o If you recall, we mentioned that there are rxns that bounce back and forth from forming products...

o If you recall, we mentioned that If you recall, we mentioned that there are rxns that bounce back there are rxns that bounce back and forth from forming products to and forth from forming products to reforming reactantsreforming reactants

• A.K.A. A.K.A. reversible rxnsreversible rxnso Reversible reactions occur Reversible reactions occur

simultan-eously in simultan-eously in both directionsboth directions• An example of a reversible rxn An example of a reversible rxn

involves reactants A and B involves reactants A and B producing C.producing C. A + B CA + B C

Reversible ReactionsReversible ReactionsReversible ReactionsReversible Reactions

o At chemical equilibrium there is At chemical equilibrium there is no no net changenet change in the actual amounts in the actual amounts of the components of the system.of the components of the system.• And although the rates of the And although the rates of the

forward & reverse rxns are forward & reverse rxns are equalequal at chemical equilibrium, the at chemical equilibrium, the concentrations of the components concentrations of the components on both sides of the chem-ical eqn on both sides of the chem-ical eqn are not necessarily the same.are not necessarily the same.

&In fact they can be dramatically In fact they can be dramatically different.different.

Chemical EquilibriumChemical Equilibrium

o Consider a set of escalators as being like the double arrows in a dynamic equilibrium.

o The # of people using the up escalator must be the same as the # of people using the down escalator for equilibrium to have been established

• However, the # of people upstairs do not have to equal the # of people downstairs

• Just the transfer between floors must be consistent

o The equilibrium position of a rxn is The equilibrium position of a rxn is given by the given by the concentrationsconcentrations of the of the system’s components at system’s components at equilibriumequilibrium

o The equilibrium position indicates The equilibrium position indicates whether the components on the whether the components on the left or right side of a reversible rxn left or right side of a reversible rxn are at a are at a higher concentrationhigher concentration..

• If A reacts to give B and the If A reacts to give B and the mixture at equilibrium contains mixture at equilibrium contains more of B – say more of B – say 1% of A vs. 99% 1% of A vs. 99% of Bof B –the formation of B is said to –the formation of B is said to be favored.be favored.

Which Direction is Which Direction is Favored?Favored?

o On the other hand, if the mixture On the other hand, if the mixture contains 99% of A and 1% of B at contains 99% of A and 1% of B at equilibrium then the formation of A equilibrium then the formation of A is favored.is favored.

Reverse Reverse direction is direction is

favored!favored!

Forward Forward direction is direction is

favored!favored!

Which Direction is Which Direction is Favored?Favored?

o In principle, almost all rxns are In principle, almost all rxns are reversible to some extent under reversible to some extent under the the right conditionsright conditions

o In practice, one set of components In practice, one set of components is often so favored at equilibrium is often so favored at equilibrium that the other set cannot be that the other set cannot be detected.detected.

• If one set of components has If one set of components has established equilibrium by established equilibrium by converting mostly into products, converting mostly into products, the rxn has the rxn has gone to completiongone to completion

• When no products can be When no products can be detected, you can say there is detected, you can say there is no rxnno rxn

Reversibility vs. RealityReversibility vs. Reality

o Reversible rxns occupy a middle Reversible rxns occupy a middle ground between the theoretical ground between the theoretical extremes of irreversibility and no rxn.extremes of irreversibility and no rxn.

o The addition of a catalyst will speed The addition of a catalyst will speed up up forward and reverse rxns equallyforward and reverse rxns equally• By reducing the energy By reducing the energy needed to needed to

activateactivate the rxn in both forward and the rxn in both forward and reverse directions.reverse directions.

• Does not effect the amount of Does not effect the amount of reactants and products present at reactants and products present at equilibrium; simply equilibrium; simply decreases the decreases the timetime it takes to establish equilibrium it takes to establish equilibrium

o Reversible rxns occupy a middle Reversible rxns occupy a middle ground between the theoretical ground between the theoretical extremes of irreversibility and no rxn.extremes of irreversibility and no rxn.

o The addition of a catalyst will speed The addition of a catalyst will speed up up forward and reverse rxns equallyforward and reverse rxns equally• By reducing the energy By reducing the energy needed to needed to

activateactivate the rxn in both forward and the rxn in both forward and reverse directions.reverse directions.

