chemical equilibrium

1

Chemical EquilibriumChemical Equilibrium

2

EquilibriumEquilibrium

Initially all liquid Gas only, produced Balance of gas and

liquid production

3

EquilibriumEquilibrium

When compounds react, they When compounds react, they eventually form a mixture of eventually form a mixture of products and unreacted reactants, in products and unreacted reactants, in a a dynamic equilibrium.dynamic equilibrium.– A dynamic equilibrium consists of a

forward reaction, in which substances react to give products, and a reverse reaction, in which products react to give the original reactants.

4

Chemical EquilibriumChemical EquilibriumH2 + I2 < -- > 2HI

– Initially only H2 and I2 are present.

– The rxn. proceeds only– As HI concentration increases, some HI is able

to decompose back into H2 and I2

– Rxn. proceeds < -- >– At some point

– The rate of H2 + I2 -- > 2HI equals

– The rate of 2HI -- > H2 + I2

Chemical equilibrium is the state reached by a reaction mixture when the rates of the forward and reverse reactions have become equal.

See Le Chatelier

6


For example, the Haber process for For example, the Haber process for producing ammonia from Nproducing ammonia from N22 and H and H22 does not go to completion. does not go to completion.

– It establishes an equilibrium state where all three species are present. (see Figure 15.3)

(g)2NH )g(H3)g(N 322

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Chemical Equilibrium is a Chemical Equilibrium is a fundamentally important concept to fundamentally important concept to master because most chemical master because most chemical reactions fail to go to completion.reactions fail to go to completion.

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A Problem to ConsiderA Problem to Consider

Applying Stoichiometry to an Applying Stoichiometry to an Equilibrium Mixture.Equilibrium Mixture.

– What is the composition of the equilibrium mixture if it contains 0.080 mol NH3?

– Suppose we place 1.000 mol N2 and 3.000 mol H2 in a reaction vessel at 450 oC and 10.0 atmospheres of pressure. The reaction is

(g)2NH )g(H3)g(N 322

9


– The equilibrium amount of NH3 was given as 0.080 mol. Therefore, 2x = 0.080 mol NH3 (x = 0.040 mol).

Using the information given, set up an ICE Using the information given, set up an ICE table.table.

(g)2NH )g(H3 )g(N 322

InitialInitial 1.0001.000 3.0003.000 00ChangeChange -x-x -3x-3x +2x+2x

EquilibriuEquilibriumm

1.000 - 1.000 - xx

3.000 - 3.000 - 3x3x

2x = 0.080 mol2x = 0.080 mol

10


Using the information given, set up the Using the information given, set up the following table.following table.

Equilibrium amount of N2 = 1.000 - 0.040 = 0.960 mol N2

Equilibrium amount of H2 = 3.000 - (3 x 0.040) = 2.880 mol H2

Equilibrium amount of NH3 = 2x = 0.080 mol NH3

(g)2NH )g(H3 )g(N 322

InitialInitial 1.0001.000 3.0003.000 00ChangeChange -x-x -3x-3x +2x+2x


1.000 - 1.000 - xx

3.000 - 3.000 - 3x3x

2x = 0.080 mol2x = 0.080 mol

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The Equilibrium ConstantThe Equilibrium Constant

Every reversible system has its own Every reversible system has its own “position of equilibrium” under any “position of equilibrium” under any given set of conditions.given set of conditions.

– The ratio of products produced to unreacted reactants for any given reversible reaction remains constant under constant conditions of pressure and temperature.

– The numerical value of this ratio is called the equilibrium constant for the given reaction.

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H2 + I2 < -- > 2HI Rate of forward rxn:

Ratef = kf [H2][I2]Rate of reverse rxn:

Rater = kr[HI]2

At equilibrium:kf [H2][I2] =kr[HI]2

Therefore:kf = [HI]2

kr [H][I]

Kc = [HI]2

[H][I]

13


The The equilibrium-constant expressionequilibrium-constant expression for a for a reaction is obtained by multiplying the reaction is obtained by multiplying the concentrations of products, dividing by the concentrations of products, dividing by the concentrations of reactants, and raising each concentrations of reactants, and raising each concentration to a power equal to its coefficient concentration to a power equal to its coefficient in the balanced chemical equation.in the balanced chemical equation.

