Chemistry: The Molecular Science
Chemistry: The Molecular Science
5th Edition
ISBN: 9781285199047
Author: John W. Moore, Conrad L. Stanitski
Publisher: Cengage Learning
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Chapter 12, Problem 63QRT
Interpretation Introduction

Interpretation:

Equilibrium reaction of CO with C(graphite)andCO2 is given, the observation has to be given when 3.5molCO and 3.5molCO2 are mixed in a 1.5L sealed graphite container.

Concept Introduction:

Equilibrium constant(Kc):

Equilibrium constant (Kc) is the ratio of the rate constants of the forward and reverse reactions at a given temperature.  In other words it is the ratio of the concentrations of the products to concentrations of the reactants.  Each concentration term is raised to a power, which is same as the coefficients in the chemical reaction.

Consider the reaction where A reacts to give B.

    aAbB

    Rate of forward reaction = Rate of reverse reactionkf[A]a=kr[B]b

On rearranging,

    [B]b[A]a=kfkr=Kc

Where,

    kf is the rate constant of the forward reaction.

    kr is the rate constant of the reverse reaction.

    Kc is the equilibrium constant.

Expert Solution & Answer
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Explanation of Solution

Given information,

  C(graphite)+CO2(g)2CO(g)KC=3.7×10-23at25oCCO=3.5molandCO2=3.5molVolumeofcontainer=1.5L

Calculate the value of QC,

  Conc.CO=Conc.CO2=3.5mol1.5L=2.3MQC=(Conc.CO)2Conc.CO2=(2.3)22.3=2.3

Compare the value of QC and KC,

  QC>KC

So the given reaction must proceeds toward reactants to reach equilibrium condition.  So first take the reaction all the way to reactants as much as stoichiometrically possible, after that take it back to equilibrium with a very small x.

Construct ICE table,

  C(s)+CO2(g)2CO(g)Initialconc.(M)solid2.32.3changegoingtoreactants(M)+some+2.3/2-2.3finalconc(M)solid+some3.450changeasreactionequilibrates(M)-some-x+2xEquilibriumconc.(M)solid3.45-x2x

At equilibrium,

  KC=[CO]2[CO2]=(2x)2(3.45+x)=3.7×10-23

Assume, 3.45>>x, so 3.45+x=3.45

Solve for x,

  (2x)23.45=3.7×10-234x2=1.3×10-23;therefore,x=5.6×10-12M

Calculate the concentration of the substances,

  [CO]=2x=2×(5.6×10-12M)=1.1×10-11MmolesofCO2=3.45M×1.5L=5.2mol[CO]=1.1×10-11M×1.5L=2.5×10-11molCO2

Almost all the CO converts to CO2, depositing some new solid graphite in the container, leaving 5.2moles of CO and 2.5×10-11molesofCO2.

