Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 12: Problem 6

WEB Write equilibrium constant \((K)\) expressions for the following reactions: (a) \(\mathrm{CH}_{4}(g)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{CO}(\mathrm{g})+3 \mathrm{H}_{2}(\mathrm{~g})\) (b) \(4 \mathrm{NH}_{3}(g)+5 \mathrm{O}_{2}(g) \rightleftharpoons 4 \mathrm{NO}(g)+6 \mathrm{H}_{2} \mathrm{O}(g)\) (c) \(\mathrm{BaCO}_{3}(s) \rightleftharpoons \mathrm{BaO}(s)+\mathrm{CO}_{2}(g)\) (d) \(\mathrm{NH}_{3}(g)+\mathrm{HCl}(g) \rightleftharpoons \mathrm{NH}_{4} \mathrm{Cl}(s)\)

Short Answer

Expert verified
Question: Write the equilibrium constant expressions for the following reactions: (a) \(\mathrm{CH}_{4}(g) + \mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{CO}(g) + 3 \mathrm{H}_{2}(g)\) (b) \(4 \mathrm{NH}_{3}(g) + 5 \mathrm{O}_{2}(g) \rightleftharpoons 4 \mathrm{NO}(g) + 6 \mathrm{H}_{2} \mathrm{O}(g)\) (c) \(\mathrm{BaCO}_{3}(s) \rightleftharpoons \mathrm{BaO}(s) + \mathrm{CO}_{2}(g)\) (d) \(\mathrm{NH}_{3}(g) + \mathrm{HCl}(g) \rightleftharpoons \mathrm{NH}_{4} \mathrm{Cl}(s)\) Answer: (a) \(K_{a}=\frac{[\mathrm{CO}]([H_{2}]^3)}{[\mathrm{CH}_{4}]}\) (b) \(K_{b}=\frac{([\mathrm{NO}]^4)([\mathrm{H}_{2}\mathrm{O}]^6)}{([\mathrm{NH}_{3}]^4)([\mathrm{O}_{2}]^5)}\) (c) \(K_{c}=[\mathrm{CO}_{2}]\) (d) \(K_{d}=\frac{1}{[\mathrm{NH}_{3}][\mathrm{HCl}]}\)

Step by step solution

01

Write an equilibrium constant expression for reaction (a)

For reaction (a), the given equation is: \(\mathrm{CH}_{4}(g) + \mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{CO}(g) + 3 \mathrm{H}_{2}(g)\) To write the expression, we'll use the law of mass action. However, since \(\mathrm{H}_{2} \mathrm{O}(l)\) is a pure liquid, it will not be included in the equilibrium constant expression. So, the equilibrium constant expression is: \(K_{a}=\frac{[\mathrm{CO}]([H_{2}]^3)}{[\mathrm{CH}_{4}]}\)
02

Write an equilibrium constant expression for reaction (b)

For reaction (b), the given equation is: \(4 \mathrm{NH}_{3}(g) + 5 \mathrm{O}_{2}(g) \rightleftharpoons 4 \mathrm{NO}(g) + 6 \mathrm{H}_{2} \mathrm{O}(g)\) Now, apply the law of mass action to write the expression: \(K_{b}=\frac{([\mathrm{NO}]^4)([\mathrm{H}_{2}\mathrm{O}]^6)}{([\mathrm{NH}_{3}]^4)([\mathrm{O}_{2}]^5)}\)
03

Write an equilibrium constant expression for reaction (c)

For reaction (c), the given equation is: \(\mathrm{BaCO}_{3}(s) \rightleftharpoons \mathrm{BaO}(s) + \mathrm{CO}_{2}(g)\) The two solids, \(\mathrm{BaCO}_{3}\) and \(\mathrm{BaO}\), will not be included in the equilibrium constant expression, so we have: \(K_{c}=[\mathrm{CO}_{2}]\)
04

Write an equilibrium constant expression for reaction (d)

For reaction (d), the given equation is: \(\mathrm{NH}_{3}(g) + \mathrm{HCl}(g) \rightleftharpoons \mathrm{NH}_{4} \mathrm{Cl}(s)\) As \(\mathrm{NH}_{4} \mathrm{Cl}(s)\) is a solid, it will not be included in the equilibrium constant expression. Thus, we have: \(K_{d}=\frac{1}{[\mathrm{NH}_{3}][\mathrm{HCl}]}\) Now all equilibrium constant expressions for the given reactions have been found.

