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Chemical Equilibrium The reversibility of reactions.

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Chemical Chemical Equilibrium Equilibrium The reversibility of reactions The reversibility of reactions
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Page 1: Chemical Equilibrium The reversibility of reactions.

Chemical Chemical EquilibriumEquilibrium

The reversibility of reactionsThe reversibility of reactions

Page 2: Chemical Equilibrium The reversibility of reactions.

EquilibriumEquilibrium

Many chemical reactions do not Many chemical reactions do not go to completion. Initially, when go to completion. Initially, when reactants are present, the forward reactants are present, the forward reaction predominates. As the reaction predominates. As the concentration of products increases, concentration of products increases, the reverse reaction begins to the reverse reaction begins to become significant.become significant.

Page 3: Chemical Equilibrium The reversibility of reactions.

EquilibriumEquilibrium

The forward reaction rate slows The forward reaction rate slows down since the concentration of down since the concentration of reactants has decreased. Since the reactants has decreased. Since the concentration of products is concentration of products is significant, the reverse reaction rate significant, the reverse reaction rate gets faster.gets faster.

Eventually, theEventually, the

forward reaction rate = reverse forward reaction rate = reverse reaction ratereaction rate

Page 4: Chemical Equilibrium The reversibility of reactions.

EquilibriumEquilibrium

forward reaction rate = reverse forward reaction rate = reverse reaction ratereaction rate

At this point, the reaction is At this point, the reaction is in in equilibriumequilibrium. The equilibrium is . The equilibrium is dynamicdynamic. Both the forward and . Both the forward and reverse reactions occur, but there is reverse reactions occur, but there is no net change in concentration.no net change in concentration.

Page 5: Chemical Equilibrium The reversibility of reactions.

Equilibrium: Equilibrium: 2NO2 ↔ N2O4

The concentrations of N2O4 and NO2 level off when the system reaches equilibrium.

Page 6: Chemical Equilibrium The reversibility of reactions.

Equilibrium: Equilibrium: 2NO2 ↔ N2O4

The system will reach equilibrium starting with either NO2 , N2O4, or a mixture of the two.

Page 7: Chemical Equilibrium The reversibility of reactions.

EquilibriumEquilibrium

Reactions that can reach Reactions that can reach equilibrium are indicated by double equilibrium are indicated by double arrows (↔ or ). arrows (↔ or ).

The extent to which a reaction The extent to which a reaction proceeds in a particular direction proceeds in a particular direction depends upon the reaction, depends upon the reaction, temperature, initial concentrations, temperature, initial concentrations, pressure (if gases), etc.pressure (if gases), etc.

Page 8: Chemical Equilibrium The reversibility of reactions.

EquilibriumEquilibrium

Scientists studying many reactions Scientists studying many reactions atat equilibriumequilibrium determined that for a general determined that for a general reaction with coefficients reaction with coefficients a, b, c a, b, c and and d d ::

a a A + A + b b B ↔ B ↔ c c C + C + d d DD

K is the K is the equilibrium constant equilibrium constant for the for the reaction.reaction.

[C]c[D]d[A]a[B]b

K=

Page 9: Chemical Equilibrium The reversibility of reactions.

EquilibriumEquilibrium

a a A + A + b b B ↔ B ↔ c c C + C + d d DD

This is the This is the equilibrium constant equilibrium constant expression expression for the reactionfor the reaction. . K is the K is the equilibrium constantequilibrium constant. The square . The square brackets indicate the concentrations of brackets indicate the concentrations of products and reactants products and reactants at equilibriumat equilibrium..

[C]c[D]d[A]a[B]b

K=

Page 10: Chemical Equilibrium The reversibility of reactions.

Equilibrium ConstantsEquilibrium Constants

The value of the equilibrium The value of the equilibrium constant depends upon temperature, constant depends upon temperature, but does not depend upon the but does not depend upon the initialinitial concentrations of reactants and concentrations of reactants and products. It is determined products. It is determined experimentally.experimentally.

Page 11: Chemical Equilibrium The reversibility of reactions.

Equilibrium ConstantsEquilibrium Constants

Page 12: Chemical Equilibrium The reversibility of reactions.

Equilibrium ConstantsEquilibrium Constants

The units of K are usually The units of K are usually omitted. The square brackets in the omitted. The square brackets in the equilibrium constant expression equilibrium constant expression indicate concentration in moles/liter.indicate concentration in moles/liter.

Some texts will use the symbol Some texts will use the symbol KKcc or K or Keqeq for equilibrium constants for equilibrium constants that use concentrations or molarity. that use concentrations or molarity.

Page 13: Chemical Equilibrium The reversibility of reactions.

Equilibrium ConstantsEquilibrium Constants

The size of the equilibrium The size of the equilibrium constant gives an indication of constant gives an indication of whether a reaction proceeds to the whether a reaction proceeds to the right. right.

Page 14: Chemical Equilibrium The reversibility of reactions.

Equilibrium ConstantsEquilibrium ConstantsIf a reaction has a small value of K, If a reaction has a small value of K,

the reverse reaction will have a large the reverse reaction will have a large value of K.value of K.

At a given temperature, K = 1.3 x 10At a given temperature, K = 1.3 x 10-2-2 for: for:

NN22((gg) + 3 H) + 3 H22((gg) ↔ 2 NH) ↔ 2 NH33((gg))

If the reaction is reversed, If the reaction is reversed,

2 NH2 NH33((gg) ↔ N) ↔ N22((gg) + 3 H) + 3 H22((gg) ) K = 1/(1.3 x 10K = 1/(1.3 x 10-2-2 ) = 7.7 x 10 ) = 7.7 x 1011

Page 15: Chemical Equilibrium The reversibility of reactions.

Equilibrium ConstantsEquilibrium Constants

Equilibrium constants for gas phase Equilibrium constants for gas phase reactions are sometimes determined reactions are sometimes determined using pressures rather than using pressures rather than concentrations. The symbol used is Kconcentrations. The symbol used is Kpp rather than K (or Krather than K (or Kcc).).

