Chapter 14Chemical EquilibriumChemical Equilibrium

““Old Chemists Never Die; they just reachOld Chemists Never Die; they just reach

EQUILIBRIUM!EQUILIBRIUM!””

All physical and chemical changes All physical and chemical changes TENDTEND toward a toward a state of equilibrium.state of equilibrium.

A (l) A (l) A (g)A (g)

A (s) A (s) A (l)A (l)

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Dynamic EquilibriumDynamic EquilibriumThe net result of a dynamic equilibrium is that no The net result of a dynamic equilibrium is that no change in the system is evident.change in the system is evident.

Le Chatelier’s Principle - Le Chatelier’s Principle - If a change is made in aIf a change is made in asystem at equilibrium, the equilibrium will shift in suchsystem at equilibrium, the equilibrium will shift in sucha way so as to reduce the effect of the change.a way so as to reduce the effect of the change.

ApplyPressure

Pressure applied to the system at equilibrium caused it toPressure applied to the system at equilibrium caused it toshift until a new equilibrium was established.shift until a new equilibrium was established.

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Dynamic EquilibriumDynamic Equilibrium

EvaporationOpen System(No Equilibrium)

Evaporation

Liquid GasLiquid Gas(No Equilibrium) (No Equilibrium)

Liquid GasLiquid Gas Liquid GasLiquid Gas(Equilibrium)(Equilibrium)

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Dynamic EquilibriumDynamic Equilibrium

AgAg + + + Cl + Cl -- AgCl (s)AgCl (s)Chemical

Equilibrium

AgAg + + Cl Cl --

Cl Cl -- Ag Ag ++

AgCl (s)

Rate of Precipitation = Rate of DissolvingRate of Precipitation = Rate of Dissolving

HCHC22HH33OO22 (aq) (aq) HH + + + C + C22HH33OO22 --

Rate of dissociation (ionization) = Rate of AssociationRate of dissociation (ionization) = Rate of Association

HCHC22HH33OO22

HH + + C C22HH33OO22 --

HCHC22HH33OO22

CC22HH33OO22 --

HH + +

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CHEM 1108 Lab ExperimentCHEM 1108 Lab Experiment

HCHC22HH33OO2 2 H H++ + C + C22HH33OO22--

RedRedOrangeOrange

[CoCl[CoCl44]]2-2- + 6 H + 6 H22O (l)O (l) Co(HCo(H22O)O)662+2+ + 4 Cl + 4 Cl--

PinkPinkBlueBlue

NHNH44Cl (s) Cl (s) NHNH 44

++ + Cl + Cl--

WhiteWhite Colorless SolutionColorless Solution

You can actually “see” the equilibrium shift!You can actually “see” the equilibrium shift!

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Reversible ReactionsReversible Reactions

NN22OO44 (g) (g) 2 NO2 NO22 (g) (g)RR11

2 NO2 NO22 (g) (g) NN22OO44 (g) (g)RR22

NN22OO44 (g) (g) 2 NO2 NO22 (g) (g)RR11

RR22

[R[R11 = R = R22]]

Homogeneous EquilibriumHomogeneous Equilibrium

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Reversible ReactionsReversible Reactions

Exp. 1Exp. 1 0.0250 M0.0250 M 0.0 M0.0 M 0.0202 M 0.009 66 M 0.0202 M 0.009 66 MExp. 2Exp. 2 0.0150 M0.0150 M 0.0125 M 0.0146 M 0.008 23 M0.0125 M 0.0146 M 0.008 23 MExp. 3Exp. 3 0.0 M0.0 M 0.0250 M 0.0923 M 0.006 54 M0.0250 M 0.0923 M 0.006 54 M

[N[N22OO44]]ii [NO [NO22]]ii [N [N22OO44]]eq eq [NO[NO22]]eqeq

NN22OO44 (g) (g) 2 NO2 NO22 (g) (g)

QQCC = [NO = [NO22]]22

[N[N22OO44]]

Reaction QuotientReaction Quotient

KKCC = [NO = [NO22]]22eqeq

[N[N22OO44]]eqeq

Equilibrium ConstantEquilibrium Constant

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Equilibrium ConstantsEquilibrium ConstantsEquilibrium Constant - Equilibrium Constant - When the rates of the forwardWhen the rates of the forwardand reverse reactions are equal, the system is “at equil-and reverse reactions are equal, the system is “at equil-ibrium” and the ibrium” and the reaction quotient reaction quotient = = equilibriumequilibriumconstantconstant..