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.

Start your free trial

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

Derive the relationship $$K=K_{\mathrm{c}} \times(R T)^{\Delta r_{\mathrm{B}}}$$ where \(K_{\mathrm{c}}\) is the equilibrium constant using molarities and \(\Delta n_{\mathrm{g}}\) is the change in the number of moles of gas in the reaction (see page 326). (Hint: Recall that \(P_{\Lambda}=n_{\Lambda} R T / V\) and \(\left.n_{A} / V=[\mathrm{A}] .\right)\)

For the system $$\mathrm{PCl}_{5}(\mathrm{~g}) \rightleftharpoons \mathrm{PCl}_{3}(g)+\mathrm{Cl}_{2}(g)$$ \(K\) is 26 at \(300^{\circ} \mathrm{C}\). In a 5.0-L flask, a gaseous mixture consists of all three gases with partial pressures as follows: \(P_{\mathrm{PCl}_{5}}=0.012 \mathrm{~atm}, P_{\mathrm{Cl}_{2}}=0.45 \mathrm{~atm}\), \(P_{\mathrm{PCl}_{3}}=0.90 \mathrm{~atm} .\) (a) Is the mixture at equilibrium? Explain. (b) If it is not at equilibrium, which way will the system shift to establish equilibrium?

Consider the statement "The equilibrium constant for a reaction at \(400 \mathrm{~K}\) is 792 . It must be a very fast reaction." What is wrong with the statement?

Consider the decomposition of ammonium hydrogen sulfide: $$\mathrm{NH}_{4} \mathrm{HS}(s) \rightleftharpoons \mathrm{NH}_{3}(g)+\mathrm{H}_{2} \mathrm{~S}(g)$$ In a sealed flask at \(25^{\circ} \mathrm{C}\) are \(10.0 \mathrm{~g}\) of \(\mathrm{NH}_{4} \mathrm{HS}\), ammonia with a partial pressure of \(0.692 \mathrm{~atm}\), and \(\mathrm{H}_{2} \mathrm{~S}\) with a partial pressure of \(0.0532 \mathrm{~atm}\). When equilibrium is established, it is found that the partial pressure of ammonia has increased by 12.4\%. Calculate \(K\) for the decomposition of \(\mathrm{NH}_{4} \mathrm{HS}\) at \(25^{\circ} \mathrm{C}\).

Write equilibrium constant \((K)\) expressions for the following reactions: (a) \(\mathrm{I}_{2}(g)+5 \mathrm{~F}_{2}(g) \rightleftharpoons 2 \mathrm{IF}_{5}(g)\) (b) \(\mathrm{CO}(\mathrm{g})+2 \mathrm{H}_{2}(g) \rightleftharpoons \mathrm{CH}_{3} \mathrm{OH}(l)\) (c) \(2 \mathrm{H}_{2} \mathrm{~S}+3 \mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{H}_{2} \mathrm{O}(l)+2 \mathrm{SO}_{2}(g)\) (d) \(\mathrm{SnO}_{2}(s)+2 \mathrm{H}_{2}(g) \rightleftharpoons \mathrm{Sn}(s)+2 \mathrm{H}_{2} \mathrm{O}(l)\)
See all solutions

Recommended explanations on Chemistry Textbooks

Nuclear Chemistry

Read Explanation

Chemical Reactions

Read Explanation

Chemistry Branches

Read Explanation

Organic Chemistry

Read Explanation

Physical Chemistry

Read Explanation

The Earths Atmosphere

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.

Sign-up for free