• Does not effect the amount of Does not effect the amount of reactants and products present at reactants and products present at equilibrium; simply equilibrium; simply decreases the decreases the timetime it takes to establish equilibrium it takes to establish equilibrium

Reversibility vs. RealityReversibility vs. Reality

Equilibrium Expressiono Chemist’s can express the Chemist’s can express the

equilibrium position in terms of a equilibrium position in terms of a numerical constantnumerical constant• The equilibrium constant shows The equilibrium constant shows

the relationship between the the relationship between the amount of product and reactant at amount of product and reactant at equilibriumequilibrium

o Consider this hypothetical rxn…Consider this hypothetical rxn…

o Chemist’s can express the Chemist’s can express the equilibrium position in terms of a equilibrium position in terms of a numerical constantnumerical constant• The equilibrium constant shows The equilibrium constant shows

the relationship between the the relationship between the amount of product and reactant at amount of product and reactant at equilibriumequilibrium

o Consider this hypothetical rxn…Consider this hypothetical rxn…aA + bB cC + dD

o We can write an expression to We can write an expression to show the ratio of product show the ratio of product concentrations to reactant concentrations to reactant concs concs called acalled a mass mass action expression action expression

[C][C]cc[D][D]dd

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

Equilibrium ExpressionEquilibrium Expressiono The conc of each substance is The conc of each substance is

raised to a power equal to the # raised to a power equal to the # of mols of that substance in the of mols of that substance in the balanced rxn eqn.balanced rxn eqn.

o The square brackets indicate The square brackets indicate concentration in Molarityconcentration in Molarity ((mol/Lmol/L))

o Molarity is a measure of how Molarity is a measure of how much “stuff” is dissolved in water.much “stuff” is dissolved in water.&The more stuff dissolved, The more stuff dissolved,

the more concentrated the the more concentrated the solutionsolution

&The higher the molarityThe higher the molarityo The quotient ratio of the The quotient ratio of the equilibrium is called theequilibrium is called the equilibrium constant or Kequilibrium constant or K• When the reactants and products When the reactants and products

amnts are in molarity the constant amnts are in molarity the constant is called a Kis called a Kcc

• When the reactants and When the reactants and products amounts are in products amounts are in pressure units is called pressure units is called a K a Kpp

K=K= [C][C]cc[D][D]dd

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

o The constant is dependent on the tempThe constant is dependent on the temp• If the temp changes so does the constantIf the temp changes so does the constant

o The constant is dependent on the tempThe constant is dependent on the temp• If the temp changes so does the constantIf the temp changes so does the constant

Equilibrium ExpressionEquilibrium Expression

NOTE: water and solid materials are NOTE: water and solid materials are not included in mass action not included in mass action expressions.expressions.Write the mass action expression for Write the mass action expression for

each of the following reactions:each of the following reactions:A.A.2SO2SO22(g) + O(g) + O22(g) <==> (g) <==>

2SO2SO33(g)(g)B.B.BiBi22SS33(s) <==> 2Bi(s) <==> 2Bi+3+3(aq) + (aq) +

3S3S-2-2(aq)(aq)

Balance and write the mass action Balance and write the mass action expression for each of the following expression for each of the following reactions:reactions:1.1. CC44HH1010(g) + O(g) + O22(g) <==> CO(g) <==> CO22(g) + (g) + H H22O(g)O(g)

2.2. Al(s) + OAl(s) + O22(g) <==> Al(g) <==> Al22OO33(s)(s)

3.3. MnMn+2+2(aq)+ BiO(aq)+ BiO33-1-1(aq) <==> (aq) <==>

MnOMnO44-1-1(aq) + Bi(aq) + Bi+3+3(aq)(aq)