For the reaction:For the reaction:

NN22((gg) + 3 H) + 3 H22((gg) ↔ 2 NH) ↔ 2 NH33((gg))

KKpp = (P = (Pammoniaammonia))22

(P(PNN22)(P)(PH2H2))33

Page 16: Chemical Equilibrium The reversibility of reactions.

Equilibrium ConstantsEquilibrium Constants

The numerical value of KThe numerical value of Kpp is usually is usually different than that for K. The values are different than that for K. The values are related, since P= related, since P= nRT nRT = MRT, where M = MRT, where M = mol/liter.= mol/liter.

For reactions involving gases:For reactions involving gases:

KKCC = K = KPP(RT)(RT)ΔnΔn

where n is the change in moles of gases where n is the change in moles of gases in the balanced chemical reaction.in the balanced chemical reaction.

V

Page 17: Chemical Equilibrium The reversibility of reactions.

Equilibrium ConstantsEquilibrium Constants

The method for solving The method for solving equilibrium problems with Kequilibrium problems with Kpp or K is or K is the same. With Kthe same. With Kpp, you use pressure , you use pressure in atmospheres. With K or Kin atmospheres. With K or Kcc, you , you use concentration in moles/liter, or use concentration in moles/liter, or molarity.molarity.

Page 18: Chemical Equilibrium The reversibility of reactions.

Heterogeneous Heterogeneous EquilibriaEquilibria

Many equilibrium reactions Many equilibrium reactions involve reactants and products where involve reactants and products where more than one phase is present. more than one phase is present. These are called These are called heterogeneous heterogeneous equilibriaequilibria..

An example is the equilibrium An example is the equilibrium between solid calcium carbonate and between solid calcium carbonate and calcium oxide and carbon dioxide.calcium oxide and carbon dioxide.

CaCOCaCO33((ss) ↔ CaO() ↔ CaO(ss) + CO) + CO22((gg))

Page 19: Chemical Equilibrium The reversibility of reactions.

Heterogeneous Heterogeneous EquilibriaEquilibria

CaCOCaCO33((ss) ↔ CaO() ↔ CaO(ss) + CO) + CO22((gg))Experiments show that the Experiments show that the

position of the equilibrium does position of the equilibrium does not depend upon the amounts of not depend upon the amounts of calcium carbonate or calcium calcium carbonate or calcium oxide. That is, adding or removing oxide. That is, adding or removing some of the CaCOsome of the CaCO33((ss) or CaO() or CaO(ss) ) does not disrupt or alter the does not disrupt or alter the concentration of carbon dioxide.concentration of carbon dioxide.

Page 20: Chemical Equilibrium The reversibility of reactions.

Heterogeneous Heterogeneous EquilibriaEquilibria

CaCOCaCO33((ss) ↔ CaO() ↔ CaO(ss) + CO) + CO22((gg))

Page 21: Chemical Equilibrium The reversibility of reactions.

Heterogeneous Heterogeneous EquilibriaEquilibria

CaCOCaCO33((ss) ↔ CaO() ↔ CaO(ss) + CO) + CO22((gg))

This is because the This is because the concentrationsconcentrations of pure solids (or of pure solids (or liquids) cannot change. As a liquids) cannot change. As a result, the equilibrium constant result, the equilibrium constant expression for the above reaction expression for the above reaction is:is:

K=[COK=[CO22] or K] or Kpp = P = PCO2CO2

Page 22: Chemical Equilibrium The reversibility of reactions.

Heterogeneous Heterogeneous EquilibriaEquilibria

The position of a heterogeneous The position of a heterogeneous equilibrium does not depend on equilibrium does not depend on the amounts of pure solids or the amounts of pure solids or liquids present.liquids present.

Page 23: Chemical Equilibrium The reversibility of reactions.

The Reaction QuotientThe Reaction Quotient

If reactants and products of a If reactants and products of a reaction are mixed together, it is reaction are mixed together, it is possible to determine whether the possible to determine whether the reaction will proceed to the right or reaction will proceed to the right or left. This is accomplished by left. This is accomplished by comparing the composition of the comparing the composition of the initial mixture to that at equilibrium.initial mixture to that at equilibrium.

Page 24: Chemical Equilibrium The reversibility of reactions.

The Reaction QuotientThe Reaction Quotient

If the concentration of one of If the concentration of one of the reactants or products is zero, the the reactants or products is zero, the reaction will proceed so as to make reaction will proceed so as to make the missing component. the missing component.

If all components are present If all components are present initially, the initially, the reaction quotient, reaction quotient, Q, is Q, is compared to K to determine which compared to K to determine which way the reaction will go.way the reaction will go.

Page 25: Chemical Equilibrium The reversibility of reactions.

The Reaction QuotientThe Reaction Quotient

Q has the same form as the Q has the same form as the equilibrium constant expression, but equilibrium constant expression, but we use we use initialinitial concentrations instead concentrations instead of equilibrium concentrations. of equilibrium concentrations. Initial concentrations are usually Initial concentrations are usually indicated with a subscript zero.indicated with a subscript zero.

Page 26: Chemical Equilibrium The reversibility of reactions.

The Reaction QuotientThe Reaction Quotient

K = 2.2 x 10K = 2.2 x 1022 for the reaction: for the reaction:

NN22((gg) + 3 H) + 3 H22((gg) ↔ 2 NH) ↔ 2 NH33((gg))

Suppose 1.00 mol NSuppose 1.00 mol N22, 2.0 mol H, 2.0 mol H22 and and 1.5 mol of NH1.5 mol of NH3 3 are placed in a 1.00L are placed in a 1.00L vessel. What will happen? In which vessel. What will happen? In which direction with the reaction proceed?direction with the reaction proceed?

Page 27: Chemical Equilibrium The reversibility of reactions.

The Reaction QuotientThe Reaction Quotient

K = 2.2 x 10K = 2.2 x 1022 for the reaction: for the reaction:

NN22((gg) + 3 H) + 3 H22((gg) ↔ 2 NH) ↔ 2 NH33((gg))

Suppose 1.00 mol NSuppose 1.00 mol N22, 2.0 mol H, 2.0 mol H22 and and 1.5 mol of NH1.5 mol of NH3 3 are placed in a 1.00L are placed in a 1.00L vessel. What will happen? In which vessel. What will happen? In which direction with the reaction proceed?direction with the reaction proceed?

