typically expressed in molarity (moles per liter). In chemical kinetics and equilibrium, the concentrations of the reactants and products at any point in time can predict the direction and extent of the reaction. In the reaction quotient \(Q_c\), the concentrations of gaseous and aqueous species are used to gauge how far a system is from reaching equilibrium.

It’s essential to recognize that in the \(Q_c\) expression, the concentration of pure solids and liquids, as shown in the reactions (a) and (c), are omitted because their concentrations remain constant. Gaseous substance concentrations, as in reaction (c) with \(\mathrm{CO}_{2}\) gas, are included in the \(Q_c\) expression. Calculating accurate concentrations of the reactive species involved is fundamental to predicting the behavior of a chemical system under various conditions and is one of the foundational concepts in the study of chemical equilibria.

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

\( \mathrm{~A} 3.00-\mathrm{L}\) reaction vessel is filled with \(0.150 \mathrm{~mol}\) \(\mathrm{CO}, 0.0900 \mathrm{~mol} \mathrm{H}_{2}\), and \(0.180 \mathrm{~mol} \mathrm{CH}_{3} \mathrm{OH}\). Equilibrium is reached in the presence of a zinc oxidechromium(III) oxide catalyst; and at \(300^{\circ} \mathrm{C}, K_{c}=\) \(1.1 \times 10^{-2}\) for the reaction, \(\mathrm{CO}(\mathrm{g})+2 \mathrm{H}_{2}(\mathrm{~g}) \rightleftharpoons\) \(\mathrm{CH}_{3} \mathrm{OH}(\mathrm{g})\). (a) As the reaction approaches equilibrium, will the molar concentration of \(\mathrm{CH}_{3} \mathrm{OH}\) increase, decrease, or remain unchanged? (b) What is the equilibrium composition of the mixture?

Given that \(K_{c}=61\) for the reaction \(\mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NH}_{3}(\mathrm{~g})\) at \(500 \mathrm{~K}\), calculate whether more ammonia will tend to form when a mixture of composition \(2.23 \times 10^{-3} \mathrm{~mol} \cdot \mathrm{L}^{-1} \mathrm{~N}_{2}\), \(1.24 \times 10^{-3} \mathrm{~mol} \cdot \mathrm{L}^{-1} \mathrm{H}_{2}\), and \(1.12 \times 10^{-4} \mathrm{~mol} \cdot \mathrm{L}^{-1}\) \(\mathrm{NH}_{3}\) is present in a container at \(500 \mathrm{~K}\).

At \(500^{\circ} \mathrm{C}, K_{c}=0.061\) for \(\mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g}) \rightleftharpoons\) \(2 \mathrm{NH}_{3}(\mathrm{~g})\). If analysis shows that the composition is \(3.00 \mathrm{~mol} \cdot \mathrm{L}^{-1} \mathrm{~N}_{2}, 2.00 \mathrm{~mol} \cdot \mathrm{L}^{-1} \mathrm{H}_{2}\) and \(0.500 \mathrm{~mol} \cdot \mathrm{L}^{-1}\) \(\mathrm{NH}_{3}\), is the reaction at equilibrium? If not, in which direction does the reaction tend to proceed to reach equilibrium?

Determine \(K_{c}\) from the value of \(K\) for cach of the following equilibria. (a) \(2 \mathrm{SO}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g})=2 \mathrm{SO}_{3}(\mathrm{~g}), K=3.4\) at \(1000 \mathrm{~K}\) (b) \(\mathrm{NH}_{4} \mathrm{HS}(\mathrm{s})=\mathrm{NH}_{3}(\mathrm{~g})+\mathrm{H}_{2} \mathrm{~S}(\mathrm{~g}), K=9.4 \times 10^{-2}\) at \(24^{\circ} \mathrm{C}\)

The equilibrium constant \(K=3.5 \times 10^{4}\) for the reaction \(\mathrm{PCl}_{3}(\mathrm{~g})+\mathrm{Cl}_{2}(\mathrm{~g})=\mathrm{PCl}_{5}(\mathrm{~g})\) at \(760^{\circ} \mathrm{C}\). At equilibrium, the partial pressure of \(\mathrm{PCl}_{5}\) was \(2.2 \times\) \(10^{4}\) bar and that of \(\mathrm{PCl}_{3}\) was \(1.33\) bar. What was the equilibrium partial pressure of \(\mathrm{Cl}_{2}\) ?

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