ba

dc

c ]B[]A[

]D[]C[K

– For the general equation above, the equilibrium-constant expression would be:

dDcC bBaA

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The The equilibrium-constant expressionequilibrium-constant expression for a for a reaction is obtained by multiplying the reaction is obtained by multiplying the concentrations of products, dividing by the concentrations of products, dividing by the concentrations of reactants, and raising each concentrations of reactants, and raising each concentration to a power equal to its coefficient concentration to a power equal to its coefficient in the balanced chemical equation.in the balanced chemical equation.

ba

dc

c ]B[]A[

]D[]C[K

– The molar concentration of a substance is denoted by writing its formula in square brackets.

dDcC bBaA

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The The equilibrium constant, Kequilibrium constant, Kcc,, is is the value obtained for the the value obtained for the equilibrium-constant expression equilibrium-constant expression when equilibrium concentrations are when equilibrium concentrations are substituted.substituted.

– A large Kc indicates large concentrations of products at equilibrium.

– A small Kc indicates large concentrations of unreacted reactants at equilibrium.

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The The law of mass actionlaw of mass action states that the states that the value of the equilibrium constant value of the equilibrium constant expression expression KKcc is constant is constant for a particular for a particular reaction at a reaction at a given temperaturegiven temperature, , whateverwhatever equilibrium concentrations are substituted.equilibrium concentrations are substituted.

– Consider the equilibrium established in the Haber process.

(g)2NH )g(H3)g(N 322

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The equilibrium-constant expression The equilibrium-constant expression would bewould be

– Note that the stoichiometric coefficients in the balanced equation have become the powers to which the concentrations are raised.

(g)2NH )g(H3)g(N 322

322

23

c ]H][N[

]NH[K

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Calculating Equilibrium Calculating Equilibrium ConcentrationsConcentrations

Once you have determined the Once you have determined the equilibrium constant for a reaction, equilibrium constant for a reaction, you can use it to calculate the you can use it to calculate the concentrations of substances in the concentrations of substances in the equilibrium mixture.equilibrium mixture.

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– Suppose a gaseous mixture contained 0.30 mol CO, 0.10 mol H2, 0.020 mol H2O, and an unknown amount of CH4 per liter.

– What is the concentration of CH4 in this mixture? The equilibrium constant Kc equals 3.92.

O(g)H (g)CH (g)H 3 )g(CO 242

– For example, consider the following equilibrium.

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First, calculate concentrations from moles of First, calculate concentrations from moles of substances.substances.

0.30 mol1.0 L

0.10 mol1.0 L

0.020 mol1.0 L??

O(g)H (g)CH (g)H 3 )g(CO 242

21


??0.30 M 0.10 M 0.020 M

– The equilibrium-constant expression is:

32

24c ]H][CO[

]OH][CH[K

O(g)H (g)CH (g)H 3 )g(CO 242


22


– Substituting the known concentrations and the value of Kc gives:

34

)M10.0)(M30.0(

)M020.0](CH[92.3

??0.30 M 0.10 M 0.020 M

O(g)H (g)CH (g)H 3 )g(CO 242


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– You can now solve for [CH4].

059.0)M020.0(

)M10.0)(M30.0)(92.3(]CH[

3

4

– The concentration of CH4 in the mixture is 0.059 mol/L.

??0.30 M 0.10 M 0.020 M


O(g)H (g)CH (g)H 3 )g(CO 242

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Suppose we begin a reaction with Suppose we begin a reaction with known amounts of starting materials known amounts of starting materials and want to calculate the quantities and want to calculate the quantities at equilibrium.at equilibrium.

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Consider the following equilibrium.Consider the following equilibrium.

• Suppose you start with 1.000 mol each of carbon monoxide and water in a 50.0 L container. Calculate the molarity of each substance in the equilibrium mixture at 1000 oC.

• Kc for the reaction is 0.58 at 1000 oC.

(g)H(g)CO )g(OH)g(CO 222

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– First, calculate the initial molarities of CO and HFirst, calculate the initial molarities of CO and H22O.O.

1.000 mol50.0 L

1.000 mol50.0 L


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• The starting concentrations of the products are 0.• We must now set up a table of concentrations (starting, change,

and equilibrium expressions in x).

0.0200 M 0.0200 M 0 M 0 M

– First, calculate the initial molarities of CO and HFirst, calculate the initial molarities of CO and H22O.O.


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– Let x be the moles per liter of product Let x be the moles per liter of product formed.formed.

InitialInitial 0.02000.0200 0.02000.0200 00 00ChangeChange -x-x -x-x +x+x +x+x


0.0200-x0.0200-x 0.0200-x0.0200-x xx xx


]OH][CO[]H][CO[

K2

22c


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– Solving for x.Solving for x.