1. Calculate Q and compare it 1. Calculate Q and compare it to K.to K.

Page 28: Chemical Equilibrium The reversibility of reactions.

The Reaction QuotientThe Reaction Quotient

A comparison of Q with K will indicate the direction the reaction will go.

Page 29: Chemical Equilibrium The reversibility of reactions.

The Reaction QuotientThe Reaction Quotient

K = 2.2 x 10K = 2.2 x 1022 for the reaction: for the reaction:

NN22((gg) + 3 H) + 3 H22((gg) ↔ 2 NH) ↔ 2 NH33((gg))

Suppose 1.00 mol NSuppose 1.00 mol N22, 2.0 mol H, 2.0 mol H22 and and 1.5 mol of NH1.5 mol of NH3 3 are placed in a 1.00L are placed in a 1.00L vessel. What will happen? In which vessel. What will happen? In which direction with the reaction proceed?direction with the reaction proceed?

1. Calculate Q and compare it 1. Calculate Q and compare it to K.to K.

Page 30: Chemical Equilibrium The reversibility of reactions.

∆∆G for Non-Standard G for Non-Standard ConditionsConditions

The thermodynamic tables are The thermodynamic tables are for standard conditions. This for standard conditions. This includes having all reactants and includes having all reactants and products present initially at a products present initially at a temperature of 25temperature of 25ooC. All gases are C. All gases are at a pressure of 1 atm, and all at a pressure of 1 atm, and all solutions are 1 M.solutions are 1 M.

Page 31: Chemical Equilibrium The reversibility of reactions.

Equilibrium and Free Equilibrium and Free EnergyEnergy

Thermodynamic tables can be used to Thermodynamic tables can be used to calculate ΔGcalculate ΔGoo, the , the standard standard Gibbs free Gibbs free energy change. Since the direction in energy change. Since the direction in which a reaction will proceed depends which a reaction will proceed depends upon the reaction quotient, Q, the value of upon the reaction quotient, Q, the value of ΔG, will also depend upon the initial ΔG, will also depend upon the initial concentrations of reactants and products.concentrations of reactants and products.

ΔG = ΔGΔG = ΔGo o + RT ln Q + RT ln Q

Page 32: Chemical Equilibrium The reversibility of reactions.

∆∆G for Non-Standard G for Non-Standard ConditionsConditions

For non-standard temperature, For non-standard temperature, concentrations or gas pressures: concentrations or gas pressures:

∆∆G = ∆GG = ∆Goo + RTlnQ + RTlnQ

Where R = 8.314 J/K-molWhere R = 8.314 J/K-mol

T is temperature in KelvinsT is temperature in Kelvins

Q is the reaction quotientQ is the reaction quotient

Page 33: Chemical Equilibrium The reversibility of reactions.

∆∆G for Non-Standard G for Non-Standard ConditionsConditions

For non-standard temperature, For non-standard temperature, concentrations or gas pressures: concentrations or gas pressures:

∆∆G = ∆GG = ∆Goo + RT ln Q + RT ln Q

For Q, gas pressures are in For Q, gas pressures are in atmospheres, and concentrations of atmospheres, and concentrations of

solutions are in molarity, M.solutions are in molarity, M.

Page 34: Chemical Equilibrium The reversibility of reactions.

∆∆GGo o and Equilibrium and Equilibrium

A large negative value of ∆GA large negative value of ∆Goo indicates that the forward reaction indicates that the forward reaction or process is spontaneous. That is, or process is spontaneous. That is, there is a large there is a large driving forcedriving force for the for the forward reaction. This also means forward reaction. This also means that the equilibrium constant for the that the equilibrium constant for the reaction will be large. reaction will be large.

Page 35: Chemical Equilibrium The reversibility of reactions.

∆∆GGo o and Equilibrium and Equilibrium

A large positive value of ∆GA large positive value of ∆Goo indicates that the reverse reaction indicates that the reverse reaction or process is spontaneous. That is, or process is spontaneous. That is, there is a large there is a large driving forcedriving force for the for the reverse reaction. This also means reverse reaction. This also means that the equilibrium constant for the that the equilibrium constant for the reaction will be small. reaction will be small.

When a reaction or process is at When a reaction or process is at equilibrium, ∆Gequilibrium, ∆Goo = zero. = zero.

Page 36: Chemical Equilibrium The reversibility of reactions.

∆∆GGo o and Equilibrium and Equilibrium

Page 37: Chemical Equilibrium The reversibility of reactions.

∆∆GGo o and Equilibrium and Equilibrium

∆∆G = ∆GG = ∆Goo + RT lnQ + RT lnQ

At equilibrium, ∆G is equal to At equilibrium, ∆G is equal to zero, and zero, and

Q = K.Q = K.

0 = ∆G0 = ∆Goo + RT lnK + RT lnK

∆∆GGoo = - RT lnK = - RT lnK

Page 38: Chemical Equilibrium The reversibility of reactions.

∆∆GGo o and Equilibrium and Equilibrium

Calculate, ∆GCalculate, ∆Go o and K at 25and K at 25ooC for:C for:

HH22O(l) ↔ HO(l) ↔ H22(g) + ½ O(g) + ½ O22(g) (g)

Page 39: Chemical Equilibrium The reversibility of reactions.

∆∆GGo o and Equilibrium and Equilibrium

Calculate, ∆GCalculate, ∆Go o and K at 25and K at 25ooC for:C for:

HH22O(l) ↔ HO(l) ↔ H22(g) + ½ O(g) + ½ O22(g) (g)

Since hydrogen and oxygen react Since hydrogen and oxygen react explosively to form water, the reverse explosively to form water, the reverse reaction should be unfavorable, and reaction should be unfavorable, and have a positive ∆Ghave a positive ∆Go o , and a small , and a small equilibrium constant.equilibrium constant.

Page 40: Chemical Equilibrium The reversibility of reactions.

