Chemistry: Principles and Practice
Chemistry: Principles and Practice
3rd Edition
ISBN: 9780534420123
Author: Daniel L. Reger, Scott R. Goode, David W. Ball, Edward Mercer
Publisher: Cengage Learning
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Chapter 14, Problem 14.61QE
Interpretation Introduction

Interpretation:

The expression for the equilibrium constant and the partial pressure of CO2 for the following reaction has to be determined.

  CaCO3(s)CaO(s)+CO2(g)

Concept Introduction:

The condition of equilibrium is a state of balance of processes that runs in opposite directions. At equilibrium, the formation of a product from the reactant balances the formation of reactant from the product. Also, the change in concentration of reaction and product seems to be negligible at equilibrium state.

The general equilibrium reaction is as follows:

  aA+bBcC+dD

Here,

A and B are the reactants.

C and D are products.

a and b are the stoichiometric coefficients of reactants.

c and d are the stoichiometric coefficients of products.

The expression of the equilibrium constant for the above reaction is as follows:

  Kp=PCcPDdPAaPBb

Here,

Kp is the equilibrium constant.

PC is the partial pressure of C.

PD is the partial pressure of D.

PA is the partial pressure of A.

PB is the partial pressure of B.

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Chapter 14 Solutions

Chemistry: Principles and Practice

Ch. 14 - Explain why terms for pure liquids and solids do...Ch. 14 - Temperature influences solubility. Does...Ch. 14 - Prob. 14.13QECh. 14 - Prob. 14.14QECh. 14 - Prob. 14.15QECh. 14 - Prob. 14.16QECh. 14 - Prob. 14.17QECh. 14 - Prob. 14.18QECh. 14 - At 2000 K, experiments show that the equilibrium...Ch. 14 - At 500 K, the equilibrium constant is 155 for...Ch. 14 - At 77 C, Kp is 1.7 104 for the formation of...Ch. 14 - Consider the following equilibria involving SO2(g)...Ch. 14 - Kc at 137 C is 4.42 for NO(g) + 12 Br2(g) NOBr(g)...Ch. 14 - Prob. 14.24QECh. 14 - Prob. 14.25QECh. 14 - Prob. 14.26QECh. 14 - Prob. 14.27QECh. 14 - Prob. 14.28QECh. 14 - Prob. 14.29QECh. 14 - Prob. 14.30QECh. 14 - Prob. 14.31QECh. 14 - Prob. 14.32QECh. 14 - Prob. 14.33QECh. 14 - Prob. 14.34QECh. 14 - Prob. 14.35QECh. 14 - Consider the system...Ch. 14 - Prob. 14.37QECh. 14 - Prob. 14.38QECh. 14 - Prob. 14.39QECh. 14 - Prob. 14.40QECh. 14 - Prob. 14.41QECh. 14 - Prob. 14.42QECh. 14 - Prob. 14.43QECh. 14 - Prob. 14.44QECh. 14 - Prob. 14.45QECh. 14 - Prob. 14.46QECh. 14 - Prob. 14.47QECh. 14 - Prob. 14.48QECh. 14 - Prob. 14.49QECh. 14 - Prob. 14.50QECh. 14 - Prob. 14.51QECh. 14 - Consider 0.200 mol phosphorus pentachloride sealed...Ch. 14 - Prob. 14.53QECh. 14 - Prob. 14.54QECh. 14 - Prob. 14.55QECh. 14 - Prob. 14.56QECh. 14 - Prob. 14.57QECh. 14 - Prob. 14.58QECh. 14 - Prob. 14.59QECh. 14 - Prob. 14.60QECh. 14 - Prob. 14.61QECh. 14 - Write the expression for the equilibrium constant...Ch. 14 - Prob. 14.63QECh. 14 - Prob. 14.64QECh. 14 - Write the expression for the solubility product...Ch. 14 - Prob. 14.66QECh. 14 - Prob. 14.67QECh. 14 - The solubility of silver iodate, AgIO3, is 1.8 ...Ch. 14 - Prob. 14.69QECh. 14 - Prob. 14.70QECh. 14 - Prob. 14.71QECh. 14 - Prob. 14.72QECh. 14 - Even though barium is toxic, a suspension of...Ch. 14 - Lead poisoning has been a hazard for centuries....Ch. 14 - Calculate the solubility of barium sulfate (Ksp =...Ch. 14 - Calculate the solubility of copper(II) iodate,...Ch. 14 - Calculate the solubility of lead fluoride, PbF2...Ch. 14 - Calculate the solubility of zinc carbonate, ZnCO3...Ch. 14 - Prob. 14.79QECh. 14 - Prob. 14.80QECh. 14 - Use the solubility product constant from Appendix...Ch. 14 - Prob. 14.82QECh. 14 - Some barium chloride is added to a solution that...Ch. 14 - Prob. 14.84QECh. 14 - Prob. 14.85QECh. 14 - Prob. 14.86QECh. 14 - Prob. 14.87QECh. 14 - Prob. 14.88QECh. 14 - Prob. 14.89QECh. 14 - Prob. 14.90QECh. 14 - Prob. 14.91QECh. 14 - At 3000 K, carbon dioxide dissociates CO2(g) ...Ch. 14 - Prob. 14.94QECh. 14 - Nitrogen, hydrogen, and ammonia are in equilibrium...Ch. 14 - The concentration of barium in a saturated...Ch. 14 - According to the Resource Conservation and...Ch. 14 - Prob. 14.98QE
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Solutions: Crash Course Chemistry #27; Author: Crash Course;https://www.youtube.com/watch?v=9h2f1Bjr0p4;License: Standard YouTube License, CC-BY