0.0200-0.0200-xx

0.0200-x0.0200-x xx xx

– Substituting the values for equilibrium concentrations, we get:

)x0200.0)(x0200.0()x)(x(

58.0


30





0.0200-0.0200-xx

0.0200-x0.0200-x xx xx

– Or:

2

2

)x0200.0(

x58.0


31





0.0200-0.0200-xx

0.0200-x0.0200-x xx xx

– Taking the square root of both sides we get:

)x0200.0(x

76.0


32





0.0200-0.0200-xx

0.0200-x0.0200-x xx xx

– Rearranging to solve for x gives:

0086.076.1

76.00200.0x


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– Solving for equilibrium concentrations.Solving for equilibrium concentrations.



0.0200-0.0200-xx

0.0200-x0.0200-x xx xx

– If you substitute for x in the last line of the table you obtain the following equilibrium concentrations.

0.0114 M CO0.0114 M H2O 0.0086 M H2

0.0086 M CO2


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The preceding example illustrates The preceding example illustrates the three steps in solving for the three steps in solving for equilibrium concentrations.equilibrium concentrations.

1. Set up a table of concentrations (starting, change, and equilibrium expressions in x).

2. Substitute the expressions in x for the equilibrium concentrations into the equilibrium-constant equation.

3. Solve the equilibrium-constant equation for the values of the equilibrium concentrations.

35


In some cases it is necessary to solve In some cases it is necessary to solve a quadratic equation to obtain a quadratic equation to obtain equilibrium concentrations. equilibrium concentrations.

The next example illustrates how to The next example illustrates how to solve such an equation.solve such an equation.

36


– Consider the following equilibrium.Consider the following equilibrium.

• Suppose 1.00 mol H2 and 2.00 mol I2 are placed in a 1.00-L vessel. How many moles per liter of each substance are in the gaseous mixture when it comes to equilibrium at 458 oC?

• Kc at this temperature is 49.7.

HI(g)2 )g(I)g(H 22

37


The concentrations of substances are as The concentrations of substances are as follows.follows.

InitialInitial 1.001.00 2.002.00 00ChangeChange -x-x -x-x +2x+2x


1.00-x1.00-x 2.00-x2.00-x 2x2x


]I][H[]HI[

K22

2

c

HI(g)2 )g(I)g(H 22

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The concentrations of substances are as The concentrations of substances are as follows.follows.

InitialInitial 1.001.00 2.002.00 00ChangeChange -x-x -x-x +2x+2x


1.00-x1.00-x 2.00-x2.00-x 2x2x

– Substituting our equilibrium concentration expressions gives:

)x00.2)(x00.1()x2(

K2

c

HI(g)2 )g(I)g(H 22

39



– Because the right side of this equation is not a perfect square, you must solve the quadratic equation.

HI(g)2 )g(I)g(H 22 InitialInitial 1.001.00 2.002.00 00

ChangeChange -x-x -x-x +2x+2xEquilibriuEquilibriu

mm1.00-x1.00-x 2.00-x2.00-x 2x2x

40



– The equation rearranges to give:



mm1.00-x1.00-x 2.00-x2.00-x 2x2x

000.2x00.3x920.0 2

41



– The two possible solutions to the quadratic equation are:



mm1.00-x1.00-x 2.00-x2.00-x 2x2x

0.93x and 33.2x

42



– However, x = 2.33 gives a negative value to 1.00 - x (the equilibrium concentration of H2), which is not possible.



mm1.00-x1.00-x 2.00-x2.00-x 2x2x

remains. 0.93Only x

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Solving for equilibrium Solving for equilibrium concentrations.concentrations.

– If you substitute 0.93 for x in the last line of the table you obtain the following equilibrium concentrations.

0.07 M H2 1.07 M I2 1.86 M HI



mm1.00-x1.00-x 2.00-x2.00-x 2x2x

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Le Chatelier’s PrincipleLe Chatelier’s Principle

Obtaining the maximum amount of Obtaining the maximum amount of product from a reaction depends on product from a reaction depends on the proper set of reaction conditions.the proper set of reaction conditions.

– Le Chatelier’s Principle states that when a system in a chemical equilibrium is disturbed by a change of temperature, pressure, or concentration, the equilibrium will shift in a way that tends to counteract this change.