Calculate, ∆GCalculate, ∆Go o and K at 25and K at 25ooC for:C for:

HH22O(l) ↔ HO(l) ↔ H22(g) + ½ O(g) + ½ O22(g) (g)

Page 41: Chemical Equilibrium The reversibility of reactions.

∆∆GGo o and Equilibrium and Equilibrium

Calculate, ∆GCalculate, ∆Go o and K at 25and K at 25ooC for:C for:

HH22O(l) ↔ HO(l) ↔ H22(g) + ½ O(g) + ½ O22(g) (g)

∆∆GGo o = = ΣΣ∆G∆Gffo o products –products –ΣΣ∆G∆Gff

o o reactants reactants

∆∆GGo o = [(1 mol) (∆G= [(1 mol) (∆Gffo o HH22(g)) +(½ mol) (g)) +(½ mol)

(∆G(∆Gffo o HH22(g))] –[(1 mol) (∆G(g))] –[(1 mol) (∆Gff

ooHH22O(l)]O(l)]

∆∆GGo o =0 –[( 1 mol)(-237.1kJ/mol)] = =0 –[( 1 mol)(-237.1kJ/mol)] = +237.1 kJ+237.1 kJ

Page 42: Chemical Equilibrium The reversibility of reactions.

Free Energy and Free Energy and EquilibriumEquilibrium

∆∆GGoo = - RT lnK = - RT lnK

where R = 8.314 J/K-molwhere R = 8.314 J/K-mol

lnK = –∆GlnK = –∆Goo /RT /RT

For HFor H22O(l) ↔ HO(l) ↔ H22(g) + ½ O(g) + ½ O22(g) :(g) :

lnK = –237.1 kJ(10lnK = –237.1 kJ(1033J/kJ)/(8.314J/mol-K) J/kJ)/(8.314J/mol-K) (298K)(298K)

ln K = –95.70ln K = –95.70

K = eK = e–95.70–95.70 = 2.7 x 10 = 2.7 x 10-42-42

Page 43: Chemical Equilibrium The reversibility of reactions.

Free Energy and Free Energy and EquilibriumEquilibrium

In summary, an unfavorable In summary, an unfavorable reaction has a positive free energy reaction has a positive free energy change, and a small equilibrium change, and a small equilibrium constant.constant.

Page 44: Chemical Equilibrium The reversibility of reactions.

Problem: Calculation of Problem: Calculation of KK

At a certain temperature, 3.00 moles At a certain temperature, 3.00 moles of ammonia were placed in a 2.00 L of ammonia were placed in a 2.00 L vessel. At equilibrium, 2.50 moles vessel. At equilibrium, 2.50 moles remain. Calculate K for the remain. Calculate K for the reaction:reaction:

NN22((gg) + 3 H) + 3 H22((gg) ↔ 2 NH) ↔ 2 NH33((gg))

Page 45: Chemical Equilibrium The reversibility of reactions.

Problem: Calculation of Problem: Calculation of KK

At a certain temperature, 3.00 moles At a certain temperature, 3.00 moles of ammonia were placed in a 2.00 L of ammonia were placed in a 2.00 L vessel. At equilibrium, 2.50 moles vessel. At equilibrium, 2.50 moles remain. Calculate K for the remain. Calculate K for the reaction:reaction:

NN22((gg) + 3 H) + 3 H22((gg) ↔ 2 NH) ↔ 2 NH33((gg))

1. Write the equilibrium constant 1. Write the equilibrium constant expression.expression.

Page 46: Chemical Equilibrium The reversibility of reactions.

Problem: Calculation of Problem: Calculation of KK

At a certain temperature, 3.00 moles At a certain temperature, 3.00 moles of ammonia were placed in a 2.00 L of ammonia were placed in a 2.00 L vessel. At equilibrium, 2.50 moles vessel. At equilibrium, 2.50 moles remain. Calculate K for the reaction:remain. Calculate K for the reaction:

NN22((gg) + 3 H) + 3 H22((gg) ↔ 2 NH) ↔ 2 NH33((gg))

1. Write the equilibrium constant 1. Write the equilibrium constant expression.expression.

K = [NHK = [NH33]]22/([N/([N22][H][H22]]33))

Page 47: Chemical Equilibrium The reversibility of reactions.

Problem: Calculation of Problem: Calculation of KK

At a certain temperature, 3.00 moles At a certain temperature, 3.00 moles of ammonia were placed in a 2.00 L of ammonia were placed in a 2.00 L vessel. At equilibrium, 2.50 moles vessel. At equilibrium, 2.50 moles remain. Calculate K for the remain. Calculate K for the reaction:reaction:

NN22((gg) + 3 H) + 3 H22((gg) ↔ 2 NH) ↔ 2 NH33((gg))

2. Make a table of concentrations.2. Make a table of concentrations.

Page 48: Chemical Equilibrium The reversibility of reactions.

Problem: Calculation of Problem: Calculation of KK

NN22((gg) + 3 H) + 3 H22((gg) ↔ ) ↔ 2NH2NH33((gg)) [N[N22]]

[H[H22

]][NH[NH33

]]

initialinitial

3.003.00molmol 2.00L2.00L

changchangee

equiliequili--

libriulibriumm

2.502.50molmol2.00 2.00 LL

Page 49: Chemical Equilibrium The reversibility of reactions.

Problem: Calculation of Problem: Calculation of KK

NN22((gg) + 3 H) + 3 H22((gg) ↔ ) ↔ 2NH2NH33((gg))

3. Complete the 3. Complete the table.table.

[N[N22]][H[H22

]][NH[NH33

]]

initialinitial

3.003.00molmol 2.00L2.00L

changchangee

equiliequili--

libriulibriumm

2.502.50molmol2.00 2.00 LL

Page 50: Chemical Equilibrium The reversibility of reactions.

Problem: Calculation of Problem: Calculation of KK

NN22((gg) + 3 H) + 3 H22((gg) ↔ ) ↔ 2NH2NH33((gg))

3. Complete the 3. Complete the table.table.

[N[N22]][H[H22

]][NH[NH33

]]

initialinitial

3.003.00molmol 2.00L2.00L

changchangee

-.50m-.50molol2.00 2.00 LL

equiliequili--

libriulibriumm

2.502.50molmol2.00 2.00 LL

Page 51: Chemical Equilibrium The reversibility of reactions.

