/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 18 Write the expression for \(K_{\m... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 12: Problem 18

Write the expression for \(K_{\mathrm{c}}\) for each reaction. (a) \(\mathrm{PCl}_{5}(\mathrm{~s}) \rightleftharpoons \mathrm{PCl}_{3}(\mathrm{~g})+\mathrm{Cl}_{2}(\mathrm{~g})\) (b) \(\mathrm{Co}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}^{2+}(\mathrm{aq})+4 \mathrm{Cl}^{-}(\mathrm{aq}) \rightleftharpoons\) \(\mathrm{CoCl}_{4}^{2-}(\mathrm{aq})+6 \mathrm{H}_{2} \mathrm{O}(\ell)\) (c) \(\mathrm{CH}_{3} \mathrm{COOH}(\mathrm{aq}) \rightleftharpoons \mathrm{CH}_{3} \mathrm{COO}^{-}(\mathrm{aq})+\mathrm{H}^{+}(\mathrm{aq})\) (d) \(2 \mathrm{~F}_{2}(\mathrm{~g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \rightleftharpoons \mathrm{OF}_{2}(\mathrm{~g})+2 \mathrm{HF}(\mathrm{g})\)

Short Answer

Expert verified
(a) \([\mathrm{PCl}_3][\mathrm{Cl}_2]\); (b) \(\frac{[\mathrm{CoCl}_4^{2-}]}{[\mathrm{Co(H}_2\mathrm{O})_6^{2+}][\mathrm{Cl}^-]^4}\); (c) \(\frac{[\mathrm{CH}_3\mathrm{COO}^-][\mathrm{H}^+]}{[\mathrm{CH}_3\mathrm{COOH}]}\); (d) \(\frac{[\mathrm{OF}_2][\mathrm{HF}]^2}{[\mathrm{F}_2]^2[\mathrm{H}_2\mathrm{O}]}\)."}

Step by step solution

01

Understand Equilibrium Constant Expression

The equilibrium constant expression, denoted as \(K_c\), is calculated using the concentrations of products raised to their stoichiometric coefficients divided by the concentrations of the reactants raised to their coefficients. Be aware of the physical states: pure solids and liquids do not appear in the expression.
02

Express K_c for Reaction (a)

The reaction is \( \mathrm{PCl}_{5}(\mathrm{s}) \rightleftharpoons \mathrm{PCl}_{3}(\mathrm{g})+\mathrm{Cl}_{2}(\mathrm{g}) \). Since \(\mathrm{PCl}_{5}(\mathrm{s})\) is a solid, it is not included. The expression for \(K_c\) is:\[ K_c = [\mathrm{PCl}_3][\mathrm{Cl}_2] \]
03

Express K_c for Reaction (b)

The reaction is \(\mathrm{Co}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}^{2+}(\mathrm{aq}) + 4\mathrm{Cl}^{-}(\mathrm{aq}) \rightleftharpoons \mathrm{CoCl}_{4}^{2-}(\mathrm{aq}) + 6 \mathrm{H}_2 \mathrm{O}(\ell)\). The water is liquid, thus omitted. The expression for \(K_c\) is:\[ K_c = \frac{[\mathrm{CoCl}_4^{2-}]}{[\mathrm{Co(H}_2\mathrm{O})_6^{2+}][\mathrm{Cl}^-]^4} \]
04

Express K_c for Reaction (c)

The reaction is \(\mathrm{CH}_{3} \mathrm{COOH}(\mathrm{aq}) \rightleftharpoons \mathrm{CH}_{3} \mathrm{COO}^{-}(\mathrm{aq}) + \mathrm{H}^{+}(\mathrm{aq})\). All species are in aqueous solution, so they all appear in the expression:\[ K_c = \frac{[\mathrm{CH}_3\mathrm{COO}^-][\mathrm{H}^+]}{[\mathrm{CH}_3\mathrm{COOH}]} \]
05

Express K_c for Reaction (d)

The reaction is \(2 \mathrm{~F}_2(\mathrm{~g})+\mathrm{H}_2 \mathrm{O}(\mathrm{g}) \rightleftharpoons \mathrm{OF}_2(\mathrm{~g})+2 \mathrm{HF}(\mathrm{g})\). All species are gases, so the expression is:\[ K_c = \frac{[\mathrm{OF}_2][\mathrm{HF}]^2}{[\mathrm{F}_2]^2[\mathrm{H}_2\mathrm{O}]} \]

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 91影视!