– See LeChatelier’s Principle animation

45

Removing Products or Adding Removing Products or Adding ReactantsReactants

Let’s refer an illustration of a U-tube.Let’s refer an illustration of a U-tube.

– It’s a simple concept to see that if we were to remove products (analogous to dipping water out of the right side of the tube) the reaction would shift to the right until equilibrium was reestablished.

“reactants” “products”

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Removing Products or Adding Removing Products or Adding ReactantsReactants

Let’s refer back to the illustration of the U-Let’s refer back to the illustration of the U-tube in the first section of this chapter.tube in the first section of this chapter.

– Likewise, if more reactant is added (analogous to pouring more water in the left side of the tube) the reaction would again shift to the right until equilibrium is reestablished.

“reactants” “products”

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Effects of Pressure ChangeEffects of Pressure Change

A pressure change caused by changing the A pressure change caused by changing the volume of the reaction vessel can affect volume of the reaction vessel can affect the yield of products in a gaseous reaction the yield of products in a gaseous reaction only if the reaction involves a change in only if the reaction involves a change in the the total moles of gastotal moles of gas present (see Figure present (see Figure 15.12).15.12).

CO CO + 3+ 3HH2 2 CHCH44++HH22OO 3 mol 9 mol 3 mol 3 mol

48


If the products in a gaseous reaction contain If the products in a gaseous reaction contain fewer moles of gas than the reactants, it is fewer moles of gas than the reactants, it is logical that they would require less space.logical that they would require less space.

• So, reducing the volume of the reaction vessel would favor the products.

• If the reactants require less volumeIf the reactants require less volume (that is, fewer moles of gaseous reactant) • decreasing the volume of the reaction vessel decreasing the volume of the reaction vessel

would shift the equilibrium to the leftwould shift the equilibrium to the left (toward reactants).

49


Literally “squeezing” the reaction will cause a Literally “squeezing” the reaction will cause a shift in the equilibrium toward the fewer shift in the equilibrium toward the fewer moles of gas.moles of gas.

• It’s a simple step to see that reducing the pressure in the reaction vessel by increasing its volume would have the opposite effect.

• In the event that the number of moles of gaseous product equals the number of moles of gaseous reactant, vessel volume will have no effect on the position of the equilibrium.

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Effect of Temperature ChangeEffect of Temperature Change

TemperatureTemperature has a significant has a significant effect on most reactions (see Figure effect on most reactions (see Figure 15.13).15.13).– Reaction rates generally increase with an increase

in temperature. Consequently, equilibrium is established sooner.

– In addition, the numerical value of the equilibrium constant Kc varies with temperature.

51


Let’s look at “heat” as if it were a Let’s look at “heat” as if it were a product in exothermic reactions product in exothermic reactions and aand a reactant in endothermic reactant in endothermic reactions.reactions.• We see that increasing the temperature is analogous to adding more product (in the case of exothermic reactions) or adding more reactant (in the case of endothermic reactions).

• This ultimately has the same effect as if heat were a physical entity.


ExothermicExothermic

A + B A + B → C + D + → C + D + heatheat

(-)∆H∆H

How would adding heat effect the equilibrium?How would adding heat effect the equilibrium?• Increasing temperature would be analogous to adding

more product, causing the equilibrium to shift left.• Since “heat” does not appear in the equilibrium-

constant expression, this change would result in a smaller numerical value for Kc.

52

53


EndothermicEndothermic

A + B + A + B + heatheat → C + D→ C + D

(+)∆H(+)∆H How would adding heat effect the equilibrium?How would adding heat effect the equilibrium?

• Increasing temperature would be analogous to adding more reactant, causing the equilibrium to shift right.

• This change results in more product at equilibrium, and a larger numerical value for Kc.

54


In summary:In summary:

– For an exothermic reaction (H is negative) the amounts of reactants are increased at equilibrium by an increase in temperature (Kc is smaller at higher temperatures).

– For an endothermic reaction (H positive) the amounts of products are increased at equilibrium by an increase in temperature (Kc is larger at higher temperatures).

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Obtaining Equilibrium Obtaining Equilibrium Constants for ReactionsConstants for Reactions

Equilibrium concentrations for a Equilibrium concentrations for a reaction must be obtained reaction must be obtained experimentallyexperimentally and then and then substitutedsubstituted into the equilibrium- into the equilibrium-constant expression in order to constant expression in order to calculate Kcalculate Kcc..