Classwork:Classwork:

o Equilibrium constants provide Equilibrium constants provide valuable chemical informationvaluable chemical information

o They show whether the products They show whether the products or the reactants or the reactants are are favored in a rxn favored in a rxn (spontaneus or (spontaneus or nonspontaneous)nonspontaneous)• always written as a ratio ofalways written as a ratio of

products over reactantsproducts over reactants • a value ofa value of K > 1K > 1 means that means that

products are favoredproducts are favored • K < 1K < 1 than reactants are favoredthan reactants are favored

o Equilibrium constants provide Equilibrium constants provide valuable chemical informationvaluable chemical information

o They show whether the products They show whether the products or the reactants or the reactants are are favored in a rxn favored in a rxn (spontaneus or (spontaneus or nonspontaneous)nonspontaneous)• always written as a ratio ofalways written as a ratio of

products over reactantsproducts over reactants • a value ofa value of K > 1K > 1 means that means that

products are favoredproducts are favored • K < 1K < 1 than reactants are favoredthan reactants are favored

Equilibrium ConstantEquilibrium Constant

Sample Problem 1Sample Problem 1Sample Problem 1Sample Problem 1Dinitrogen tetroxide (NDinitrogen tetroxide (N22OO44), a colorless ), a colorless

gas, and nitrogen dioxide (NOgas, and nitrogen dioxide (NO22), a brown ), a brown gas, exist in equilibrium with each other gas, exist in equilibrium with each other

according to the following eqn:according to the following eqn:

Dinitrogen tetroxide (NDinitrogen tetroxide (N22OO44), a colorless ), a colorless gas, and nitrogen dioxide (NOgas, and nitrogen dioxide (NO22), a brown ), a brown gas, exist in equilibrium with each other gas, exist in equilibrium with each other

according to the following eqn:according to the following eqn:

NN22OO44(g)(g) 2NO 2NO22(g)(g)

A 1.0 liter of gas mixture at 10A 1.0 liter of gas mixture at 10C at C at equilibrium contains .0045 mol equilibrium contains .0045 mol

NN22OO44 & .030 mol NO & .030 mol NO22. Write the . Write the mass action expression and mass action expression and

calculate K for the rxn.calculate K for the rxn.

o Known:Known:• [N[N22OO44] =] = .0045 mol/1.0 L.0045 mol/1.0 L

• [NO[NO22] =] = .030 mol/1.0 L.030 mol/1.0 L

o UnknownUnknown::• Mass action expression = ?Mass action expression = ?• K = ?K = ?

o Known:Known:• [N[N22OO44] =] = .0045 mol/1.0 L.0045 mol/1.0 L

• [NO[NO22] =] = .030 mol/1.0 L.030 mol/1.0 L

o UnknownUnknown::• Mass action expression = ?Mass action expression = ?• K = ?K = ?

Analyze: list what we knowAnalyze: list what we know

o At equil, there is no net change in At equil, there is no net change in the amount of Nthe amount of N22OO44 or NO or NO22 at any at any given instantgiven instant

• The only product of the rxn is NOThe only product of the rxn is NO22, , which has a coefficient of 2 in the which has a coefficient of 2 in the balanced eqnbalanced eqn

• The only reactant NThe only reactant N22OO44 has a has a coefficient of 1 in the balanced coefficient of 1 in the balanced eqneqn

o The mass action expression is:The mass action expression is:

• The only product of the rxn is NOThe only product of the rxn is NO22, , which has a coefficient of 2 in the which has a coefficient of 2 in the balanced eqnbalanced eqn

• The only reactant NThe only reactant N22OO44 has a has a coefficient of 1 in the balanced coefficient of 1 in the balanced eqneqn

o The mass action expression is:The mass action expression is:

Calculate: solve for Calculate: solve for unknownsunknowns

[.030M][.030M]22

[.0045M][.0045M]11KK ==

[NO[NO22]]22

[N[N22OO44]]11KK ==

o K is equal to: KK is equal to: K == 0.200.20o KK < 1< 1,, therefore rxn doesn’t favor therefore rxn doesn’t favor

productsproducts

1.1. Find the equilibrium constant if [SOFind the equilibrium constant if [SO22] = ] = 1.0 M; [O 1.0 M; [O22]=1.0 M; [SO]=1.0 M; [SO33]=2.0 M; using ]=2.0 M; using the mass action expression written in the mass action expression written in the examples the examples

2.2. Find the equilibrium constant if [BiFind the equilibrium constant if [Bi+3+3] = ] = 0.00058 M; [S0.00058 M; [S-2-2] = 0.00087 M; using ] = 0.00087 M; using the mass action expression written in the mass action expression written in the examples the examples