Key Concepts

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

Chemical Equilibrium
In chemistry, equilibrium refers to a state in which the rates of the forward and reverse reactions are equal. At this point, the concentrations of the reactants and products remain constant over time, even though both reactions continue to occur. This balance results from the dynamic nature of chemical reactions, where reactions are ongoing but occur at the same rate in both directions.
  • The concept of equilibrium is fundamental in understanding how reactions progress and reach a state of balance.
  • It is important to note that equilibrium does not mean that the quantities of reactants and products are equal; rather, it means their ratios remain constant.
One important aspect of chemical equilibrium is the equilibrium constant, often represented as \(K_c\). This constant is derived from the concentrations of products and reactants at equilibrium.
In the expressions for \(K_c\):
  • It includes only the species that are present in a gaseous or aqueous state.
  • Solids and pure liquids do not appear in the equilibrium expression because their concentrations are constant and do not affect the equilibrium position.
Reaction Quotient
The reaction quotient, denoted as \(Q\), is a measure used to determine the direction in which a reaction will proceed to reach equilibrium. The expression for \(Q\) is similar to that of the equilibrium constant \(K_c\), but \(Q\) can be calculated at any point during a reaction and not just at equilibrium.
  • If \(Q = K_c\), the system is at equilibrium.
  • If \(Q < K_c\), the forward reaction is favored, meaning reactants are converted into products.
  • If \(Q > K_c\), the reverse reaction is favored, meaning products are converted back into reactants.
Calculating the reaction quotient involves using the same formula as the equilibrium constant but substituting the current concentrations of the reactants and products.
This helps predict how the reaction will shift to achieve equilibrium, providing valuable insight into the reaction's progress.
Le Chatelier's Principle
Le Chatelier's Principle helps us understand how a system at equilibrium responds to disturbances such as changes in concentration, temperature, or pressure. This principle states that if a dynamic equilibrium is disturbed by changing the conditions, the system shifts its position to counteract the change and re-establish equilibrium.
  • Adding more reactants will typically cause the equilibrium to shift toward the product side to reduce the disturbance.
  • Increasing the temperature in an exothermic reaction will shift the equilibrium toward the reactants to absorb the excess heat.
  • Conversely, decreasing the temperature in an exothermic reaction shifts the equilibrium toward the products.
  • Increased pressure will cause the equilibrium to favor the side of the reaction with fewer moles of gas, if gases are involved.
Understanding Le Chatelier's Principle is crucial for controlling chemical reactions in industrial processes, where maximizing yields and minimizing costs are critical. It provides a predictive tool for how conditions can be manipulated to drive the equilibrium position in a desired direction.

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

The atmosphere consists of about \(80 \% \mathrm{~N}_{2}\) and \(20 \% \mathrm{O}_{2}\), yet there are many oxides of nitrogen that are stable and can be isolated in the laboratory. (a) Is the atmosphere at chemical equilibrium with respect to forming NO? (b) If not, why doesn't NO form? If so, how is it that \(\mathrm{NO}\) can be made and kept in the laboratory for long periods?

The decomposition of \(\mathrm{NH}_{4} \mathrm{HS}\) is endothermic. $$ \mathrm{NH}_{4} \mathrm{HS}(\mathrm{s}) \rightleftharpoons \mathrm{NH}_{3}(\mathrm{~g})+\mathrm{H}_{2} \mathrm{~S}(\mathrm{~g}) $$ (a) Using Le Chatelier's principle, explain how increasing the temperature would affect the equilibrium. (b) If more \(\mathrm{NH}_{4} \mathrm{HS}\) is added to a sealed flask in which this equilibrium exists, how is the equilibrium affected? (c) What if some additional \(\mathrm{NH}_{3}\) is placed in a sealed flask containing an equilibrium mixture? (d) What will happen to the partial pressure of \(\mathrm{NH}_{3}\) if some \(\mathrm{H}_{2} \mathrm{~S}\) is removed from the flask?

For the equilibrium $$ \mathrm{Co}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}^{2+}(\mathrm{aq})+4 \mathrm{Cl}^{-}(\mathrm{aq}) \rightleftharpoons \mathrm{CoCl}_{4}^{2-}(\mathrm{aq})+6 \mathrm{H}_{2} \mathrm{O}(\ell) $$ pink blue \(K_{\mathrm{c}}\) is somewhat greater than 1 . If water is added to a blue solution of \(\mathrm{CoCl}_{4}^{2-}(\mathrm{aq}),\) the color changes from blue to pink. (a) Does water appear in the equilibrium constant expression for this reaction? (b) How can adding water shift the equilibrium to the left? (c) Is this shift in the equilibrium in accord with Le Chatelier's principle? Why or why not?

The hydrocarbon \(\mathrm{C}_{4} \mathrm{H}_{10}\) can exist in two gaseous forms: butane and 2 -methylpropane. The value of \(K_{\mathrm{c}}\) for conversion of butane to 2 -methylpropane is 2.5 at \(25^{\circ} \mathrm{C}\). CCCCC=CC(C)(C)CO (a) Suppose that the initial concentrations of butane and 2-methylpropane are each \(0.100 \mathrm{~mol} / \mathrm{L}\). Make up a table of initial concentrations, change in concentrations, and equilibrium concentrations for this reaction. (b) Write the equilibrium constant expression in terms of \(x\), the change in the concentration of butane, and then solve for \(x\). (c) If you place 0.017 mol butane in a \(0.50-\mathrm{L}\) sealed flask at \(25^{\circ} \mathrm{C},\) calculate the equilibrium concentration of each isomer.

At \(503 \mathrm{~K}\) the equilibrium constant \(K_{\mathrm{c}}\) for the dissociation of \(\mathrm{N}_{2} \mathrm{O}_{4}\) $$ \mathrm{N}_{2} \mathrm{O}_{4}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NO}_{2}(\mathrm{~g}) $$ has the value 40.0 . (a) Calculate the fraction of \(\mathrm{N}_{2} \mathrm{O}_{4}\) left undissociated when \(1.00 \mathrm{~mol}\) of this gas is heated to \(503 \mathrm{~K}\) in a \(10.0-\mathrm{L}\) sealed container. (b) If the volume is now reduced to \(2.0 \mathrm{~L},\) calculate the new fraction of \(\mathrm{N}_{2} \mathrm{O}_{4}\) that is undissociated. (c) Calculate all three equilibrium concentrations.
See all solutions

Recommended explanations on Chemistry Textbooks

Making Measurements

Read Explanation

Kinetics

Read Explanation

Organic Chemistry

Read Explanation

The Earths Atmosphere

Read Explanation

Chemical Reactions

Read Explanation

Nuclear Chemistry

Read Explanation
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.