56


Consider the reaction belowConsider the reaction below

O(g)H (g)CH (g)H 3 )g(CO 242

– Suppose we started with initial concentrations of CO and H2 of 0.100 M and 0.300 M, respectively.

57

– When the system finally settled into equilibrium we determined the equilibrium concentrations to be as follows.

Reactants Products[CO] = 0.0613 M[H2] = 0.1893 M

[CH4] = 0.0387 M[H2O] = 0.0387 M


Consider the reaction below (see Figure Consider the reaction below (see Figure 15.5).15.5).

O(g)H (g)CH (g)H 3 )g(CO 242

58

– The equilibrium-constant expression for this reaction is:

32

24c ]H][CO[

]OH][CH[K



O(g)H (g)CH (g)H 3 )g(CO 242

59

– If we substitute the equilibrium concentrations, we obtain:

93.3)M1839.0)(M0613.0(

)M0387.0)(M0387.0(K 3c



O(g)H (g)CH (g)H 3 )g(CO 242

60

– Regardless of the initial concentrations (whether they be reactants or products), the law of mass action dictates that the reaction will always settle into an equilibrium where the equilibrium-constant expression equals Kc.



O(g)H (g)CH (g)H 3 )g(CO 242

61

– As an example, let’s repeat the previous experiment, only this time starting with initial concentrations of products:

[CH4]initial = 0.1000 M and [H2O]initial = 0.1000 M



O(g)H (g)CH (g)H 3 )g(CO 242

62

– We find that these initial concentrations result in the following equilibrium concentrations.

Reactants Products

[CO] = 0.0613 M

[H2] = 0.1893 M

[CH4] = 0.0387 M

[H2O] = 0.0387 M



O(g)H (g)CH (g)H 3 )g(CO 242

63

– Substituting these values into the equilibrium-constant expression, we obtain the same result.

93.3)M1839.0)(M0613.0(

)M0387.0)(M0387.0(K 3c

– Whether we start with reactants or products, the system establishes the same ratio.

(see Figure 15.5).



O(g)H (g)CH (g)H 3 )g(CO 242

64

The Equilibrium Constant, KThe Equilibrium Constant, Kpp

In discussing gas-phase equilibria, it is In discussing gas-phase equilibria, it is often more convenient to express often more convenient to express concentrations in terms of partial concentrations in terms of partial pressures rather than molaritiespressures rather than molarities

– It can be seen from the ideal gas equation,

PV = nRT that the partial pressure of a gas is proportional to its molarity.

MRTRTVn

P )(

65


If we express a gas-phase equilibria If we express a gas-phase equilibria in terms of partial pressures, we in terms of partial pressures, we obtain Kobtain Kpp..– Consider the reaction below.

O(g)H (g)CH (g)H 3 )g(CO 242 – The equilibrium-constant expression in terms of

partial pressures becomes:

3HCO

OHCHp

2

24

K

PP

PP

66


In general, the numerical value of KIn general, the numerical value of Kpp differs from that of Kdiffers from that of Kcc..

– From the relationship n/V=P/RT, we can show that

where n is the sum of the moles of gaseous products in a reaction minus the sum of the moles of gaseous reactants.

Animation: Pressure and Concentration of a Gas).

ncp )RT(KK

67


Consider the reactionConsider the reaction

– Kc for the reaction is 2.8 x 102 at 1000 oC. Calculate Kp for the reaction at this temperature.

(g)SO 2 )g(O)g(SO2 322

68


– We know thatn

cp )RT(KK

From the equation we see that Δn = -1. We can simply substitute the given reaction temperature and the value of R (0.08206 L.atm/mol.K) to obtain Kp.


(g)SO 2 )g(O)g(SO2 322

69


– Sincen

cp )RT(KK

3.4 K) 1000 08206.0( 108.2K 1-Kmol

atmL2p


(g)SO 2 )g(O)g(SO2 322

70

Equilibrium Constant for the Equilibrium Constant for the Sum of ReactionsSum of Reactions

Similar to the method of combining Similar to the method of combining reactions that we saw using Hess’s law in reactions that we saw using Hess’s law in Chapter 6, we can combine equilibrium Chapter 6, we can combine equilibrium reactions whose Kreactions whose Kcc values are known to values are known to obtain Kobtain Kcc for the overall reaction. for the overall reaction.

– With Hess’s law, when we reversed reactions or multiplied them prior to adding them together, we had to manipulate the H’s values to reflect what we had done.

– The rules are a bit different for manipulating Kc.

