Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 12: Problem 7

Write equilibrium constant expressions ( \(K\) ) for the following reactions: (a) \(2 \mathrm{NO}_{3}^{-}(a q)+8 \mathrm{H}^{+}(a q)+3 \mathrm{Cu}(s) \rightleftharpoons\) \(2 \mathrm{NO}(g)+3 \mathrm{Cu}^{2+}(a q)+4 \mathrm{H}_{2} \mathrm{O}(l)\) (b) \(2 \mathrm{PbS}(s)+3 \mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{PbO}(s)+2 \mathrm{SO}_{2}(g)\) (c) \(\mathrm{Ca}^{2+}(a q)+\mathrm{CO}_{3}{ }^{2-}(a q) \rightleftharpoons \mathrm{CaCO}_{3}(s)\)

Short Answer

Expert verified
Question: Write the equilibrium constant expressions for the following reactions: a) \(2 \mathrm{NO}_{3}^{-}(a q)+8 \mathrm{H}^{+}(a q)+3 \mathrm{Cu}(s) \rightleftharpoons 2 \mathrm{NO}(g)+3 \mathrm{Cu}^{2+}(a q)+4 \mathrm{H}_{2} \mathrm{O}(l)\) b) \(2 \mathrm{PbS}(s)+3 \mathrm{O}_{2}(g)\rightleftharpoons2\mathrm{PbO}(s)+2 \mathrm{SO}_{2}(g)\) c) \(\mathrm{Ca}^{2+}(a q)+\mathrm{CO}_{3}{ }^{2-}(a q) \rightleftharpoons \mathrm{CaCO}_{3}(s)\) Answer: a) \(K = \frac{[\mathrm{NO}]^{2}[\mathrm{Cu}^{2+}]^{3}}{[\mathrm{NO}_{3}^{-}]^{2}[\mathrm{H}^{+}]^{8}}\) b) \(K = \frac{[\mathrm{SO}_{2}]^{2}}{[\mathrm{O}_{2}]^{3}}\) c) \(K = \frac{1}{[\mathrm{Ca}^{2+}][\mathrm{CO}_{3}{ }^{2-}]}\)

Step by step solution

01

Identify reactants and products

From the balanced chemical equation, we have: - Reactants: \(2 \mathrm{NO}_{3}^{-}(a q)\), \(8 \mathrm{H}^{+}(a q)\), and \(3 \mathrm{Cu}(s)\). - Products: \(2 \mathrm{NO}(g)\), \(3 \mathrm{Cu}^{2+}(a q)\), and \(4 \mathrm{H}_{2} \mathrm{O}(l)\).
02

Write the equilibrium expression

According to the definition of \(K\) explained earlier, the equilibrium expression for this reaction is: \(K = \frac{[\mathrm{NO}]^{2}[\mathrm{Cu}^{2+}]^{3}}{[\mathrm{NO}_{3}^{-}]^{2}[\mathrm{H}^{+}]^{8}}\) _b)# Equilibrium Constant Expression for Reaction (b): Here is the reaction: \(2 \mathrm{PbS}(s)+3 \mathrm{O}_{2}(g)\rightleftharpoons2\mathrm{PbO}(s)+2 \mathrm{SO}_{2}(g)\) Now, let's find the equilibrium constant expression by following the same steps:
03

Identify reactants and products

From the balanced chemical equation, we have: - Reactants: \(2 \mathrm{PbS}(s)\) and \(3 \mathrm{O}_{2}(g)\). - Products: \(2\mathrm{PbO}(s)\) and \(2 \mathrm{SO}_{2}(g)\).
04

Write the equilibrium expression

According to the definition of \(K\) explained earlier, the equilibrium expression for this reaction is: \(K = \frac{[\mathrm{SO}_{2}]^{2}}{[\mathrm{O}_{2}]^{3}}\) _c)# Equilibrium Constant Expression for Reaction (c): Here is the reaction: \(\mathrm{Ca}^{2+}(a q)+\mathrm{CO}_{3}{ }^{2-}(a q) \rightleftharpoons \mathrm{CaCO}_{3}(s)\) Now, let's find the equilibrium constant expression by following the same steps:
05