Problem: Calculation of Problem: Calculation of KK

NN22((gg) + 3 H) + 3 H22((gg) ↔ ) ↔ 2NH2NH33((gg))

3. Complete the 3. Complete the table.table.

[N[N22]][H[H22

]][NH[NH33

]]

initialinitial 00 00

3.003.00molmol 2.00L2.00L

changchangee

-.50m-.50molol2.00 2.00 LL

equiliequili--

libriulibriumm

2.502.50molmol2.00 2.00 LL

Page 52: Chemical Equilibrium The reversibility of reactions.

[N[N22]] [H[H22]][NH[NH33

]]

initiainitiall

00 00

3.003.00molmol 2.00L2.00L

chanchangege

+.25m+.25molol2.00 L2.00 L

+.75m+.75molol2.00L2.00L

-.50m-.50molol2.00 2.00 LL

equiliequili--

libriulibriumm

2.502.50molmol2.00 2.00 LL

NN22((gg) + 3H) + 3H22((gg) ) ↔2NH↔2NH33((gg))

Page 53: Chemical Equilibrium The reversibility of reactions.

[N[N22]] [H[H22]][NH[NH33

]]

initiainitiall

00 00

3.003.00molmol 2.00L2.00L

chanchangege

+.25m+.25molol2.00 L2.00 L

+.75m+.75molol2.00L2.00L

-.50m-.50molol2.00 2.00 LL

equiliequili--

libriulibriumm

2.502.50molmol2.00 2.00 LL

NN22((gg) + 3H) + 3H22((gg) ) ↔2NH↔2NH33((gg))

The equilibrium values are the sum of the initial concentrations plus any changes that occur.

Page 54: Chemical Equilibrium The reversibility of reactions.

[N[N22]] [H[H22]] [NH[NH33]]

initiainitiall

00 003.00m3.00molol 2.00L2.00L

chanchangege

+.25m+.25molol2.00 L2.00 L

+.75m+.75molol2.00L2.00L

-.50mo-.50moll2.00 L2.00 L

equiliequili--

libriulibriumm

+.13+.13MM

+.38+.38MM

+1.25+1.25MM

NN22((gg) + 3H) + 3H22((gg) ) ↔2NH↔2NH33((gg))

The equilibrium values are the sum of the initial concentrations plus any changes that occur.

Page 55: Chemical Equilibrium The reversibility of reactions.

4. Substitute and 4. Substitute and solve for K.solve for K.

K K == [NH [NH33]]22/([N/([N22]][H[H22]]33))

[N[N22]] [H[H22]] [NH[NH33]]

equiliequili--

libriulibriumm

+.13+.13MM

+.38+.38MM

+1.25+1.25MM

NN22((gg) + 3H) + 3H22((gg) ) ↔2NH↔2NH33((gg))

K=(1.25)2/[(.13)(.38)3] = 2.2 x 102

Page 56: Chemical Equilibrium The reversibility of reactions.

ProblemProblem

HH22((gg) + I) + I22((gg) ↔ 2HI() ↔ 2HI(gg) K) Kpp = 1.00 = 1.00 x 10x 1022

Initially, PInitially, Phydrogenhydrogen = P = Piodineiodine = 0.500 atm. = 0.500 atm. Calculate the equilibrium partial Calculate the equilibrium partial pressures of all three gases.pressures of all three gases.

Page 57: Chemical Equilibrium The reversibility of reactions.

ProblemProblem

HH22((gg) + I) + I22((gg) ↔ 2HI() ↔ 2HI(gg) K) Kpp = 1.00 = 1.00 x 10x 1022

Initially, PInitially, Phydrogenhydrogen = P = Piodineiodine = 0.500 atm. = 0.500 atm. Calculate the equilibrium partial Calculate the equilibrium partial pressures of all three gases.pressures of all three gases.

1. Write the K1. Write the Kpp expression.expression.

Page 58: Chemical Equilibrium The reversibility of reactions.

ProblemProblem

HH22((gg) + I) + I22((gg) ↔ 2HI() ↔ 2HI(gg) K) Kpp = 1.00 = 1.00 x 10x 1022

Initially, PInitially, Phydrogenhydrogen = P = Piodineiodine = 0.500 atm. = 0.500 atm. Calculate the equilibrium partial Calculate the equilibrium partial pressures of all three gases.pressures of all three gases.

1. Write the K1. Write the Kpp expression.expression.

KKpp = (P = (PHIHI))22

(P(PH2H2)(P)(PI2I2))

Page 59: Chemical Equilibrium The reversibility of reactions.

ProblemProblem

HH22((gg) + I) + I22((gg) ↔ 2HI() ↔ 2HI(gg) K) Kpp = 1.00 = 1.00 x 10x 1022

Initially, PInitially, Phydrogenhydrogen = P = Piodineiodine = 0.500 atm. = 0.500 atm. Calculate the equilibrium partial Calculate the equilibrium partial pressures of all three gases.pressures of all three gases.

2. Make a table of initial, change 2. Make a table of initial, change and and equilibrium pressures.equilibrium pressures.

Page 60: Chemical Equilibrium The reversibility of reactions.

Problem: HProblem: H22((gg) + I) + I22((gg) ↔ ) ↔ 2HI(2HI(gg) )

HH22 II22 HIHI

initialinitial .500 .500 atmatm

.500 .500 atmatm 00

changchangee

equili-equili-

briumbrium

Page 61: Chemical Equilibrium The reversibility of reactions.

Problem: HProblem: H22((gg) + I) + I22((gg) ↔ ) ↔ 2HI(2HI(gg) )

HH22 II22 HIHI

initialinitial .500 .500 atmatm

.500 .500 atmatm 00

changchangee

-x-x -x-x +2x+2x

equili-equili-

briumbrium

Page 62: Chemical Equilibrium The reversibility of reactions.