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2. If you multiply each of the coefficients in an equation by the same factor (2, 3, …), raise Kc to the same power (2, 3, …).

3. If you divide each coefficient in an equation by the same factor (2, 3, …), take the corresponding root of Kc (i.e., square root, cube root, …).

4. When you finally combine (that is, add) the individual equations together, take the product of the equilibrium constants to obtain the overall Kc.

1.1. If you If you reversereverse a reaction, a reaction, invertinvert the value of Kthe value of Kcc..


72

Building Equations with K valuesBuilding Equations with K values

Weak acids and bases are assigned a Weak acids and bases are assigned a Ka or Kb values based on the degree Ka or Kb values based on the degree to which they ionize in water.to which they ionize in water.

Larger K values indicate a greater Larger K values indicate a greater degree of ionization (strength).degree of ionization (strength).

Ka and Kb, along with other K values Ka and Kb, along with other K values that we will study later (Ksp, KD, Kf) that we will study later (Ksp, KD, Kf) are all manipulated in the same are all manipulated in the same manner.manner.

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When equations are added When equations are added K valuesK values are are multiplied.multiplied.

MnS ↔ Mn+2 + S-2MnS ↔ Mn+2 + S-2 K= 5.1 x 10K= 5.1 x 10-15-15

SS-2-2 + H2O ↔HS- + OH- + H2O ↔HS- + OH- K= 1.0 x 10K= 1.0 x 10-19-19

2H+ + HS- OH- ↔ H2S +H2O2H+ + HS- OH- ↔ H2S +H2O K= 1.0 x 10 K= 1.0 x 10-7-7

MnS + 2H+ ↔ Mn+2+ H2MnS + 2H+ ↔ Mn+2+ H2 K= 5.1 x 10K= 5.1 x 10-41-41


74

When equations are reversed the When equations are reversed the K K values values are are reciprocated.reciprocated.

Al(OH)Al(OH)33 ↔ Al ↔ Al+3+3 + 3OH + 3OH-- K = 1.9 x K = 1.9 x 10-3310-33

3OH- + Al3OH- + Al+3+3 ↔ Al(OH) ↔ Al(OH)33 K = 1/1.9 x K = 1/1.9 x 1010-33-33

= 5.2 x 10= 5.2 x 103232


75

When equations are multiplied the When equations are multiplied the K K values values are are raised to the power.raised to the power.

NHNH33 + H + H22O ↔ NHO ↔ NH44+++ OH-+ OH- K = 1.8 x K = 1.8 x 10-510-5

2NH2NH33 + 2H + 2H22O ↔ 2NHO ↔ 2NH44+++ 2OH-+ 2OH-

K = (1.8 x 10K = (1.8 x 10--

55))22

= 3.24 x 10= 3.24 x 10--

1010


76

When equations are divided theWhen equations are divided the root root of of the the K values K values areare taken.taken.

2HPO2HPO44-2-2 ↔ 2H ↔ 2H++ + 2PO + 2PO44

3-3- K = 1.3 x 10 K = 1.3 x 10-25-25

HPOHPO44-2-2 ↔ H ↔ H++ + PO + PO44

3-3- K = √1.3 x K = √1.3 x 1010-25-25

= 3.6 x 10= 3.6 x 10-13-13


77

For example, nitrogen and oxygen can For example, nitrogen and oxygen can combine to form either NO(g) or Ncombine to form either NO(g) or N22O (g) O (g) according to the following equilibria.according to the following equilibria.

NO(g) 2 )g(O)g(N 22

O(g)N )g(O)g(N 2221

2

Kc = 4.1 x 10-31

Kc = 2.4 x 10-18

(1)

(2)

Kc = ?

– Using these two equations, we can obtain Kc for the formation of NO(g) from N2O(g):

NO(g) 2 )g(O)g(ON 221

2 (3)


78

To combine equations (1) and (2) to obtain To combine equations (1) and (2) to obtain equation (3), we must first reverse equation (3), we must first reverse equation (2). When we do we must also equation (2). When we do we must also take the reciprocal of its Ktake the reciprocal of its Kcc value. value.