1.1. Find the equilibrium constant if [SOFind the equilibrium constant if [SO22] = ] = 1.0 M; [O 1.0 M; [O22]=1.0 M; [SO]=1.0 M; [SO33]=2.0 M; using ]=2.0 M; using the mass action expression written in the mass action expression written in the examples the examples

2.2. Find the equilibrium constant if [BiFind the equilibrium constant if [Bi+3+3] = ] = 0.00058 M; [S0.00058 M; [S-2-2] = 0.00087 M; using ] = 0.00087 M; using the mass action expression written in the mass action expression written in the examples the examples


o Another type of equilibrium is Another type of equilibrium is the equilibrium of the equilibrium of dissolvingdissolving

o This is the equilibrium of a solid This is the equilibrium of a solid and its aqueous form and its aqueous form

o Even the most insoluble salts will Even the most insoluble salts will dissolve to some extent in waterdissolve to some extent in water

• For example, when AgCl is mixed with For example, when AgCl is mixed with water a tiny amount of Agwater a tiny amount of Ag+1+1 and Cl and Cl-1-1 exist exist

o Another type of equilibrium is Another type of equilibrium is the equilibrium of the equilibrium of dissolvingdissolving

o This is the equilibrium of a solid This is the equilibrium of a solid and its aqueous form and its aqueous form

o Even the most insoluble salts will Even the most insoluble salts will dissolve to some extent in waterdissolve to some extent in water

• For example, when AgCl is mixed with For example, when AgCl is mixed with water a tiny amount of Agwater a tiny amount of Ag+1+1 and Cl and Cl-1-1 exist exist

Solubility ProductSolubility Product

AgCl(s) AgAgCl(s) Ag++(aq) + Cl(aq) + Cl--

(aq)(aq)

[Ag+][Cl-] [Ag+][Cl-]

[AgCl][AgCl]Ksp= Ksp=

ο If we exclude the If we exclude the solidsolid “reactant” “reactant” because of its constant conc. because of its constant conc. • We get a special kind of We get a special kind of

equilibrium constant called the equilibrium constant called the solubility product constant (solubility product constant (KKspsp))

• The The lowerlower the solubility of a the solubility of a substance the substance the smallersmaller the K the Kspsp


ο If there are If there are coefficientscoefficients in the in the dissociation equation (from dissociation equation (from balancing) they become balancing) they become exponentsexponents that the conc are that the conc are raised to.raised to.KKspsp = [Ag = [Ag++]]11[Cl[Cl--]]11

ο The solubility product (KThe solubility product (Kspsp) for ) for AgCl at 25AgCl at 25C is 1.8x10C is 1.8x10-10 -10 MM22

• 1010-10-10 indicates a very small conc indicates a very small conc of silver and chloride ionsof silver and chloride ions


The equilibrium concentration of The equilibrium concentration of hydroxide ions in a saturated solution hydroxide ions in a saturated solution of iron(II) hydroxide is 1.2 x 10of iron(II) hydroxide is 1.2 x 10 -5 -5 M at M at a certain temperature. Calculate the a certain temperature. Calculate the KKspsp of Fe(OH) of Fe(OH)22 at this temperature. at this temperature.

The equilibrium concentration of The equilibrium concentration of hydroxide ions in a saturated solution hydroxide ions in a saturated solution of iron(II) hydroxide is 1.2 x 10of iron(II) hydroxide is 1.2 x 10 -5 -5 M at M at a certain temperature. Calculate the a certain temperature. Calculate the KKspsp of Fe(OH) of Fe(OH)22 at this temperature. at this temperature.

