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Chapter 15: Problem 4

Write equilibrium constant expressions, \(K_{\mathrm{p}},\) for the reactions (a) \(\mathrm{CS}_{2}(\mathrm{g})+4 \mathrm{H}_{2}(\mathrm{g}) \rightleftharpoons \mathrm{CH}_{4}(\mathrm{g})+2 \mathrm{H}_{2} \mathrm{S}(\mathrm{g})\) (b) \(\mathrm{Ag}_{2} \mathrm{O}(\mathrm{s}) \rightleftharpoons 2 \mathrm{Ag}(\mathrm{s})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g})\) (c) \(2 \mathrm{NaHCO}_{3}(\mathrm{s}) \rightleftharpoons\) \(\mathrm{Na}_{2} \mathrm{CO}_{3}(\mathrm{s})+\mathrm{CO}_{2}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g})\)

Short Answer

Expert verified
The equilibrium constant expressions for the given reactions are: \n(a) \(K_{\mathrm{p}} = \frac{(P_{CH4}) \cdot (P_{H2S})^2}{(P_{CS2}) \cdot (P_{H2})^4}\) \n(b) \(K_{\mathrm{p}}=(P_{O2})^{1/2}\) \n(c) \(K_{\mathrm{p}}=(P_{CO2}) \cdot (P_{H2O})\)

Step by step solution

01

Writing the equilibrium constant for reaction (a).

The reaction is: \(\mathrm{CS}_{2}(\mathrm{g})+4 \mathrm{H}_{2}(\mathrm{g}) \rightleftharpoons \mathrm{CH}_{4}(\mathrm{g})+2 \mathrm{H}_{2}\mathrm{S}(\mathrm{g})\). Apply the formula for \(K_{\mathrm{p}}\): \(K_{\mathrm{p}} = \frac{(P_{CH4})^1 \cdot (P_{H2S})^2}{(P_{CS2})^1 \cdot (P_{H2})^4}\)
02

Writing the equilibrium constant for reaction (b).

The reaction is: \(\mathrm{Ag}_{2}\mathrm{O}(\mathrm{s}) \rightleftharpoons 2\mathrm{Ag}(\mathrm{s})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g})\). Notice that solids are omitted. Thus, \(K_{\mathrm{p}}\) expression is: \(K_{\mathrm{p}}=(P_{O2})^{1/2}\)
03

Writing the equilibrium constant for reaction (c).

The reaction is: \(2\mathrm{NaHCO}_{3}(\mathrm{s}) \rightleftharpoons\mathrm{Na}_{2}\mathrm{CO}_{3}(\mathrm{s})+\mathrm{CO}_{2}(\mathrm{g})+\mathrm{H}_{2}\mathrm{O}(\mathrm{g})\). Notice how we can omit the substances that are solid. Thus, \(K_{\mathrm{p}}\) expression is: \(K_{\mathrm{p}}=(P_{CO2})^1 \cdot (P_{H2O})^1\)

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Key Concepts

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

Chemical Equilibrium
Chemical equilibrium is an essential concept in chemistry that describes a state in which the concentrations of reactants and products remain constant over time. This happens when the forward and reverse reactions occur at the same rate.

At equilibrium, no net change is visible, although individual molecules continuously react. The use of equilibrium constants helps in quantifying the ratio of concentrations of products to reactants.

  • It allows chemists to predict the direction of reactions
  • Enables understanding of reaction conditions
  • Helps calculate the concentrations of different species at equilibrium

Chemical equilibrium is dynamic, not static. Though it looks like nothing is happening, reactions happen both ways. It's useful to remember that different reactions reach equilibrium at different concentrations and conditions.
Pressure-based Equilibrium Constants
In chemical reactions involving gases, equilibrium constants can be expressed in terms of pressure, referred to as pressure-based equilibrium constants, or \( K_{p} \).

These constants are derived from the concentrations of gaseous reactants and products, converted using the ideal gas law. Understanding \( K_{p} \) is crucial for reactions in which gases play a significant role.

  • \( K_{p} \) is calculated using partial pressures
  • Expresses the ratio, similar to concentration-based \( K_{c} \)
  • \( K_{p} \) is useful in predicting the behavior of gas-phase reactions

To solve for \( K_{p} \), use the equilibrium expression by inserting each species' partial pressure raised to the power of its stoichiometric coefficient. This approach provides insight into how pressure changes impact reaction equilibria.
Phase Exclusion in Equilibrium Expressions
In equilibrium expressions, only gases and aqueous solutions are generally included, while solids and liquids are excluded. Phase exclusion, specifically the exclusion of solids and liquids, is due to their constant concentrations.

