General Chemistry
General Chemistry
7th Edition
ISBN: 9780073402758
Author: Chang, Raymond/ Goldsby
Publisher: McGraw-Hill College
bartleby

Videos

Question
Book Icon
Chapter 15, Problem 15.36QP
Interpretation Introduction

Interpretation:

The equilibrium concentration (Kc) should be calculated to given the statement of equilibrium reaction at 686οC.

Concept Introduction:

Chemical equilibrium: The term applied to reversible chemical reactions. It is the point at which the rate of the forward reaction is equal to the rate of the reverse reaction. The equilibrium is achieved; the concentrations of reactant and products become constant.

Kp and Kc: This equilibrium constants of gaseous mixtures, these difference between the two constants is that Kc is defined by molar concentrations, whereas Kp is defined by the partial pressures of the gasses inside a closed system.

Equilibrium constant: Concentration of the products to the respective molar concentration of reactants it is called equilibrium constant. If the K value is less than one the reaction will move to the left side and the K values is higher (or) greater than one the reaction will move to the right side of reaction.

Heterogeneous equilibrium: This equilibrium reaction does not depend on the amounts of pure solid and liquid present, in other words heterogeneous equilibrium, substances are in different phases.

Expert Solution & Answer
Check Mark

Explanation of Solution

To find: Calculate the (Kc) values for given the statement of equilibrium reaction (a).

Calculate and analyze the (Kc) values at 6860C.

The consider the following expression of (Kc)

CO2(g)+H2(g)CO2(g)+H2O(g)Kc=[H2O][CO2][CO2][H2][1]Thegivenstatemntofmolarvaluessubstitutedforaboveequation=(0.040)(0.050)(0.086)(0.045)=0.52

The simple equilibrium constant is derived showed above equation (1).

To find: Calculate the molar values for given the statement of equilibrium reaction (b).

Calculate and analyze the molar values at 6860C.

First we derived for reaction quotient (Qc) values for given equilibrium reaction.

CO2(g)+H2(g)CO2(g)+H2O(g)Qc=[H2O][CO2][CO2][H2]The concentartion of CO2 =0.50mol/LThis values are substitued for above (Kc) equlibrium reactionThegivenstatemntofmolarvaluessubstitutedforaboveequationQc=(0.040)(0.050)(0.50)(0.045)=0.089Qc=0.089

Calculate the molar concentration of the given equilibrium reaction.

CO2(g)+H2(g)CO(g)H2O(g)Initial (M): 0.500.0450.0500.040Change (M):   xx+x+xEqilibrium (M):(0.50x)(0.045x)(0.050+x)(0.040+x)Theequlibriumconstant(Kc)followedbyKc=[H2O][CO][H2][CO2]=[Product][Reactant][1]Calculatetheequlibriumvalueof(Kc)Given the respactive molar values Kc=[H2O][CO][H2][CO2]=(0.040+x)(0.050+x)(0.50x)(0.045x)=0.52[2]Rewrite the equation (2)0.52 (x2-0.545x+0.0225)=x2+(0.090x+0.0020)0.48x2+0.373x(9.7×103)=0Herex=(0.853)(9.7×103)x=0.025The equlibrium concentration areReactant [CO2]=(0.50-0.025)M=0.48M[H2]=(0.045-0.025)M=0.020MProduct[CO]=(0.050+0.025)M=0.075M[H2O]=(0.040+0.025)M=0.065M

The given equilibrium process, the equal moles of H2 and CO2 reacted with in gas phase conditions to give a H2O and CO molecule, the reactant moles 0.50m/L. Then the number of moles calculating (Kc) molar concentration of reacting procedure as fallows, first construct an above equilibrium table and fill in the initial concentration and zero also include the respective table than we use the initial concentration to calculate the reaction quotient (Qr) and compare (K) values to identified the direction in which the reaction will proceeds. Further calculate (x) as the amount of particular species consumed and use the stoichiometry of the reaction to define in terms of (x) the amount of the species produced. Finally the each species in the equilibrium add the change in concentration to the initial concentration to get the equilibrium concentration. The derived molar concentration values are showed above.

Conclusion

The product molecule molar concentration is derived given the equilibrium concentration (Kc) and respective equilibrium reaction at 686οC.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Chapter 15 Solutions