Experiment 1 KExperiment 1 KCC = [0.009 66] = [0.009 66]22/[0.0202] = /[0.0202] = 0.004 62 M0.004 62 M

Experiment 2 KExperiment 2 KCC = [0.008 23] = [0.008 23]22/[0.0146] = /[0.0146] = 0.004 64 M0.004 64 M

Experiment 3 KExperiment 3 KCC = [0.006 54] = [0.006 54]22/[0.009 23] = /[0.009 23] = 0.004 63 M0.004 63 M

aA + bB aA + bB cC + dDcC + dD

KKCC = = [C][C]cc[D][D]dd

[A][A]aa[B][B]bb

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Equilibrium ConstantsEquilibrium Constants

HH22 (g) + I (g) + I22 (g) (g) 2 HI (g)2 HI (g)

KKCC = = [HI][HI]22eqeq

[H[H22]]eqeq[I[I22]]eqeq

QQCC = = [HI][HI]22

[H [H22][I][I22]]Exp.Exp. [H[H22]]eqeq [I [I22]]eqeq [HI] [HI]eqeq KKCC

11 0.00291 0.001710.00291 0.00171 0.01648 0.01648 54.5854.5822 0.00356 0.001250.00356 0.00125 0.01559 0.01559 54.6254.6233 0.00225 0.002340.00225 0.00234 0.01685 0.01685 53.9353.9344 0.00183 0.003130.00183 0.00313 0.01767 0.01767 54.5154.5155 0.00114 0.001140.00114 0.00114 0.00841 0.00841 54.4254.4266 0.00050 0.000500.00050 0.00050 0.00366 0.00366 53.5853.58

Median KMedian KCC = = 54.4754.47

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Equilibrium ConstantsEquilibrium Constants

4 NH4 NH33 (g) + 3 O (g) + 3 O22 (g) (g) 2 N2 N22 (g) + 6 H (g) + 6 H22O (g)O (g)

QQCC = =

KKCC = =

[NH[NH33]]44[O[O22]]33

[N[N22]]22[H[H22O]O]66

[N[N22]]eqeq22[H[H22O]O]66

[NH[NH33]]eqeq44[O[O22]]eqeq

33

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Reaction Quotient vs. Reaction Quotient vs. Equilibrium ConstantEquilibrium Constant

Class Problem 14.1 -Class Problem 14.1 - The concentration of NThe concentration of N22OO44 = =

concentration of NOconcentration of NO22 = 0.0125 M in a reaction vessel. = 0.0125 M in a reaction vessel.

The equilibrium constant for The equilibrium constant for NN22OO44 (g) = 2 NO (g) = 2 NO22 (g) (g) isis

0.004 63. Calculate Q0.004 63. Calculate QCC and state which direction the and state which direction the

reaction will go.reaction will go.

Class Problem 14.2 -Class Problem 14.2 - If [OIf [O22] = 0.21 M and [O] = 0.21 M and [O33] = ] =

6.0 x 106.0 x 10-8-8 M, what is the value of the K M, what is the value of the KCC for the for the

equilibrium, equilibrium, 2 O2 O33 (g) = 3 O (g) = 3 O22 (g) (g)??

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Chemical EquilibriumChemical Equilibrium

Heterogeneous Reaction -Heterogeneous Reaction - A reaction that takes place A reaction that takes place in more than one phase or state. These reactions occur in more than one phase or state. These reactions occur at the interface between phases - on the surface of at the interface between phases - on the surface of liquids and solids.liquids and solids.