Problem: HProblem: H22((gg) + I) + I22((gg) ↔ ) ↔ 2HI(2HI(gg) )

HH22 II22 HIHI

initialinitial .500 .500 atmatm

.500 .500 atmatm 00

changchangee

-x-x -x-x +2x+2x

equili-equili-

briumbrium.500-x.500-x .500-x.500-x 2x2x

Page 63: Chemical Equilibrium The reversibility of reactions.

Problem: HProblem: H22((gg) + I) + I22((gg) ↔ ) ↔ 2HI(2HI(gg) ) HH22 II22 HIHI

equili-equili-

briumbrium.500-x.500-x .500-x.500-x 2x2x

3. Substitute and solve.

Page 64: Chemical Equilibrium The reversibility of reactions.

Problem: HProblem: H22((gg) + I) + I22((gg) ↔ ) ↔ 2HI(2HI(gg) ) HH22 II22 HIHI

equili-equili-

briumbrium.500-x.500-x .500-x.500-x 2x2x

3.Substitute and solve.

Kp= (PHI)2/(PH2)(PI2) = 1.00 x 10= 1.00 x 1022

(2x)2 = 1.00 x = 1.00 x 101022

(.500-x) (.500-x)

Page 65: Chemical Equilibrium The reversibility of reactions.

Problem: HProblem: H22((gg) + I) + I22((gg) ↔ ) ↔ 2HI(2HI(gg) )

3. Substitute and solve.Kp= (PHI)2/(PH2)(PI2) = 1.00 x 10= 1.00 x 1022

(2x)2 = 1.00 x 10= 1.00 x 1022

(.500-x) (.500-x)

Take the square root of both sides: (2x) = 10.0= 10.0

(.500-x)

Page 66: Chemical Equilibrium The reversibility of reactions.

Problem: HProblem: H22((gg) + I) + I22((gg) ↔ ) ↔ 2HI(2HI(gg) )

3. Substitute and solve. (2x) = 10.0= 10.0

(.500-x)

2x = 10.0 (.500-x) = 5.00 -10.0x

12.0 x = 5.00

x = .417

Page 67: Chemical Equilibrium The reversibility of reactions.

Problem: HProblem: H22((gg) + I) + I22((gg) ↔ ) ↔ 2HI(2HI(gg) )

4. Answer the question: Calculate the Calculate the

equilibrium partial pressures of all three equilibrium partial pressures of all three

gases.gases.

x = .417

HH22 II22 HIHI

equili-equili-

briumbrium .500-x.500-x .500-x.500-x 2x2x

Page 68: Chemical Equilibrium The reversibility of reactions.

Problem: HProblem: H22((gg) + I) + I22((gg) ↔ ) ↔ 2HI(2HI(gg) )

4. Answer the question: Calculate the Calculate the

equilibrium partial pressures of all three equilibrium partial pressures of all three

gases.gases.

x = .417

HH22 II22 HIHI

equili-equili-

briumbrium.500-x.500-x

0.083 atm0.083 atm.500-x.500-x

0.083 atm0.083 atm2x2x

.834 atm.834 atm

Page 69: Chemical Equilibrium The reversibility of reactions.

Problem: HProblem: H22((gg) + I) + I22((gg) ↔ ) ↔ 2HI(2HI(gg) )

5. Check your answer, if possible. Does (PHI)2/(PH2)(PI2) = 1.00 x 10= 1.00 x 102 2 ??

(.834)(.834)22/(.083)(.083) = (.696)/(.0069) = 1.0 x 10/(.083)(.083) = (.696)/(.0069) = 1.0 x 1022

HH22 II22 HIHI

equili-equili-

briumbrium.500-x.500-x

0.083 atm0.083 atm.500-x.500-x

0.083 atm0.083 atm2x2x

.834 atm.834 atm

Page 70: Chemical Equilibrium The reversibility of reactions.

Problem: NProblem: N22((gg) + O) + O22((gg) ↔ ) ↔ 2NO(2NO(gg))

At 2200 At 2200 ooC, K = 0.050 for the above C, K = 0.050 for the above reaction. Initially, 1.60 mol of reaction. Initially, 1.60 mol of nitrogen and 0.400 mol of oxygen nitrogen and 0.400 mol of oxygen are sealed in a 2.00 liter vessel. are sealed in a 2.00 liter vessel. Calculate the concentration of all Calculate the concentration of all species at equilibrium.species at equilibrium.

Page 71: Chemical Equilibrium The reversibility of reactions.

Problem: NProblem: N22((gg) + O) + O22((gg) ↔ ) ↔ 2NO(2NO(gg))

At 2200 At 2200 ooC, K = 0.050 for the above C, K = 0.050 for the above reaction. Initially, 1.60 mol of nitrogen reaction. Initially, 1.60 mol of nitrogen and 0.400 mol of oxygen are sealed in a and 0.400 mol of oxygen are sealed in a 2.00 liter vessel. Calculate the 2.00 liter vessel. Calculate the concentration of all species at equilibrium.concentration of all species at equilibrium.

1. Write the equilibrium constant 1. Write the equilibrium constant expression.expression.

K = 0.050 = [NO]K = 0.050 = [NO]22/[N/[N22][O][O22]]

Page 72: Chemical Equilibrium The reversibility of reactions.

Problem: NProblem: N22((gg) + O) + O22((gg) ↔ ) ↔ 2NO(2NO(gg))

At 2200 At 2200 ooC, K = 0.050 for the above C, K = 0.050 for the above reaction. Initially, 1.60 mol of reaction. Initially, 1.60 mol of nitrogen and 0.400 mol of oxygen nitrogen and 0.400 mol of oxygen are sealed in a 2.00 liter vessel. are sealed in a 2.00 liter vessel. Calculate the concentration of all Calculate the concentration of all species at equilibrium.species at equilibrium.

2. Make a table of initial, change 2. Make a table of initial, change and and equilibrium concentrations.equilibrium concentrations.

Page 73: Chemical Equilibrium The reversibility of reactions.