Identify reactants and products

From the balanced chemical equation, we have: - Reactants: \(\mathrm{Ca}^{2+}(a q)\) and \(\mathrm{CO}_{3}{ }^{2-}(a q)\). - Product: \(\mathrm{CaCO}_{3}(s)\).
06

Write the equilibrium expression

According to the definition of \(K\) explained earlier, the equilibrium expression for this reaction is: \(K = \frac{1}{[\mathrm{Ca}^{2+}][\mathrm{CO}_{3}{ }^{2-}]}\) Note that the solid product \(\mathrm{CaCO}_{3}(s)\) is not included in the equilibrium expression.

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

Benzaldehyde, a flavoring agent, is obtained by the dehydrogenation of benzyl alcohol. $$\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{2} \mathrm{OH}(g) \rightleftharpoons \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CHO}(g)+\mathrm{H}_{2}(g)$$ \(K\) for the reaction at \(250^{\circ} \mathrm{C}\) is \(0.56\). If \(1.50 \mathrm{~g}\) of benzyl alcohol is placed in a 2.0-L flask and heated to \(250^{\circ} \mathrm{C}\), (a) what is the partial pressure of the benzaldehyde when equilibrium is established? (b) how many grams of benzyl alcohol remain at equilibrium?

Solid ammonium iodide decomposes to ammonia and hydrogen iodide gases at sufficiently high temperatures. $$\mathrm{NH}_{4} \mathrm{I}(s) \rightleftharpoons \mathrm{NH}_{3}(g)+\mathrm{HI}(g)$$ The equilibrium constant for the decomposition at \(673 \mathrm{~K}\) is \(0.215\). Fifteen grams of ammonium iodide are sealed in a \(5.0\) -L flask and heated to \(673 \mathrm{~K}\). (a) What is the total pressure in the flask at equilibrium? (b) How much ammonium iodide decomposes?

Hemoglobin (Hb) binds to both oxygen and carbon monoxide. When the carbon monoxide replaces the oxygen in an organism, the following reaction occurs: $$\mathrm{HbO}_{2}(a q)+\mathrm{CO}(g) \rightleftharpoons \mathrm{HbCO}(a q)+\mathrm{O}_{2}(g)$$ At \(37^{\circ} \mathrm{C}, K\) is about 200 . When equal concentrations of \(\mathrm{HbO}_{2}\) and \(\mathrm{HbCO}\) are present, the effect of CO inhalation is fatal. Assuming \(\mathrm{P}_{\mathrm{O}_{2}}=0.21 \mathrm{~atm}\), what is \(\mathrm{P}_{\mathrm{CO}}\) when \(\left[\mathrm{HbO}_{2}\right]=[\mathrm{HbCO}] ?\)

For the following reactions, predict whether the pressure of the reactants or products increases or remains the same when the volume of the reaction vessel is increased. (a) \(\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{H}_{2} \mathrm{O}(g)\) (b) \(\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g)\) (c) \(\mathrm{C}_{2} \mathrm{H}_{4}(g)+\mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(g)\)

At a certain temperature, nitrogen and oxygen gases combine to form nitrogen oxide gas. $$\mathrm{N}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NO}(g)$$ When equilibrium is established, the partial pressures of the gases are: \(P_{\mathrm{N}_{2}}=\) \(1.2 \mathrm{~atm}, P_{\mathrm{O}_{2}}=0.80 \mathrm{~atm}, P_{\mathrm{NO}}=0.022 \mathrm{~atm} .\) (a) Calculate \(K\) at the temperature of the reaction. (b) After equilibrium is reached, more oxygen is added to make its partial pressure \(1.2\) atm. Calculate the partial pressure of all gases when equilibrium is reestablished.
See all solutions

Recommended explanations on Chemistry Textbooks

Chemical Reactions

Read Explanation

Chemical Analysis

Read Explanation

Kinetics

Read Explanation

Physical Chemistry

Read Explanation

Chemistry Branches

Read Explanation

Organic 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.

Sign-up for free