Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 15: Problem 27

The following equilibria were attained at \(823 \mathrm{~K}\) : $$\begin{aligned} \mathrm{CoO}(s)+\mathrm{H}_{2}(g) & \rightleftharpoons \mathrm{Co}(s)+\mathrm{H}_{2} \mathrm{O}(g) & K_{c} &=67 \\ \mathrm{CoO}(s)+\mathrm{CO}(g) & \rightleftharpoons \mathrm{Co}(s)+\mathrm{CO}_{2}(g) & K_{c} &=490 \end{aligned}$$ Based on these equilibria, calculate the equilibrium constant $$\text { for } \mathrm{H}_{2}(g)+\mathrm{CO}_{2}(g) \rightleftharpoons \mathrm{CO}(g)+\mathrm{H}_{2} \mathrm{O}(g) \text { at } 823 \mathrm{~K} \text { . }$$

Short Answer

Expert verified
The equilibrium constant for the reaction \(\mathrm{H}_{2}(g)+\mathrm{CO}_{2}(g) \rightleftharpoons \mathrm{CO}(g)+\mathrm{H}_{2} \mathrm{O}(g)\) at 823 K can be calculated by combining the given equilibria and their constants: \(K_{c} = \frac{K_{c_{2}}}{K_{c_{1}}} = \frac{490}{67} \approx 7.31\).

Step by step solution

01

Write down the given equilibria with their equilibrium constants

We are given two equilibria with their equilibrium constants at 823 K: Equilibrium 1: \[\mathrm{CoO}(s)+\mathrm{H}_{2}(g) \rightleftharpoons \mathrm{Co}(s)+\mathrm{H}_{2} \mathrm{O}(g) \quad K_{c_{1}}=67\] Equilibrium 2: \[\mathrm{CoO}(s)+\mathrm{CO}(g) \rightleftharpoons \mathrm{Co}(s)+\mathrm{CO}_{2}(g) \quad K_{c_{2}}=490\] We want to find the equilibrium constant for the following reaction: \[\mathrm{H}_{2}(g)+\mathrm{CO}_{2}(g) \rightleftharpoons \mathrm{CO}(g)+\mathrm{H}_{2} \mathrm{O}(g)\]
02

Combine the given equilibria to obtain the desired reaction

In order to get the desired reaction, we need to subtract the first equilibrium from the second equilibrium: \[(\mathrm{CoO}(s)+\mathrm{CO}(g) \rightleftharpoons \mathrm{Co}(s)+\mathrm{CO}_{2}(g)) - (\mathrm{CoO}(s)+\mathrm{H}_{2}(g) \rightleftharpoons \mathrm{Co}(s)+\mathrm{H}_{2} \mathrm{O}(g))\] When we subtract the two equilibria, we get: \[\mathrm{H}_{2}(g)+\mathrm{CO}_{2}(g) \rightleftharpoons \mathrm{CO}(g)+\mathrm{H}_{2} \mathrm{O}(g)\]
03

Combine the equilibrium constants for the combined reaction

Now that we have combined the reactions, we need to combine their equilibrium constants. When we subtract two chemical reactions, we divide their equilibrium constants: \[K_{c} = \frac{K_{c_{2}}}{K_{c_{1}}}\] Using the given equilibrium constants, we calculate the equilibrium constant for the desired reaction: \[K_{c} = \frac{490}{67} \approx 7.31\]
04

Report the answer

The equilibrium constant for the reaction \(\mathrm{H}_{2}(g)+\mathrm{CO}_{2}(g) \rightleftharpoons \mathrm{CO}(g)+\mathrm{H}_{2} \mathrm{O}(g)\) at 823 K is approximately 7.31.

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

Water molecules in the atmosphere can form hydrogenbonded dimers, \(\left(\mathrm{H}_{2} \mathrm{O}\right)_{2} .\) The presence of these dimers is thought to be important in the nucleation of ice crystals in the atmosphere and in the formation of acid rain. (a) Using VSEPR theory, draw the structure of a water dimer, using dashed lines to indicate intermolecular interactions. (b) What kind of intermolecular forces are involved in water dimer formation? (c) The \(K_{p}\) for water dimer formation in the gas phase is 0.050 at \(300 \mathrm{~K}\) and 0.020 at \(350 \mathrm{~K}\). Is water dimer formation endothermic or exothermic?

Consider the following equilibrium, for which \(K_{p}=0.0752\) at $$\begin{array}{l} 480{ }^{\circ} \mathrm{C}: \\ \quad 2 \mathrm{Cl}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons 4 \mathrm{HCl}(g)+\mathrm{O}_{2}(g) \end{array}$$ (a) What is the value of \(K_{p}\) for the reaction \(4 \mathrm{HCl}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{Cl}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(g) ?\) (b) What is the value of \(K_{p}\) for the reaction \(\mathrm{Cl}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons\) \(2 \mathrm{HCl}(g)+\frac{1}{2} \mathrm{O}_{2}(g) ?(\mathrm{c})\) What is the value of \(K_{c}\) for the reac- tion in part (b)?

Mercury(I) oxide decomposes into elemental mercury and elemental oxygen: \(2 \mathrm{Hg}_{2} \mathrm{O}(s) \rightleftharpoons 4 \mathrm{Hg}(l)+\mathrm{O}_{2}(g) .\) Write the equilibrium-constant expression for this reaction in terms of partial pressures. (b) Suppose you run this reaction in a solvent that dissolves elemental mercury and elemental oxygen. Rewrite the equilibrium- constant expression in terms of molarities for the reaction, using (solv) to indicate solvation.

The equilibrium constant \(K_{c}\) for \(\mathrm{C}(s)+\mathrm{CO}_{2}(g) \rightleftharpoons\) \(2 \mathrm{CO}(g)\) is 1.9 at \(1000 \mathrm{~K}\) and 0.133 at \(298 \mathrm{~K}\). (a) If excess \(\mathrm{C}\) is allowed to react with \(25.0 \mathrm{~g}\) of \(\mathrm{CO}_{2}\) in a \(3.00-\mathrm{L}\) vessel at \(1000 \mathrm{~K}\), how many grams of \(\mathrm{CO}\) are produced? (b) How many grams of \(\mathrm{C}\) are consumed? \((\mathrm{c})\) If a smaller vessel is used for the reaction, will the yield of CO be greater or smaller? (d) Is the reaction endothermic or exothermic?

At \(373 \mathrm{~K}, K_{p}=0.416\) for the equilibrium $$2 \mathrm{NOBr}(g) \rightleftharpoons 2 \mathrm{NO}(g)+\mathrm{Br}_{2}(g)$$ If the pressures of \(\mathrm{NOBr}(g)\) and \(\mathrm{NO}(g)\) are equal, what is the equilibrium pressure of \(\mathrm{Br}_{2}(g)\) ?
See all solutions

Recommended explanations on Chemistry Textbooks

Nuclear Chemistry

Read Explanation

The Earths Atmosphere

Read Explanation

Making Measurements

Read Explanation

Kinetics

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Chemical Reactions

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