Problem: NProblem: N22((gg) + O) + O22((gg) ↔ ) ↔ 2NO(2NO(gg))

NN22 OO22 NONO

initialinitial1.60mo1.60mo

ll2.00 L2.00 L

.400 .400 molmol

2.00 L2.00 L00

changechange -x-x -x-x +2x+2x

equili-equili-

briumbrium.800 - x.800 - x .200 - x.200 - x 2x2x

Page 74: Chemical Equilibrium The reversibility of reactions.

Problem: NProblem: N22((gg) + O) + O22((gg) ↔ ) ↔ 2NO(2NO(gg))

NN22 OO22 NONO

equili-equili-

briumbrium.800 - x.800 - x .200 - x.200 - x 2x2x

3.Substitute and solve.

K = 0.050 = [NO]K = 0.050 = [NO]22/[N/[N22][O][O22]]

Page 75: Chemical Equilibrium The reversibility of reactions.

Approaches to Solving Approaches to Solving ProblemsProblems

Some equilibrium problems Some equilibrium problems require use of the quadratic require use of the quadratic equation to obtain an accurate equation to obtain an accurate solution. In cases where the solution. In cases where the equilibrium constant is quite small equilibrium constant is quite small (roughly 10(roughly 10-5-5 or smaller), it is often or smaller), it is often possible to make assumptions that possible to make assumptions that will simplify the mathematics.will simplify the mathematics.

Page 76: Chemical Equilibrium The reversibility of reactions.

Problem: 2NOCl(g)↔2NO(g) + Problem: 2NOCl(g)↔2NO(g) + ClCl22(g)(g)

At 35At 35ooC, K = 1.6 x 10C, K = 1.6 x 10-5-5 for the above for the above reaction. Calculate the equilibrium reaction. Calculate the equilibrium concentrations of all species present concentrations of all species present if 2.0 moles of NOCl and 1.0 mole of if 2.0 moles of NOCl and 1.0 mole of ClCl22 are placed in a 1.0 liter flask. are placed in a 1.0 liter flask.

1. K = 1.6 x 101. K = 1.6 x 10-5-5 =[NO]=[NO]22[Cl[Cl22]/[NOCl]]/[NOCl]22

Page 77: Chemical Equilibrium The reversibility of reactions.

Problem: 2NOCl(g)↔2NO(g) + Problem: 2NOCl(g)↔2NO(g) + ClCl22(g)(g)

At 35At 35ooC, K = 1.6 x 10C, K = 1.6 x 10-5-5 for the above for the above reaction. Calculate the equilibrium reaction. Calculate the equilibrium concentrations of all species present concentrations of all species present if 2.0 moles of NOCl and 1.0 mole of if 2.0 moles of NOCl and 1.0 mole of ClCl22 are placed in a 1.0 liter flask. are placed in a 1.0 liter flask.

2. Make a table of concentrations.2. Make a table of concentrations.

Page 78: Chemical Equilibrium The reversibility of reactions.

Problem: 2NOCl(g)↔2NO(g) + Problem: 2NOCl(g)↔2NO(g) + ClCl22(g)(g)

NOClNOCl NONO ClCl22

initiainitiall

2.0mo2.0moll

1.00 1.00 LL

001.0 1.0 molmol

1.00 L1.00 L

chanchangege -2x-2x +2x+2x +x+x

equiliequili--

briubriumm

2.0 - 2.0 - 2x2x 2x2x 1.0+x1.0+x

Page 79: Chemical Equilibrium The reversibility of reactions.

Problem: 2NOCl(g)↔2NO(g) + Problem: 2NOCl(g)↔2NO(g) + ClCl22(g)(g)

NOClNOCl NONO ClCl22

equiliequili--

briubriumm

2.0 - 2.0 - 2x2x 2x2x 1.0+x1.0+x

3.Substitute and solve.

K = 1.6 x 10K = 1.6 x 10-5-5 =[NO] =[NO]22[Cl[Cl22]/[NOCl]]/[NOCl]22

1.6 x 101.6 x 10-5-5 =[2x] =[2x]22[1.0 + x]/[2.0-2x][1.0 + x]/[2.0-2x]22

Page 80: Chemical Equilibrium The reversibility of reactions.

Problem: 2NOCl(g)↔2NO(g) + Problem: 2NOCl(g)↔2NO(g) + ClCl22(g)(g)

3.Substitute and solve.

1.6 x 101.6 x 10-5-5 = [2x] = [2x]2 2 [1.0 + x] [1.0 + x] [2.0-2x][2.0-2x]22

Since K is small, x, the amount of Since K is small, x, the amount of product formed, will be a small product formed, will be a small number. As a result, 1.0 + x ≈1.0, number. As a result, 1.0 + x ≈1.0, and 2.0-2x≈2.0and 2.0-2x≈2.0

Page 81: Chemical Equilibrium The reversibility of reactions.

Problem: 2NOCl(g)↔2NO(g) + Problem: 2NOCl(g)↔2NO(g) + ClCl22(g)(g)

3.Substitute and solve.

1.6 x 101.6 x 10-5-5 = [2x] = [2x]2 2 [1.0 + x] [1.0 + x] [2.0-2x][2.0-2x]22

The expression simplifies to:The expression simplifies to:1.6 x 101.6 x 10-5-5 = [2x] = [2x]22 [1.0] [1.0]

[2.0][2.0]22

0

0

Page 82: Chemical Equilibrium The reversibility of reactions.

Problem: 2NOCl(g)↔2NO(g) + Problem: 2NOCl(g)↔2NO(g) + ClCl22(g)(g)

3.Solve for x.

1.6 x 101.6 x 10-5-5 = [2x] = [2x]22 [1.0] [1.0] [2.0][2.0]22

4x4x22 = 1.6 x 10 = 1.6 x 10-5-5 (4)/1 = 6.4 x 10-5

xx22 = 1.6 x 10 = 1.6 x 10-5 -5 ; x =0.0040 ; x =0.0040

Page 83: Chemical Equilibrium The reversibility of reactions.

Problem: 2NOCl(g)↔2NO(g) + Problem: 2NOCl(g)↔2NO(g) + ClCl22(g)(g)

4. Check your assumption.

x =0.0040x =0.0040

Does 1.0 + x ≈1.0, and 2.0-Does 1.0 + x ≈1.0, and 2.0-2x≈2.0?2x≈2.0?