Solubility Product: Solubility Product: example 1example 1

-122 2OH Fe Fe(OH)

M 1.2x10][OH 51

2-6

1-

21-

5

FeLiter 1

mols 6.0x10

OH moles 2Fe mole 1

OHLiter 1

mols 1.2x10


-122 2OH Fe Fe(OH)

M 1.2x10][OH 51 M 6.0x10][Fe 62

2-12sp ]][OH[Fe K 2-5-6

sp ]][1.2x10.0x10[ K 6

-16sp 10 x 8.64 K

What is the concentration of silver ions What is the concentration of silver ions in a saturated solution of silver in a saturated solution of silver

carbonate? The Kcarbonate? The Kspsp of Ag of Ag22COCO33 is 8.1 x 10 is 8.1 x 10-12-12

What is the concentration of silver ions What is the concentration of silver ions in a saturated solution of silver in a saturated solution of silver

carbonate? The Kcarbonate? The Kspsp of Ag of Ag22COCO33 is 8.1 x 10 is 8.1 x 10-12-12


-23

132 CO 2Ag COAg

][CO][Ag 8.1x10 -23

211 2 2x

[x][2x] 8.1x10 212

x

3212 4x x4x 8.1x10

][CO 1.27x10x -23

4 M 2.53x10 ][Ag2x -41

o We can also determine if a reaction We can also determine if a reaction has reached equilibrium by calculating has reached equilibrium by calculating a reaction quotient (a reaction quotient (QQ). ). • It’s like taking a snapshot of a reaction It’s like taking a snapshot of a reaction

at a given time and interpreting how at a given time and interpreting how far along the reaction is.far along the reaction is.

o Once the reaction quotient is solved, Once the reaction quotient is solved, it is compared to the equilibrium it is compared to the equilibrium constantconstant• The following picture helps us decide The following picture helps us decide

how to interpret the direction the how to interpret the direction the reaction will continue.reaction will continue.

o We can also determine if a reaction We can also determine if a reaction has reached equilibrium by calculating has reached equilibrium by calculating a reaction quotient (a reaction quotient (QQ). ). • It’s like taking a snapshot of a reaction It’s like taking a snapshot of a reaction

at a given time and interpreting how at a given time and interpreting how far along the reaction is.far along the reaction is.

o Once the reaction quotient is solved, Once the reaction quotient is solved, it is compared to the equilibrium it is compared to the equilibrium constantconstant• The following picture helps us decide The following picture helps us decide

how to interpret the direction the how to interpret the direction the reaction will continue.reaction will continue.

Reaction QuotientReaction Quotient

Sample Problem 3Sample Problem 3Sample Problem 3Sample Problem 3Will a precipitate of PbSOWill a precipitate of PbSO44 form form

when 400.0 ml of 0.0050M MgSOwhen 400.0 ml of 0.0050M MgSO44 is mixed with 600.0 ml of is mixed with 600.0 ml of

0.0020M Pb(NO0.0020M Pb(NO33))22? The K? The Kspsp of of PbSOPbSO44 = 6.3x10 = 6.3x10-7-7..

Will a precipitate of PbSOWill a precipitate of PbSO44 form form when 400.0 ml of 0.0050M MgSOwhen 400.0 ml of 0.0050M MgSO44

is mixed with 600.0 ml of is mixed with 600.0 ml of 0.0020M Pb(NO0.0020M Pb(NO33))22? The K? The Kspsp of of

PbSOPbSO44 = 6.3x10 = 6.3x10-7-7.. 234234 )Mg(NO PbSO )Pb(NOMgSO -2

42

4 SO Pb PbSO ]][SO[Pb Q -24

2

Sample Problem 3Sample Problem 3Sample Problem 3Sample Problem 3-2

42

4 SO Pb PbSO ]][SO[Pb Q -24

2

mols 0.002 M 0.0050 x L 0.4 :Pb mols 2

mols 0.0012 M 0.0020 x L 0.6 :SO mols 24

M 0.0012 L mols/1 0.0012 ][SO 24

M 0.002 L 1 /mols 0.002 ]Pb [ 2

)K (the6.3x102.4x10) )(0.002 (0.0012 Q sp7-6

out eprecipitat will PbSO4

1.1. 0.035 moles of SO0.035 moles of SO22, 0.500 moles of , 0.500 moles of SOSO22ClCl22, and 0.080 moles of Cl, and 0.080 moles of Cl22 are are combined in an evacuated 5.00 L combined in an evacuated 5.00 L flask and heated to 100°C. What is flask and heated to 100°C. What is Q before the reaction begins? Q before the reaction begins? Which direction will the reaction Which direction will the reaction proceed in order to establish proceed in order to establish equilibrium? equilibrium? SOSO22ClCl22(g) <==> SO(g) <==> SO22(g) + Cl(g) + Cl22(g) (g)