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Chapter 12 Solutions

Chemistry: The Molecular Science

Ch. 12.5 - For the equilibrium 2 SO2(g) + O2(g) 2 SO3(g) Kc...Ch. 12.5 - Prob. 12.7CECh. 12.5 - Prob. 12.6PSPCh. 12.5 - Prob. 12.7PSPCh. 12.6 - Prob. 12.8CECh. 12.6 - Prob. 12.9ECh. 12.6 - Prob. 12.10CECh. 12.6 - Prob. 12.8PSPCh. 12.7 - For the ammonia synthesis reaction ⇌ Does the...Ch. 12.8 - Prob. 12.13CECh. 12 - Prob. 1QRTCh. 12 - Prob. 2QRTCh. 12 - Prob. 3QRTCh. 12 - Decomposition of ammonium dichromate is shown in...Ch. 12 - For the equilibrium reaction in Question 4, write...Ch. 12 - Indicate whether each statement below is true or...Ch. 12 - Prob. 7QRTCh. 12 - Prob. 8QRTCh. 12 - Prob. 9QRTCh. 12 - Prob. 10QRTCh. 12 - The atmosphere consists of about 80% N2 and 20%...Ch. 12 - Prob. 12QRTCh. 12 - Prob. 13QRTCh. 12 - Prob. 14QRTCh. 12 - Prob. 15QRTCh. 12 - Prob. 16QRTCh. 12 - Prob. 17QRTCh. 12 - Prob. 18QRTCh. 12 - Prob. 19QRTCh. 12 - Prob. 20QRTCh. 12 - Prob. 21QRTCh. 12 - Prob. 22QRTCh. 12 - Prob. 23QRTCh. 12 - Prob. 24QRTCh. 12 - Prob. 25QRTCh. 12 - Prob. 26QRTCh. 12 - Prob. 27QRTCh. 12 - Prob. 28QRTCh. 12 - Prob. 29QRTCh. 12 - Prob. 30QRTCh. 12 - Given these data at a certain temperature,...Ch. 12 - The vapor pressure of water at 80. C is 0.467 atm....Ch. 12 - Prob. 33QRTCh. 12 - Prob. 34QRTCh. 12 - Prob. 35QRTCh. 12 - Prob. 36QRTCh. 12 - Carbon dioxide reacts with carbon to give carbon...Ch. 12 - Prob. 38QRTCh. 12 - Prob. 39QRTCh. 12 - Prob. 40QRTCh. 12 - Nitrosyl chloride, NOC1, decomposes to NO and Cl2...Ch. 12 - Suppose 0.086 mol Br2 is placed in a 1.26-L flask....Ch. 12 - Prob. 43QRTCh. 12 - Prob. 44QRTCh. 12 - Prob. 45QRTCh. 12 - Using the data of Table 12.1, predict which of...Ch. 12 - Prob. 47QRTCh. 12 - The equilibrium constants for dissolving silver...Ch. 12 - Prob. 49QRTCh. 12 - Prob. 50QRTCh. 12 - At room temperature, the equilibrium constant Kc...Ch. 12 - Prob. 52QRTCh. 12 - Consider the equilibrium N2(g)+O2(g)2NO(g) At 2300...Ch. 12 - The equilibrium constant, Kc, for the reaction...Ch. 12 - Prob. 55QRTCh. 12 - Prob. 56QRTCh. 12 - Prob. 57QRTCh. 12 - At 503 K the equilibrium constant Kc for the...Ch. 12 - Prob. 59QRTCh. 12 - Prob. 60QRTCh. 12 - Prob. 61QRTCh. 12 - Prob. 62QRTCh. 12 - Prob. 63QRTCh. 12 - Prob. 64QRTCh. 12 - Prob. 65QRTCh. 12 - Prob. 66QRTCh. 12 - Prob. 67QRTCh. 12 - Hydrogen, bromine, and HBr in the gas phase are in...Ch. 12 - Prob. 69QRTCh. 12 - Prob. 70QRTCh. 12 - Prob. 71QRTCh. 12 - Prob. 72QRTCh. 12 - Prob. 73QRTCh. 12 - Prob. 74QRTCh. 12 - Consider the system 4 NH3(g) + 3 O2(g) ⇌ 2 N2(g) +...Ch. 12 - Prob. 76QRTCh. 12 - Predict whether the equilibrium for the...Ch. 12 - Prob. 78QRTCh. 12 - Prob. 79QRTCh. 12 - Prob. 80QRTCh. 12 - Prob. 81QRTCh. 12 - Prob. 82QRTCh. 12 - Prob. 83QRTCh. 12 - Prob. 84QRTCh. 12 - Prob. 85QRTCh. 12 - Prob. 86QRTCh. 12 - Prob. 87QRTCh. 12 - Consider the decomposition of ammonium hydrogen...Ch. 12 - Prob. 89QRTCh. 12 - Prob. 90QRTCh. 12 - Prob. 91QRTCh. 12 - Prob. 92QRTCh. 12 - Prob. 93QRTCh. 12 - Prob. 94QRTCh. 12 - Prob. 95QRTCh. 12 - Prob. 96QRTCh. 12 - Prob. 97QRTCh. 12 - Prob. 98QRTCh. 12 - Prob. 99QRTCh. 12 - Prob. 100QRTCh. 12 - Two molecules of A react to form one molecule of...Ch. 12 - Prob. 102QRTCh. 12 - In Table 12.1 (←Sec. 12-3a) the equilibrium...Ch. 12 - Prob. 104QRTCh. 12 - Prob. 105QRTCh. 12 - Prob. 106QRTCh. 12 - Prob. 107QRTCh. 12 - Which of the diagrams for Questions 107 and 108...Ch. 12 - Draw a nanoscale (particulate) level diagram for...Ch. 12 - The diagram represents an equilibrium mixture for...Ch. 12 - The equilibrium constant, Kc, is 1.05 at 350 K for...Ch. 12 - For the reaction in Question 111, which diagram...Ch. 12 - Prob. 113QRTCh. 12 - Prob. 114QRTCh. 12 - Prob. 115QRTCh. 12 - For the equilibrium...Ch. 12 - Prob. 117QRTCh. 12 - Prob. 119QRTCh. 12 - Prob. 120QRTCh. 12 - When a mixture of hydrogen and bromine is...Ch. 12 - Prob. 122QRTCh. 12 - Prob. 123QRTCh. 12 - Prob. 124QRTCh. 12 - Prob. 125QRTCh. 12 - Prob. 12.ACPCh. 12 - Prob. 12.BCPCh. 12 - Prob. 12.CCPCh. 12 - Prob. 12.DCPCh. 12 - Prob. 12.ECPCh. 12 - Prob. 12.FCP
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