Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Write the equilibrium expression for each of the following heterogeneous equilibria. a. \(2 \mathrm{HgO}(s) \rightleftharpoons 2 \mathrm{Hg}(l)+\mathrm{O}_{2}(g)\) b. \(2 \mathrm{KClO}_{3}(s) \rightleftharpoons 2 \mathrm{KCl}(s)+3 \mathrm{O}_{2}(g)\) c. \(C(s)+C O_{2}(g) \rightleftharpoons 2 C O(g)\)

Short Answer

Expert verified
a. \(K_\mathrm{a} = [\mathrm{O}_2]\) b. \(K_\mathrm{b} = [\mathrm{O}_2]^3\) c. \(K_\mathrm{c} = \frac{[\mathrm{CO}]^2}{[\mathrm{CO}_2]}\)

Step by step solution

01

Identify the gaseous components

In this equilibrium, the gaseous component is O2.
02

Write the equilibrium expression

Since only the gaseous components are included in the equilibrium expression and their stoichiometric coefficients become the exponents, the equilibrium expression for this reaction is: \[K_\mathrm{a} = [\mathrm{O}_2]\] b. \(2 \mathrm{KClO}_{3}(s) \rightleftharpoons 2 \mathrm{KCl}(s)+3 \mathrm{O}_{2}(g)\)
03

Identify the gaseous components

In this equilibrium, the gaseous component is O2.
04

Write the equilibrium expression

Since only the gaseous components are included in the equilibrium expression and their stoichiometric coefficients become the exponents, the equilibrium expression for this reaction is: \[K_\mathrm{b} = [\mathrm{O}_2]^3\] c. \(C(s)+C O_{2}(g) \rightleftharpoons 2 C O(g)\)
05

Identify the gaseous components

In this equilibrium, the gaseous components are CO2 and CO.
06

Write the equilibrium expression

Since only the gaseous components are included in the equilibrium expression and their stoichiometric coefficients become the exponents, the equilibrium expression for this reaction is: \[K_\mathrm{c} = \frac{[\mathrm{CO}]^2}{[\mathrm{CO}_2]}\]

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.

heterogeneous equilibria
Heterogeneous equilibria occur when a chemical reaction involves substances in different phases, such as solids, liquids, and gases. These reactions are essential for understanding chemical processes in nature and industry. When writing the equilibrium expression for such reactions, only the concentrations of gaseous components are typically included. This is because only gases and solutions have measurable concentrations.
For example, in the reaction \(2 \mathrm{HgO}(s) \rightleftharpoons 2 \mathrm{Hg}(l) + \mathrm{O}_2(g)\), HgO is a solid, and Hg is a liquid. Therefore, they are not included in the equilibrium expression. The resulting expression focuses only on the gaseous oxygen \(\mathrm{O}_2\).
Remember, in heterogeneous equilibria, the pure solids and liquids are considered as having constant activity and do not appear in the equilibrium expression.
stoichiometric coefficients
Stoichiometric coefficients are the numbers placed in front of compounds in a balanced chemical equation. They indicate the ratio in which substances react and are essential in determining the relative amounts of reactants and products.
In equilibrium expressions, these coefficients are used as exponents. For instance, in the reaction \(2 \mathrm{KClO}_3(s) \rightleftharpoons 2 \mathrm{KCl}(s) + 3 \mathrm{O}_2(g)\), the stoichiometric coefficient of \(\mathrm{O}_2\) is 3. As such, the equilibrium expression \(K_b = [\mathrm{O}_2]^3\) includes this coefficient as an exponent, emphasizing the influence of the amount of oxygen produced.
Understanding stoichiometric coefficients helps ensure accurate calculations of equilibrium expressions and can predict how changes in concentrations of reactants or products will affect the reaction's position.
gaseous components
Gaseous components in chemical reactions are primarily considered when writing equilibrium expressions, especially in heterogeneous equilibria. This is because gases can expand to fill their container, and their concentration can be conveniently measured and controlled.
Take the reaction \(C(s) + CO_2(g) \rightleftharpoons 2 CO(g)\). Here, \(CO_2\) and \(CO\) are the gaseous components. The equilibrium expression derived from this reaction includes only these gases: \(K_c = \frac{[\mathrm{CO}]^2}{[\mathrm{CO}_2]}\). This highlights the importance of understanding which components of a reaction are in the gaseous state.
Pay attention to gaseous components as they often play a crucial role in the equilibrium of a reaction, and knowing how to manipulate them can significantly influence the outcome of chemical processes.
chemical reactions
Chemical reactions involve the transformation of reactants into products through breaking and forming chemical bonds. This process is essential for numerous natural and industrial processes.
Equilibrium in chemical reactions is achieved when the rate of the forward reaction equals the rate of the reverse reaction. At this point, the concentrations of reactants and products remain constant over time, though they are not necessarily equal.
In heterogeneous equilibriums, it is vital to distinguish between different phases of reactants and products because only the concentrations of gaseous substances influence the equilibrium state. Recognizing the nature of chemical reactions helps in formulating correct equilibrium expressions and in understanding how various factors can shift the equilibrium, such as changes in concentration, pressure, or temperature.
This foundational concept lays the groundwork for studying advanced chemical processes and developing new reactions with desired outcomes.

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

A saturated solution of nickel(II) sulfide contains approximately \(3.6 \times 10^{-4} \mathrm{g}\) of dissolved NiS per liter at \(20^{\circ} \mathrm{C}\). Calculate the solubility product \(K_{\mathrm{sp}}\) for NiS at \(20^{\circ} \mathrm{C}\).

The reaction \(\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g)\) is exothermic as written. For the maximum production of ammonia, should this reaction be performed at a lower or a higher temperature? Explain.

Write the balanced chemical equation describing the dissolving of the following solids in water. Write the expression for \(K_{\mathrm{sp}}\) for each process. a. \(\mathrm{Co}_{2} \mathrm{S}_{3}(\mathrm{s})\) b. \(\operatorname{Ag} \mathrm{I}(s)\) c. \(\mathrm{MgCO}_{3}(s)\) d. \(\operatorname{Fe}(\mathrm{OH})_{3}(s)\)

For a given reaction at a given temperature, the special ratio of products to reactants defined by the equilibrium constant is always equal to the same number. Explain why this is true, no matter what initial concentrations of reactants (or products) may have been taken in setting up an experiment.

For the reaction $$2 \mathrm{CO}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{CO}_{2}(g)$$ it is found at equilibrium at a certain temperature that the concentrations are \([\mathrm{CO}(g)]=2.7 \times 10^{-4} \mathrm{M}\) \(\left[\mathrm{O}_{2}(g)\right]=1.9 \times 10^{-3} \mathrm{M},\) and \(\left[\mathrm{CO}_{2}(g)\right]=1.1 \times 10^{-1} \mathrm{M}\) Calculate \(K\) for the reaction at this temperature.

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