KKcc = 0.078 at 100°C = 0.078 at 100°C

1.1. 0.035 moles of SO0.035 moles of SO22, 0.500 moles of , 0.500 moles of SOSO22ClCl22, and 0.080 moles of Cl, and 0.080 moles of Cl22 are are combined in an evacuated 5.00 L combined in an evacuated 5.00 L flask and heated to 100°C. What is flask and heated to 100°C. What is Q before the reaction begins? Q before the reaction begins? Which direction will the reaction Which direction will the reaction proceed in order to establish proceed in order to establish equilibrium? equilibrium? SOSO22ClCl22(g) <==> SO(g) <==> SO22(g) + Cl(g) + Cl22(g) (g)

KKcc = 0.078 at 100°C = 0.078 at 100°C


Manipulating the Manipulating the Equilibrium…Equilibrium…Manipulating the Manipulating the Equilibrium…Equilibrium…o There is a principle that can be There is a principle that can be studied to govern changes in studied to govern changes in equilibrium equilibrium Le Le Chatelier’sChatelier’s PrinciplePrinciple..

o Le Chatelier’s Principle states:Le Chatelier’s Principle states:• ““If a stress is applied to a system If a stress is applied to a system

in dynamic equilibrium, in dynamic equilibrium, thethe system changes to relieve the system changes to relieve the stress.”stress.”

• Stresses are changes in Stresses are changes in temperature, pressure, temperature, pressure, concentration of reactants, or concentration of reactants, or concentration of productsconcentration of products

o There is a principle that can be There is a principle that can be studied to govern changes in studied to govern changes in equilibrium equilibrium Le Le Chatelier’sChatelier’s PrinciplePrinciple..

o Le Chatelier’s Principle states:Le Chatelier’s Principle states:• ““If a stress is applied to a system If a stress is applied to a system

in dynamic equilibrium, in dynamic equilibrium, thethe system changes to relieve the system changes to relieve the stress.”stress.”

• Stresses are changes in Stresses are changes in temperature, pressure, temperature, pressure, concentration of reactants, or concentration of reactants, or concentration of productsconcentration of products

o Adjusting the concentrations of either Adjusting the concentrations of either reactants or products can have reactants or products can have dramatic impact on the equilibriumdramatic impact on the equilibrium• If we add more of reactant A to a If we add more of reactant A to a

system at equilibrium the system system at equilibrium the system will strive to reestablish equilibrium will strive to reestablish equilibrium at a new equilibrium position.at a new equilibrium position.

• The reaction will push to use up the The reaction will push to use up the extra A and generate more Cextra A and generate more C

o Adjusting the concentrations of either Adjusting the concentrations of either reactants or products can have reactants or products can have dramatic impact on the equilibriumdramatic impact on the equilibrium• If we add more of reactant A to a If we add more of reactant A to a


• The reaction will push to use up the The reaction will push to use up the extra A and generate more Cextra A and generate more C

Concentration & EquilibriumConcentration & Equilibrium

A + 2B C H= - 250 kJ[A]↑, rxn will shift toward [A]↑, rxn will shift toward productsproducts[A]↑, rxn will shift toward [A]↑, rxn will shift toward productsproducts

o Adjusting the concentrations of Adjusting the concentrations of either reactants or products can either reactants or products can have dramatic impact on the have dramatic impact on the equilibriumequilibrium• If we add more of product C to a If we add more of product C to a


• The reaction will push to use up the The reaction will push to use up the extra C and generate more A and Bextra C and generate more A and B

• If we add more of product C to a If we add more of product C to a system at equilibrium the system system at equilibrium the system will strive to reestablish equilibrium will strive to reestablish equilibrium at a new equilibrium position.at a new equilibrium position.