At a constant temperature, the concentration of a solid At a constant temperature, the concentration of a solid or liquid component remains constant in a or liquid component remains constant in a heterogeneous equilibrium. heterogeneous equilibrium. WHY?WHY?Since the concentration is constant, it can be Since the concentration is constant, it can be considered a part of the equilibrium constant and, considered a part of the equilibrium constant and, thus, does NOT appear in the Kthus, does NOT appear in the KCC expression. expression.

C (s, graphite) + COC (s, graphite) + CO22 (g) = 2 CO (g) (g) = 2 CO (g)KKCC = [CO] = [CO]22

[C][CO[C][CO22]]KKeqeq = [CO] = [CO]22

[CO[CO22]]

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Chemical EquilibriumChemical Equilibrium

Class Problem 14.3 -Class Problem 14.3 - A mixture that was initially 0.005 00 M A mixture that was initially 0.005 00 M in Hin H22 (g) and 0.012 50 M in I (g) and 0.012 50 M in I22 (g), and contained no HI (g), (g), and contained no HI (g),

was heated at 425.4was heated at 425.4ooC until equilibrium was reached. The C until equilibrium was reached. The resulting equilibrium concentration of Iresulting equilibrium concentration of I22 (g) was found to be (g) was found to be

0.007 72 M. What is the value of the K0.007 72 M. What is the value of the KCC for this equilibrium for this equilibrium

at 425.4at 425.4ooC?C?Construct an “ICE” Table:Construct an “ICE” Table:

425.4425.4ooCCEquation:Equation: HH22 (g) + I (g) + I22 (g) (g) 2 HI (g) 2 HI (g)

Initial (I) conc., M Initial (I) conc., M 0.005 000.005 00 0.012 50 0.000 00 0.012 50 0.000 00

Change (C) in conc., MChange (C) in conc., M

Equil. (E) conc., MEquil. (E) conc., M 0.007 720.007 72

- 0.004 78- 0.004 78- 0.004 78- 0.004 78 + 0.009 56+ 0.009 56

- 0.000 22- 0.000 22 + 0.009 56+ 0.009 56

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Chemical EquilibriumChemical EquilibriumCalculate KCalculate KCC::

[HI][HI]22 (0.009 56) (0.009 56)22 [H [H22][I][I22] (0.000 22)] (0.000 22)

(0.007 72)(0.007 72)

==KKCC = = = = 5454

Class Problem 14.4a. -Class Problem 14.4a. - When 1.000 mol each of HWhen 1.000 mol each of H22O (g) O (g)

and CO (g) are introduced into an empty 1.000 L vessel at and CO (g) are introduced into an empty 1.000 L vessel at 959 K and allowed to come to equilibrium, the equilibrium 959 K and allowed to come to equilibrium, the equilibrium mixture contains 0.422 mol Hmixture contains 0.422 mol H22O (g). Find KO (g). Find KCC for for

HH22O (g) + CO (g) O (g) + CO (g) H H22 (g) + CO (g) + CO22 (g) (g)

Construct an “ICE” Table:Construct an “ICE” Table:

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959959ooCCEquation:Equation: HH22O (g) + CO (g) O (g) + CO (g) H H22 (g) + CO (g) + CO22 (g) (g)

I I 1.0001.000 1.000 0.000 1.000 0.000 0.0000.000

CC - 0.578- 0.578 - 0.578 - 0.578 + 0.578 + 0.578 + 0.578+ 0.578

EE 0.4220.422 0.4220.422 0.578 0.578 0.578 0.578

Chemical EquilibriumChemical Equilibrium

KKcc = = [H[H22][CO][CO22]]

[H[H22O][CO]O][CO]= = 1.881.88

Class Problem 14.4b. -Class Problem 14.4b. - Suppose that Suppose that [H[H22O]O]II = 2.00 M and [CO] = 2.00 M and [CO]II = 4.00 M? = 4.00 M?