1.0 + .004 = 1.01.0 + .004 = 1.0

2.0 – 2(.004) = 2.0 - .008=2.02.0 – 2(.004) = 2.0 - .008=2.0

Page 84: Chemical Equilibrium The reversibility of reactions.

Validity of the AssumptionValidity of the Assumption

These assumptions are generally considered valid if they cause an insignificant error. In this case, due to significant figures, no error is introduced. Usually, if the difference is within 5%, the error is considered acceptable.

Page 85: Chemical Equilibrium The reversibility of reactions.

Problem: 2NOCl(g)↔2NO(g) + Problem: 2NOCl(g)↔2NO(g) + ClCl22(g)(g)

NOClNOCl NONO ClCl22

equiliequili--

briubriumm

2.0 - 2.0 - 2x2x 2x2x 1.0+x1.0+x

5.Answer the question.

[NOCl] = 2.0M; [NO]= 0.0080M; [NOCl] = 2.0M; [NO]= 0.0080M;

[Cl[Cl22] = 1.0M] = 1.0M

Page 86: Chemical Equilibrium The reversibility of reactions.

Problem: 2NOCl(g)↔2NO(g) + Problem: 2NOCl(g)↔2NO(g) + ClCl22(g)(g)

5.Check your answer.

[NOCl] = 2.0M; [NO]= 0.0080M; [NOCl] = 2.0M; [NO]= 0.0080M;

[Cl[Cl22] = 1.0M] = 1.0M

Does [NO]Does [NO]22[Cl[Cl22]/[NOCl]]/[NOCl]2 2 = 1.6 x 10= 1.6 x 10--

55 ? ?

(.0080)(.0080)22(1.0)/(2.0)(1.0)/(2.0)22 = 1.6 x 10 = 1.6 x 10-5-5

Page 87: Chemical Equilibrium The reversibility of reactions.

Le ChatelierLe Chatelier’’s Principles Principle

If a stress is applied to a system at If a stress is applied to a system at equilibrium, the position of the equilibrium, the position of the equilibrium will shift so as to equilibrium will shift so as to counteract the stress.counteract the stress.

Page 88: Chemical Equilibrium The reversibility of reactions.

Le ChatelierLe Chatelier’’s Principles Principle

If a stress is applied to a system at If a stress is applied to a system at equilibrium, the position of the equilibrium, the position of the equilibrium will shift so as to counteract equilibrium will shift so as to counteract the stress.the stress.

Types of stresses: Types of stresses: Addition or removal of reactants or Addition or removal of reactants or

productsproducts Changes in pressure or volume (for gases)Changes in pressure or volume (for gases) Changes in temperatureChanges in temperature

Page 89: Chemical Equilibrium The reversibility of reactions.

FeFe3+3+(aq)(aq) + SCN + SCN-- ↔ ↔ FeSCNFeSCN2+2+

Page 90: Chemical Equilibrium The reversibility of reactions.

Heat + NHeat + N22OO44(g) ↔ 2 (g) ↔ 2 NONO22(g)(g)

Page 91: Chemical Equilibrium The reversibility of reactions.

CoClCoCl442- 2- + 6 H+ 6 H22OOCo(HCo(H22O)O)66

2+2++ + 4Cl4Cl1-1- + +

heatheat

Page 92: Chemical Equilibrium The reversibility of reactions.

Le ChatelierLe Chatelier’’s Principles Principle

For reactions involving gaseous For reactions involving gaseous products or reactants, changes in products or reactants, changes in pressure of volume pressure of volume maymay shift the shift the equilibrium. The equilibrium will shift equilibrium. The equilibrium will shift only if there is an unequal number of only if there is an unequal number of moles of gas on either side of the moles of gas on either side of the reaction. Reactions involving liquids reaction. Reactions involving liquids or solids are not significantly affected or solids are not significantly affected by changes in pressure or volume.by changes in pressure or volume.

Page 93: Chemical Equilibrium The reversibility of reactions.

Le ChatelieLe Chatelierr ’’s Principles Principle

Page 94: Chemical Equilibrium The reversibility of reactions.

Pressure ChangesPressure Changes

Page 95: Chemical Equilibrium The reversibility of reactions.

Le ChatelierLe Chatelier’’s Principles Principle

For the reaction For the reaction

C(C(ss) + 2 H) + 2 H22(g) ↔ CH(g) ↔ CH44(g) ∆H(g) ∆Hoo=-75kJ=-75kJ

Assuming the reaction is initially at Assuming the reaction is initially at equilibrium, predict the shift in equilibrium, predict the shift in equilibrium (if any) for each of the equilibrium (if any) for each of the following changes. following changes.

a) add carbona) add carbon

b) remove methaneb) remove methane

c) increase the temperaturec) increase the temperature

Page 96: Chemical Equilibrium The reversibility of reactions.

Le ChatelierLe Chatelier’’s Principles Principle

The value of K will change with The value of K will change with temperature. The effect of a change in temperature. The effect of a change in temperature on K depends on whether temperature on K depends on whether the reaction is exothermic or the reaction is exothermic or endothermic.endothermic.

The easiest way to predict the result The easiest way to predict the result when temperature is changed is to write when temperature is changed is to write in heat as a product (for exothermic in heat as a product (for exothermic reactions) or a reactant (for reactions) or a reactant (for endothermic reactions).endothermic reactions).

Page 97: Chemical Equilibrium The reversibility of reactions.

Le ChatelierLe Chatelier’’s Principles Principle For the reaction For the reaction

C(C(ss) + 2 H) + 2 H22(g) ↔ CH(g) ↔ CH44(g) ∆H(g) ∆Hoo=-75kJ=-75kJ Assuming the reaction is initially at Assuming the reaction is initially at equilibrium, predict the shift in equilibrium, predict the shift in equilibrium (if any) for each of the equilibrium (if any) for each of the following changes. following changes.

d) add a catalystd) add a catalyste) decrease the volumee) decrease the volumef) remove some carbonf) remove some carbong) add hydrogeng) add hydrogen


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