Based on these descriptions, write a balanced equation and the corresponding \(K_{c}\) expression for each reversible reaction. (a) Carbonyl fluoride, \(\mathrm{COF}_{2}(\mathrm{g}),\) decomposes into gaseous carbon dioxide and gaseous carbon tetrafluoride. (b) Copper metal displaces silver(I) ion from aqueous solution, producing silver metal and an aqueous solution of copper(II) ion. (c) Peroxodisulfate ion, \(\mathrm{S}_{2} \mathrm{O}_{8}^{2-}\), oxidizes iron(II) ion to iron(III) ion in aqueous solution and is itself reduced to sulfate ion.

Short Answer

Expert verified
The balanced chemical equations and their corresponding \(K_{c}\) expressions are: (a) \(2 \, \mathrm{COF}_{2}(\mathrm{g}) \leftrightarrow \mathrm{CO}_{2}(\mathrm{g}) + \mathrm{CF}_{4}(\mathrm{g})\), \(K_{c} = \frac{[\mathrm{CO}_{2}][\mathrm{CF}_{4}]}{[\mathrm{COF}_{2}]^{2}}\); (b) \(\mathrm{Cu}(\mathrm{s}) + 2\, \mathrm{Ag}^{+}(\mathrm{aq}) \leftrightarrow \mathrm{Cu}^{2+}(\mathrm{aq}) + 2\, \mathrm{Ag}(\mathrm{s})\), \(K_{c} = \frac{[\mathrm{Cu}^{2+}]}{[\mathrm{Ag}^{+}]^{2}}\); (c) \(2 \, \mathrm{S}_{2} \mathrm{O}_{8}^{2-} + 2 \, \mathrm{Fe}^{2+} \leftrightarrow 4 \, \mathrm{SO}_{4}^{2-} + 2 \, \mathrm{Fe}^{3+}\), \(K_{c} = \frac{[\mathrm{SO}_{4}^{2-}]^{4}[\mathrm{Fe}^{3+}]^{2}}{[\mathrm{S}_{2} \mathrm{O}_{8}^{2-}]^{2}[\mathrm{Fe}^{2+}]^{2}}\)

Step by step solution

01

Balance Equation and Compute \(K_{c}\) for (a)

Firstly, the balanced equation for the decomposition of Carbonyl fluoride into Carbon dioxide and Carbon tetrafluoride is: \[2 \, \mathrm{COF}_{2}(\mathrm{g}) \leftrightarrow \mathrm{CO}_{2}(\mathrm{g}) + \mathrm{CF}_{4}(\mathrm{g})\]. For the \(K_{c}\) expression, it is given by the ratio of the concentrations of the products over the reactants: \[K_{c} = \frac{[\mathrm{CO}_{2}][\mathrm{CF}_{4}]}{[\mathrm{COF}_{2}]^{2}}\]
02

Balance Equation and Compute \(K_{c}\) for (b)

The balanced equation for the displacement of silver(I) ion by copper metal is: \[\mathrm{Cu}(\mathrm{s}) + 2\, \mathrm{Ag}^{+}(\mathrm{aq}) \leftrightarrow \mathrm{Cu}^{2+}(\mathrm{aq}) + 2\, \mathrm{Ag}(\mathrm{s})\]. The \(K_{c}\) expression becomes: \[K_{c} = \frac{[\mathrm{Cu}^{2+}]}{[\mathrm{Ag}^{+}]^{2}}\]. Notice that Cu(s) and Ag(s) are not included in the \(K_{c}\) expression, because pure solids and liquids are not included.
03

Balance Equation and Compute \(K_{c}\) for (c)

The balanced equation for the oxidation of iron(II) ion by peroxodisulfate ion is: \[2 \, \mathrm{S}_{2} \mathrm{O}_{8}^{2-} + 2 \, \mathrm{Fe}^{2+} \leftrightarrow 4 \, \mathrm{SO}_{4}^{2-} + 2 \, \mathrm{Fe}^{3+}\]. The \(K_{c}\) expression in this case is: \[K_{c} = \frac{[\mathrm{SO}_{4}^{2-}]^{4}[\mathrm{Fe}^{3+}]^{2}}{[\mathrm{S}_{2} \mathrm{O}_{8}^{2-}]^{2}[\mathrm{Fe}^{2+}]^{2}}\]

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!

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

Continuous removal of one of the products of a chemical reaction has the effect of causing the reaction to go to completion. Explain this fact in terms of Le Châtelier's principle.

Cadmium metal is added to \(0.350 \mathrm{L}\) of an aqueous solution in which \(\left[\mathrm{Cr}^{3+}\right]=1.00 \mathrm{M} .\) What are the concentrations of the different ionic species at equilibrium? What is the minimum mass of cadmium metal required to establish this equilibrium? $$\begin{array}{r} 2 \mathrm{Cr}^{3+}(\mathrm{aq})+\mathrm{Cd}(\mathrm{s}) \rightleftharpoons 2 \mathrm{Cr}^{2+}(\mathrm{aq})+\mathrm{Cd}^{2+}(\mathrm{aq}) \\ K_{\mathrm{c}}=0.288 \end{array}$$

In the reversible reaction \(\mathrm{H}_{2}(\mathrm{g})+\mathrm{I}_{2}(\mathrm{g}) \rightleftharpoons\) \(2 \mathrm{HI}(\mathrm{g}),\) an initial mixture contains \(2 \mathrm{mol} \mathrm{H}_{2}\) and 1 mol I \(_{2} .\) The amount of HI expected at equilibrium is (a) \(1 \mathrm{mol} ;\) (b) \(2 \mathrm{mol} ;\) (c) less than \(2 \mathrm{mol}\); (d) more than 2 mol but less than 4 mol.

A sample of \(\mathrm{NH}_{4} \mathrm{HS}(\mathrm{s})\) is placed in a \(2.58 \mathrm{L}\) flask containing 0.100 mol \(\mathrm{NH}_{3}(\mathrm{g}) .\) What will be the total gas pressure when equilibrium is established at \(25^{\circ} \mathrm{C} ?\) $$\begin{aligned} \mathrm{NH}_{4} \mathrm{HS}(\mathrm{s}) \rightleftharpoons \mathrm{NH}_{3}(\mathrm{g})+\mathrm{H}_{2} \mathrm{S}(\mathrm{g}) & \\ K_{\mathrm{p}} &=0.108 \text { at } 25^{\circ} \mathrm{C} \end{aligned}$$

One important reaction in the citric acid cycle is citrate(aq) \(\rightleftharpoons\) aconitate \((\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \quad K=0.031\) Write the equilibrium constant expression for the above reaction. Given that the concentrations of \([\text { citrate }(\mathrm{aq})]=0.00128 \mathrm{M},[\text { aconitate }(\mathrm{aq})]=4.0 \times\) \(10^{-5} \mathrm{M},\) and \(\left[\mathrm{H}_{2} \mathrm{O}\right]=55.5 \mathrm{M},\) calculate the reaction quotient. Is this reaction at equilibrium? If not, in which direction will it proceed?

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