Solids and liquids don't affect the position of equilibrium, as their activities (or effective concentrations) are considered to be constant.
  • Solids have constant concentration under given conditions
  • Liquids are also omitted if they are pure solvents
  • Gas and aqueous phases primarily determine the equilibrium
The simplification from phase exclusion results in easier calculations. Focusing on gaseous and aqueous species narrows the aspects affecting equilibrium. This principle was crucial in the steps of the original solutions, allowing \( K_{p} \) to focus on the relevant measurable quantities.

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Most popular questions from this chapter

Briefly describe each of the following ideas or phenomena: (a) dynamic equilibrium; (b) direction of a net chemical change; (c) Le Ch芒telier's principle; (d) effect of a catalyst on equilibrium.

Lead metal is added to \(0.100 \mathrm{M} \mathrm{Cr}^{3+}(\mathrm{aq}) .\) What are \(\left[\mathrm{Pb}^{2+}\right],\left[\mathrm{Cr}^{2+}\right],\) and \(\left[\mathrm{Cr}^{3+}\right]\) when equilibrium is established in the reaction? $$\begin{aligned} \mathrm{Pb}(\mathrm{s})+2 \mathrm{Cr}^{3+}(\mathrm{aq}) \rightleftharpoons \mathrm{Pb}^{2+}(\mathrm{aq})+2 \mathrm{Cr}^{2+}(\mathrm{aq}) & \\ K_{\mathrm{c}}=3.2 \times 10^{-10} & \end{aligned}$$

The following data are given at \(1000 \mathrm{K}: \mathrm{CO}(\mathrm{g})+\) \(\mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \rightleftharpoons \mathrm{CO}_{2}(\mathrm{g})+\mathrm{H}_{2}(\mathrm{g}) ; \quad \Delta H^{\circ}=-42 \mathrm{k} \mathrm{J}\) \(K_{\mathrm{c}}=0.66 .\) After an initial equilibrium is established in a \(1.00 \mathrm{L}\) container, the equilibrium amount of \(\mathrm{H}_{2}\) can be increased by (a) adding a catalyst; (b) increasing the temperature; (c) transferring the mixture to a 10.0 L container; (d) in some way other than (a), (b), Or (c).

Refer to Example \(15-4 . \mathrm{H}_{2} \mathrm{S}(\mathrm{g})\) at \(747.6 \mathrm{mmHg}\) pressure and a \(1.85 \mathrm{g}\) sample of \(\mathrm{I}_{2}(\mathrm{s})\) are introduced into a \(725 \mathrm{mL}\) flask at \(60^{\circ} \mathrm{C} .\) What will be the total pressure in the flask at equilibrium? $$\begin{aligned} \mathrm{H}_{2} \mathrm{S}(\mathrm{g})+\mathrm{I}_{2}(\mathrm{s}) \rightleftharpoons 2 \mathrm{HI}(\mathrm{g})+\mathrm{S}(\mathrm{s}) & \\ K_{\mathrm{p}}=& 1.34 \times 10^{-5} \mathrm{at} 60^{\circ} \mathrm{C} \end{aligned}$$

A classic experiment in equilibrium studies dating from 1862 involved the reaction in solution of ethanol \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\right)\) and acetic acid \(\left(\mathrm{CH}_{3} \mathrm{COOH}\right)\) to produce ethyl acetate and water. $$\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}+\mathrm{CH}_{3} \mathrm{COOH} \rightleftharpoons \mathrm{CH}_{3} \mathrm{COOC}_{2} \mathrm{H}_{5}+\mathrm{H}_{2} \mathrm{O}$$ The reaction can be followed by analyzing the equilibrium mixture for its acetic acid content. $$\begin{array}{r} 2 \mathrm{CH}_{3} \mathrm{COOH}(\mathrm{aq})+\mathrm{Ba}(\mathrm{OH})_{2}(\mathrm{aq}) \rightleftharpoons \\ \mathrm{Ba}\left(\mathrm{CH}_{3} \mathrm{COO}\right)_{2}(\mathrm{aq})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \end{array}$$ In one experiment, a mixture of 1.000 mol acetic acid and 0.5000 mol ethanol is brought to equilibrium. A sample containing exactly one-hundredth of the equilibrium mixture requires \(28.85 \mathrm{mL} 0.1000 \mathrm{M}\) \(\mathrm{Ba}(\mathrm{OH})_{2}\) for its titration. Calculate the equilibrium constant, \(K_{c}\), for the ethanol-acetic acid reaction based on this experiment.
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