NO(g) 2 )g(O)g(ON 221

2

NO(g) 2 )g(O)g(N 22 Kc = 4.1 x 10-31(1)

)g(O (g)N O(g)N 221

22 Kc = (2)

(3)

18-10 4.2

1

1318

31c 107.1

104.2

1)101.4()overall(K )(


79

At your seats build the equation: At your seats build the equation: CaCOCaCO33 + H + H33OO++ ↔ Ca ↔ Ca2+2+ + HCO + HCO33

- - + H+ H22O O from the following,from the following, 2H2H22O ↔ HO ↔ H33OO+ + + OH-+ OH- K = 1 x 10K = 1 x 10-14-14

COCO332-2- + H + H22O ↔HCOO ↔HCO33

-- + OH- + OH- K = 2.1 x K = 2.1 x 1010-4-4‑‑

3CaCO3CaCO33 ↔ 3Ca ↔ 3Ca2+2+ + 3CO + 3CO332-2- K = 5.5 x 10K = 5.5 x 102626


80


HH33OO+ + + + OHOH- ↔ - ↔ 22HH22O K = 1/1 x 10O K = 1/1 x 10-14-14

COCO332-2- + + HH22OO ↔HCO ↔HCO33

-- + + OHOH-- K = 2.1 x 10K = 2.1 x 10-4-4

CaCOCaCO33 ↔ Ca ↔ Ca2+2+ + + COCO332-2- K = K = 33√√5.5 x 5.5 x

10102626

CaCOCaCO33 + H + H33OO++ ↔ Ca ↔ Ca2+2+ + HCO + HCO33- - + H+ H22OO

K = (1x10K = (1x101414)(2.1x10)(2.1x10-4-4)(8.19x10)(8.19x1088))

K = 1.7x10K = 1.7x101919

81

Equilibrium: A Kinetics Equilibrium: A Kinetics ArgumentArgument

If the forward and reverse reaction rates in If the forward and reverse reaction rates in a system at equilibrium are equal, then it a system at equilibrium are equal, then it follows that their rate laws would be equal.follows that their rate laws would be equal.

– If we start with some dinitrogen tetroxide and heat it, it begins to decompose to produce NO2.

– Consider the decomposition of N2O4, dinitrogen tetroxide.

(g)2NO )g(ON 242

82


If the forward and reverse reaction rates in If the forward and reverse reaction rates in a system at equilibrium are equal, then it a system at equilibrium are equal, then it follows that their rate laws would be equal.follows that their rate laws would be equal.

– Consider the decomposition of N2O4, dinitrogen tetroxide.

– However, once some NO2 is produced it can recombine to form N2O4.

(See Animation: Equilibrium Decomposition of N2O4)

(g)2NO )g(ON 242

83


Call the decomposition of NCall the decomposition of N22OO44 the forward the forward reaction and the formation of Nreaction and the formation of N22OO44 the the reverse reaction.reverse reaction.

(g)2NO )g(ON 242

kf

kr

– These are elementary reactions, and you can immediately write the rate law for each.

]ON[kRate 42f(forward) 2

2r(reverse) ]NO[kRate

Here kf and kr represent the forward and reverse rate constants.

84


Ultimately, this reaction reaches an Ultimately, this reaction reaches an equilibrium state where the rates of equilibrium state where the rates of the forward and reverse reactions are the forward and reverse reactions are equal. Therefore,equal. Therefore,

(g)2NO )g(ON 242

22r42f ]NO[k]ON[k

kf

kr

85


Combining the constants you can identify Combining the constants you can identify the equilibrium constant, Kc, as the ratio of the equilibrium constant, Kc, as the ratio of the forward and reverse rate constants.the forward and reverse rate constants.

(g)2NO )g(ON 242

]ON[]NO[

kk

K42

22

r

fc

kf

kr

86

Heterogeneous EquilibriaHeterogeneous Equilibria

A A heterogeneous equilibriumheterogeneous equilibrium is is an equilibrium that involves an equilibrium that involves reactants and products in more than reactants and products in more than one phase.one phase.

– The equilibrium of a heterogeneous system is unaffected by the amounts of pure solids or liquids present, as long as some of each is present.

– The concentrations of pure solids and liquids are always considered to be “1” and therefore, do not appear in the equilibrium expression.

87

Heterogeneous EquilibriaHeterogeneous Equilibria

Consider the reaction below.Consider the reaction below.

(g)H CO(g) )g(OH)s(C 22

– The equilibrium-constant expression contains terms for only those species in the homogeneous gas phase…H2O, CO, and H2.

]OH[]H][CO[

K2

2c

88

Predicting the Direction of Predicting the Direction of ReactionReaction

How could we predict the direction in How could we predict the direction in which a reaction at non-equilibrium which a reaction at non-equilibrium conditions will shift to reestablish conditions will shift to reestablish equilibrium?equilibrium?