What are the equilibrium concentrations of the four species?What are the equilibrium concentrations of the four species?

(0.578)(0.578)22

(0.422)(0.422)22==

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959959ooCCEquation:Equation: HH22O (g) + CO (g) O (g) + CO (g) H H22 (g) + CO (g) + CO22 (g) (g)

I I 2.002.00 4.00 0.00 4.00 0.00 0.000.00

CC - x- x - x - x + x + x + x+ x

EE 2.00 - x2.00 - x 4.00 - x 4.00 - x x x x x

Chemical EquilibriumChemical Equilibrium

[H[H22][CO][CO22]]

[H[H22O][CO]O][CO]KKcc = = = = 1.881.88(2.00 – x)(4.00 – x)(2.00 – x)(4.00 – x)

xx22

= =

xx22 = 1.88(2.00 – x)(4.00 – x) = 1.88(8.00 - 6x – x = 1.88(2.00 – x)(4.00 – x) = 1.88(8.00 - 6x – x22))

xx22 = 15.0 – 11.3x +1.88x = 15.0 – 11.3x +1.88x22 0 = 0.88x0 = 0.88x22 – 11.3x +15.0 – 11.3x +15.0

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0 = 0.88 x0 = 0.88 x22 – 11.3 x + 15.0 – 11.3 x + 15.0 (ax(ax22 + bx + c) + bx + c)

Dust off the old Quadratic Formula:Dust off the old Quadratic Formula:

http://www.freemathhelp.com/algebra-help.html

-(-11.3) -(-11.3) ± [(-11.3)± [(-11.3)22 – 4(0.88)(15.0)] – 4(0.88)(15.0)]1/21/2

2(0.88)2(0.88)

= 11 = 11 andand 1.5 ! 1.5 ! Which is RIGHT?Which is RIGHT?

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Chemical EquilibriumChemical Equilibrium

What is ‘x’? It is the concentration of HWhat is ‘x’? It is the concentration of H22 and CO and CO22

at equilibrium! But…you can’t have more hydrogen gas thanat equilibrium! But…you can’t have more hydrogen gas thanyou have of reactants to begin with!you have of reactants to begin with!

Thus, 11 M can’t be right!Thus, 11 M can’t be right!

1.5 M is the only sensible answer!1.5 M is the only sensible answer!

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Chemical EquilibriumChemical Equilibrium

x = 1.5 M x = 1.5 M Therefore:Therefore:[H[H22O]O]eqeq = 0.5 M = 0.5 M [CO] [CO]eqeq = 2.5 M = 2.5 M

[H[H22]]eqeq = [CO = [CO22]]eqeq = 1.5 M = 1.5 M

Check: KCheck: Kcc = (1.5M) = (1.5M)22/(0.5M)(2.5M) = /(0.5M)(2.5M) = 1.81.8

There are no units in this case!There are no units in this case!

What if you don’t remember the quadraticWhat if you don’t remember the quadraticformula??formula??

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Use SuccessiveUse SuccessiveApproximation!!Approximation!!

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Class Exercise 14.5: Class Exercise 14.5: Consider the following reaction Consider the following reaction for the decomposition of hydrogen sulfide:for the decomposition of hydrogen sulfide:

2 H2 H22S S 2 H 2 H22 (g) + S (g) + S22 (g) (g) K KCC = 1.67 x 10 = 1.67 x 10-7-7 800o C

A 0.500-L vessel initially contains 1.25 x 10A 0.500-L vessel initially contains 1.25 x 10 -1-1 mol of H mol of H22S.S.

Find the equilibrium concentrations of HFind the equilibrium concentrations of H22 and S and S22..