General Chemistry

Ch. 15.4 - Practice Exercise At 430°C. the equilibrium...Ch. 15.4 - Prob. 2PECh. 15.4 - Prob. 1RCCh. 15.4 - Prob. 2RCCh. 15.4 - Prob. 3PECh. 15 - Prob. 15.1QPCh. 15 - 15.2 Explain the difference between physical...Ch. 15 - 15.3 Briefly describe the importance of...Ch. 15 - 15.4 Consider the equilibrium system 3A ⇌ B....Ch. 15 - 15.5 Define homogeneous equilibrium and...Ch. 15 - Prob. 15.6QPCh. 15 - 15.7 Write equilibrium constant expressions for...Ch. 15 - 15.8 Write the expressions for the equilibrium...Ch. 15 - 15.9 Write the equilibrium constant expressions...Ch. 15 - 15.10 Write the equation relating Kc and KP and...Ch. 15 - Prob. 15.11QPCh. 15 - Prob. 15.12QPCh. 15 - 15.13 The equilibrium constant (Kc) for the...Ch. 15 - Prob. 15.14QPCh. 15 - 15.15 What is the KP at 1273°C for the...Ch. 15 - 15.16 The equilibrium constant KP for the...Ch. 15 - Prob. 15.17QPCh. 15 - Prob. 15.18QPCh. 15 - Prob. 15.19QPCh. 15 - Prob. 15.20QPCh. 15 - Prob. 15.21QPCh. 15 - 15.22 Ammonium carbamate, NH4CO2NH2, decomposes...Ch. 15 - Prob. 15.23QPCh. 15 - 15.24 Pure phosgene gas (COCl2), 3.00 × 10−2...Ch. 15 - Prob. 15.25QPCh. 15 - 15.26 A 2.50-mol quantity of NOCl was initially...Ch. 15 - 15.27 Define reaction quotient. How does it...Ch. 15 - Prob. 15.28QPCh. 15 - Prob. 15.29QPCh. 15 - Prob. 15.30QPCh. 15 - Prob. 15.31QPCh. 15 - Prob. 15.32QPCh. 15 - Prob. 15.33QPCh. 15 - Prob. 15.34QPCh. 15 - Prob. 15.35QPCh. 15 - Prob. 15.36QPCh. 15 - Prob. 15.37QPCh. 15 - Prob. 15.38QPCh. 15 - Prob. 15.39QPCh. 15 - Prob. 15.40QPCh. 15 - Prob. 15.41QPCh. 15 - Prob. 15.42QPCh. 15 - Prob. 15.43QPCh. 15 - Prob. 15.44QPCh. 15 - Prob. 15.45QPCh. 15 - 15.46 What effect does an increase in pressure...Ch. 15 - Prob. 15.47QPCh. 15 - Prob. 15.48QPCh. 15 - 15.49 Consider the reaction Comment on the...Ch. 15 - Prob. 15.50QPCh. 15 - Prob. 15.51QPCh. 15 - Prob. 15.53QPCh. 15 - Prob. 15.54QPCh. 15 - Prob. 15.55QPCh. 15 - Prob. 15.56QPCh. 15 - Prob. 15.57QPCh. 15 - 15.58 Baking soda (sodium bicarbonate) undergoes...Ch. 15 - 15.59 Consider the following reaction at...Ch. 15 - Prob. 15.60QPCh. 15 - Prob. 15.61QPCh. 15 - Prob. 15.62QPCh. 15 - Prob. 15.64QPCh. 15 - Prob. 15.65QPCh. 15 - Prob. 15.66QPCh. 15 - Prob. 15.67QPCh. 15 - Prob. 15.68QPCh. 15 - Prob. 15.69QPCh. 15 - Prob. 15.70QPCh. 15 - Prob. 15.71QPCh. 15 - Prob. 15.72QPCh. 15 - Prob. 15.73QPCh. 15 - Prob. 15.74QPCh. 15 - Prob. 15.75QPCh. 15 - Prob. 15.76QPCh. 15 - Prob. 15.78QPCh. 15 - Prob. 15.79QPCh. 15 - Prob. 15.81QPCh. 15 - Prob. 15.82QPCh. 15 - Prob. 15.83QPCh. 15 - Prob. 15.84QPCh. 15 - Prob. 15.85QPCh. 15 - Prob. 15.86QPCh. 15 - Prob. 15.89QPCh. 15 - Prob. 15.90QPCh. 15 - Prob. 15.91QPCh. 15 - Prob. 15.92QPCh. 15 - Prob. 15.93QPCh. 15 - 15.94 Consider the decomposition of ammonium...Ch. 15 - Prob. 15.95QPCh. 15 - 15.96 In 1899 the German chemist Ludwig Mond...Ch. 15 - Prob. 15.98QPCh. 15 - Prob. 15.99QPCh. 15 - Prob. 15.100QPCh. 15 - Prob. 15.101QPCh. 15 - Prob. 15.102QPCh. 15 - Prob. 15.103SPCh. 15 - Prob. 15.104SPCh. 15 - Prob. 15.105SPCh. 15 - Prob. 15.106SPCh. 15 - Prob. 15.107SPCh. 15 - Prob. 15.110SPCh. 15 - Prob. 15.111SPCh. 15 - Prob. 15.112SP
Knowledge Booster
Background pattern image
Chemistry
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, chemistry and related others by exploring similar questions and additional content below.
Recommended textbooks for you
Text book image
Chemistry
Chemistry
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Cengage Learning
Text book image
Chemistry
Chemistry
ISBN:9781259911156
Author:Raymond Chang Dr., Jason Overby Professor
Publisher:McGraw-Hill Education
Text book image
Principles of Instrumental Analysis
Chemistry
ISBN:9781305577213
Author:Douglas A. Skoog, F. James Holler, Stanley R. Crouch
Publisher:Cengage Learning
Text book image
Organic Chemistry
Chemistry
ISBN:9780078021558
Author:Janice Gorzynski Smith Dr.
Publisher:McGraw-Hill Education
Text book image
Chemistry: Principles and Reactions
Chemistry
ISBN:9781305079373
Author:William L. Masterton, Cecile N. Hurley
Publisher:Cengage Learning
Text book image
Elementary Principles of Chemical Processes, Bind...
Chemistry
ISBN:9781118431221
Author:Richard M. Felder, Ronald W. Rousseau, Lisa G. Bullard
Publisher:WILEY
Chemical Equilibria and Reaction Quotients; Author: Professor Dave Explains;https://www.youtube.com/watch?v=1GiZzCzmO5Q;License: Standard YouTube License, CC-BY