– To answer this question, substitute the current concentrations into the reaction quotient expression and compare it to Kc.

– The reaction quotient, Qc, is an expression that has the same form as the equilibrium-constant expression but whose concentrations are not necessarily at equilibrium.

89


For the general reactionFor the general reaction

dDcC bBaA the Qc expresssion would be:

ba

dc

c ]B[]A[

]D[]C[Q

ii

ii

90


For the general reactionFor the general reaction

– If Kc < Qc, the reaction will shift left…toward reactants.

– If Kc = Qc, then the reaction is at equilibrium.

dDcC bBaA – If Kc > Qc, the reaction will shift right …toward

products.

91


Consider the following equilibrium.Consider the following equilibrium.

– A 50.0 L vessel contains 1.00 mol N2, 3.00 mol H2, and 0.500 mol NH3. In which direction (toward reactants or toward products) will the system shift to reestablish equilibrium at 400 oC?

– Kc for the reaction at 400 oC is 0.500.

(g)2NH )g(H3 )g(N 322

92


First, calculate concentrations from First, calculate concentrations from moles of substances.moles of substances.

1.00 mol50.0 L

3.00 mol50.0 L

0.500 mol50.0 L

(g)2NH )g(H3 )g(N 322

93

0.0100 M0.0600 M0.0200 M



(g)2NH )g(H3 )g(N 322

– The Qc expression for the system would be:

322

23

c ]H][N[

]NH[Q

94

0.0100 M0.0600 M0.0200 M



(g)2NH )g(H3 )g(N 322

– Substituting these concentrations into the reaction quotient gives:

1.23)0600.0)(0200.0(

)0100.0(Q 3

2

c

95

0.0100 M0.0600 M0.0200 M



(g)2NH )g(H3 )g(N 322

– Because Qc = 23.1 is greater than Kc = 0.500, the reaction will go to the left (toward reactants) as it approaches equilibrium.

96

Effect of a CatalystEffect of a Catalyst

A A catalystcatalyst is a substance that is a substance that increases the rate of a reaction but is increases the rate of a reaction but is not consumed by it.not consumed by it.

– It is important to understand that a catalyst has no effect on the equilibrium composition of a reaction mixture (see Figure 15.15).

– A catalyst merely speeds up the attainment of equilibrium.

97

Operational SkillsOperational Skills

Applying stoichiometry to an equilibrium Applying stoichiometry to an equilibrium mixturemixture

Writing equilibrium-constant expressionsWriting equilibrium-constant expressions Obtaining the equilibrium constant from Obtaining the equilibrium constant from

reaction compositionreaction composition Using the reaction quotientUsing the reaction quotient Obtaining one equilibrium concentration Obtaining one equilibrium concentration

given the othersgiven the others

98

Operational SkillsOperational Skills

Solving equilibrium problemsSolving equilibrium problems

• Applying Le Chatelier’s principle

99

Figure 15.3: Catalytic methanation Figure 15.3: Catalytic methanation reaction approaches equilibrium. reaction approaches equilibrium.

Return to Slide 6

100

Animation: Equilibrium Decomposition of Animation: Equilibrium Decomposition of NN22OO4 4

Return to Slide 15

(Click here to open QuickTime animation)

101

Figure 15.5: Some equilibrium Figure 15.5: Some equilibrium compositions for the methanation compositions for the methanation

reaction. reaction.

Return to Slide 35

102Return to Slide 21


reaction. reaction.



reaction. reaction.

109

Figure 15.6: The concentration of a gas at Figure 15.6: The concentration of a gas at a given temperature is proportional to the a given temperature is proportional to the

pressure. pressure.

Return to Slide 43

110

Animation: Pressure and Concentration Animation: Pressure and Concentration of a Gas of a Gas

Return to Slide 28

(Click here to open QuickTime animation)

111

Figure 15.12 A-CFigure 15.12 A-C

Slide 21

112

Figure Figure 15.13: 15.13: The effect The effect of changing of changing the the temperaturtemperature on e on chemical chemical equilibrium. equilibrium. Photo courtesy Photo courtesy of American of American

Color.Color.

Return to Slide 24

113

Figure Figure 15.15: 15.15: Oxidation Oxidation of of ammonia ammonia using a using a copper copper catalyst.catalyst.

Photo Photo courtesy of courtesy of James James Scherer.Scherer.

Return to Slide 85

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