Equation:Equation: 2 H 2 H22S (g) S (g) 2 H 2 H22 (g) + S (g) + S22 (g) (g)

Initial (M) Initial (M) 2.50 x 102.50 x 10-1-1 0.00 0.00 0.00 0.00

Change (M)Change (M) - 2x- 2x + 2x + 2x + x + x

Equilibrium (M) Equilibrium (M) (2.50 x 10(2.50 x 10-1-1 – 2x – 2x)) 2x 2x x x

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KKcc = = [H[H22]]22[S[S22]]

HH22SS==

(2x)(2x)22xx(2.50 x 10(2.50 x 10-1-1 – 2x) – 2x)22

= 4x4x33

(2.50 x 10(2.50 x 10-1-1 – 2x) – 2x)22

Assume Assume xx is NEGLIGIBLE compared to 2.50 x 10 is NEGLIGIBLE compared to 2.50 x 10-1-1 M. M.Then:Then: 4x4x33

(2.50 x 10(2.50 x 10-1-1))22

= 1.67 x 10-7

~ 1.67 x 10-7 4x4x33

6.25 x 106.25 x 10-2-2=

(1.67 x 10-7)(6.25 x 10(6.25 x 10-2-2)) = 4x4x33 = 1.04 x 101.04 x 10-8-8

xx33 = 2.61 x 102.61 x 10-9-9 x = 1.38 x 10x = 1.38 x 10-3-3 M M

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Is Is xx NEGLIGIBLE compared to 1.38 x 10 NEGLIGIBLE compared to 1.38 x 10-3-3 M? M?Plug it back in to check:Plug it back in to check:

4x4x33

(2.50 x 10(2.50 x 10-1-1 – 2x) – 2x)22 [(2.50 x 10[(2.50 x 10-1-1) - 2(1.38 x 10) - 2(1.38 x 10-3-3)])]22

4x4x33

=

4x4x33

4.95 x 10-4= 1.67 x 10-7

x3 = 2.07 x 10-11 >>>>> x x = 2.74 x 10= 2.74 x 10-4-4

There appears to be a mistake in these calculations!There appears to be a mistake in these calculations!Please check carefully and see if you can see where it is!Please check carefully and see if you can see where it is!

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Complex example of Successive Approximation

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Do NOT Do NOT Panic!Panic!This is NOT a typical Problem! This is NOT a typical Problem!

It is a It is a Worst Case ScenarioWorst Case Scenario!!!!!!

Any Exam Problem will Any Exam Problem will be MUCH Shorter!!be MUCH Shorter!!

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Class Exercise 14.5: Class Exercise 14.5: In an experiment starting withIn an experiment starting with[N[N22OO44]]II = 0.020 00 M and [NO = 0.020 00 M and [NO22]]II = 0.000 00 M, [N = 0.000 00 M, [N22OO44]]eqeq

= 0.004 52 M. (a) What is [NO= 0.004 52 M. (a) What is [NO22]]eqeq? (b) What is the? (b) What is the

value for Kvalue for Kcc??

Equation:Equation: N N22OO44 (g) (g) 2 NO 2 NO22 (g) (g)

I (M)I (M) 2.000 x 102.000 x 10-2-2 0.000 00 0.000 00

C (M)C (M)

E (M) E (M) 4.52 x 104.52 x 10-3-3

-0.015 48-0.015 48 +0.030 96+0.030 96

0.030 960.030 96

[NO[NO22]]22

[N[N22OO44]]KC = = (0.030 96)= (0.030 96)22/(0.004 52) = /(0.004 52) = 0.2120.212

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What does the value of KWhat does the value of Kcc MEAN? MEAN?

The larger KThe larger KCC is, the closer to completion the rxn is! is, the closer to completion the rxn is!

NN22 (g) + O (g) + O22 (g) (g) 2 NO (g) 2 NO (g) KKCC = 1 x 10 = 1 x 10-30-30

2 NH2 NH33 (g) (g) N N22 (g) + 3 H (g) + 3 H22 (g) (g) KKCC = 9.5 = 9.5

HH22 (g) + Cl (g) + Cl22 (g) (g) 2 HCl (g) 2 HCl (g) KKCC = 1.33 x 10 = 1.33 x 103434