• The reaction will push to use up the The reaction will push to use up the extra C and generate more A and Bextra C and generate more A and B

Concentration & EquilibriumConcentration & Equilibrium

A + 2B C H= - 250 kJ [C]↑, rxn will shift [C]↑, rxn will shift toward reactantstoward reactants [C]↑, rxn will shift [C]↑, rxn will shift toward reactantstoward reactants

o The impact of temperature The impact of temperature changes on an equilibrium is changes on an equilibrium is dependent on if the process is dependent on if the process is endothermic or exothermicendothermic or exothermic• Endothermic processes use energy Endothermic processes use energy

as a reactant, while exothermic as a reactant, while exothermic processes produce energyprocesses produce energy

• KKeqeq is temperature dependent is temperature dependent

Temp effects on Temp effects on EquilibriumEquilibrium

A + 2B C H= - 250 kJ

250 kJ is a product250 kJ is a product

If T↑, the equilibrium shifts If T↑, the equilibrium shifts left left

o The impact of temperature The impact of temperature changes on an equilibrium is changes on an equilibrium is dependent on if the process is dependent on if the process is endothermic or exothermicendothermic or exothermic• Endothermic processes use energy Endothermic processes use energy

as a reactant, while exothermic as a reactant, while exothermic processes produce energyprocesses produce energy

Temp effects on Temp effects on EquilibriumEquilibrium

A + 2B H= + 250 kJC

energy is a reactantenergy is a reactant

If T↑, the equilibrium shifts If T↑, the equilibrium shifts right right

o If A, B, and C are all gases, then the equil If A, B, and C are all gases, then the equil they establish is pressure dependentthey establish is pressure dependent

o When the pressure is increased, When the pressure is increased, the system relieves the pressure by the system relieves the pressure by favoring the direction thatfavoring the direction that produces produces fewer gas molecules.fewer gas molecules.• Pressure is # of particles dependent, Pressure is # of particles dependent,

the more particles the higher the pressurethe more particles the higher the pressure• Fewer gas molecules will exertFewer gas molecules will exert less less

pressure.pressure.o So, more product is formed, which overall So, more product is formed, which overall

reduces the pressure, this is a reduces the pressure, this is a shift rightshift right

o If A, B, and C are all gases, then the equil If A, B, and C are all gases, then the equil they establish is pressure dependentthey establish is pressure dependent

o When the pressure is increased, When the pressure is increased, the system relieves the pressure by the system relieves the pressure by favoring the direction thatfavoring the direction that produces produces fewer gas molecules.fewer gas molecules.• Pressure is # of particles dependent, Pressure is # of particles dependent,

the more particles the higher the pressurethe more particles the higher the pressure• Fewer gas molecules will exertFewer gas molecules will exert less less

pressure.pressure.o So, more product is formed, which overall So, more product is formed, which overall

reduces the pressure, this is a reduces the pressure, this is a shift rightshift right

Pressure & EquilibriumPressure & Equilibrium

o Conversely, a decrease in Conversely, a decrease in pressure will favor the rxn that pressure will favor the rxn that produces the most moleculesproduces the most molecules• So we have a shift to theSo we have a shift to the leftleft

o Conversely, a decrease in Conversely, a decrease in pressure will favor the rxn that pressure will favor the rxn that produces the most moleculesproduces the most molecules• So we have a shift to theSo we have a shift to the leftleft

Pressure & EquilibriumPressure & Equilibrium

A + 2B C H= - 250 kJ

P↑, this equilibrium shifts P↑, this equilibrium shifts rightright

If P↓, this equilibrium shifts If P↓, this equilibrium shifts left left

Predict the effect of the following changes on the reaction in which SO3 decomposes to form SO2 and O2. 2SO3(g) <=> 2SO2 (g) + O2 (g) Ho = 197.78

kJ (a) Increasing the temperature of the reaction. (b) Increasing the pressure on the reaction. (c) Adding more O2 when the reaction is at equilibrium. (d) Removing O2 from the system when the reaction is at equilibrium.

Predict the effect of the following changes on the reaction in which SO3 decomposes to form SO2 and O2. 2SO3(g) <=> 2SO2 (g) + O2 (g) Ho = 197.78

kJ (a) Increasing the temperature of the reaction. (b) Increasing the pressure on the reaction. (c) Adding more O2 when the reaction is at equilibrium. (d) Removing O2 from the system when the reaction is at equilibrium.

ClassworkClassworkClassworkClasswork

o If you recall, we mentioned that there are rxns that bounce back and forth from forming products...

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Transcript of o If you recall, we mentioned that there are rxns that bounce back and forth from forming products...