Equilibrium PhaseSolutionChemical. Reversible Reactions a number of chemical reactions have a ΔH...

EquilibriumEquilibrium

PhasePhase

SolutionSolution

ChemicalChemical

Reversible ReactionsReversible Reactions

a number of chemical reactions have a number of chemical reactions have a ΔH and ΔS that are both positive or a ΔH and ΔS that are both positive or both negative; one force favours the both negative; one force favours the reaction, while the other opposes it:reaction, while the other opposes it:

energy + Henergy + H22OO(l)(l) H H22OO(g)(g)

ΔH = +ΔH = + ΔS = + ΔS = +

for the opposite reaction the ΔH and ΔS for the opposite reaction the ΔH and ΔS are both negative:are both negative:

HH22OO(g)(g) → energy + H → energy + H22OO(l)(l)

ΔH = -ΔH = - ΔS = -ΔS = - as a result this reaction is as a result this reaction is reversiblereversible

EquilibriumEquilibrium

occurs when a reaction can go in occurs when a reaction can go in either direction.either direction.

the majority of reactions can be the majority of reactions can be equilibrium situations.equilibrium situations.

there are 3 main types of equilibria.there are 3 main types of equilibria.

Phase EquilibriumPhase Equilibrium

occurs between the phases of a substance:occurs between the phases of a substance:


even at room temperature water evaporates even at room temperature water evaporates until the air above is until the air above is saturatedsaturated. .

((relative humidityrelative humidity is the amount of water is the amount of water vapour in the air, in %, relative to the amount vapour in the air, in %, relative to the amount the air can hold at that temperature. The air the air can hold at that temperature. The air is saturated if the relative humidity is 100%.)is saturated if the relative humidity is 100%.)

as a water molecule evaporates, another as a water molecule evaporates, another molecule of water vapour condenses. The molecule of water vapour condenses. The rate of the evaporation reaction equals the rate of the evaporation reaction equals the rate of the condensation reaction; the rate of the condensation reaction; the reactions are at reactions are at equilibriumequilibrium::


Solution EquilibriumSolution Equilibrium

Occurs when a soluble non-volative Occurs when a soluble non-volative substance is dissolved in a solvent.substance is dissolved in a solvent.

If the solution is If the solution is saturatedsaturated and there is and there is still solid solute at the bottom of the still solid solute at the bottom of the beaker the system is at equilibrium.beaker the system is at equilibrium.

For example:For example:

NaClNaCl(s)(s) Na Na1+1+(aq)(aq) + Cl + Cl1-1-

(aq)(aq)

CC1212HH2222OO11 (s)11 (s) C C1212HH2222OO11 (aq)11 (aq)

Chemical EquilibriumChemical Equilibrium

occurs in a chemical reaction system occurs in a chemical reaction system (the other two were examples of (the other two were examples of physical reactions)physical reactions)

Examples:Examples:

2 NO2 NO2 (g)2 (g) N N22OO4 (g)4 (g)

PClPCl5 (g)5 (g) PCl PCl3 (g)3 (g) + Cl + Cl2 (g)2 (g)

Equilibrium vs Steady StateEquilibrium vs Steady State

Equilibrium is a Equilibrium is a closed systemclosed system; no ; no inputs or outputs.inputs or outputs.

Steady state is an Steady state is an open systemopen system; ; inputs and outputs.inputs and outputs.

Dynamic EquilibriumDynamic Equilibrium

5 conditions are necessary for 5 conditions are necessary for equilibrium to occur:equilibrium to occur:• the reaction must be the reaction must be reversiblereversible• the system must be the system must be closedclosed• both reactant and product must be presentboth reactant and product must be present• concentrations of reactant and product do concentrations of reactant and product do

not changenot change• conditions of temperature and pressure conditions of temperature and pressure

must be held constantmust be held constant

Dynamic EquilibriumDynamic Equilibrium

the forward reaction continuesthe forward reaction continues the reverse reaction continuesthe reverse reaction continues the the raterate of the forward reaction of the forward reaction

equals the equals the raterate of the reverse of the reverse reaction.reaction.

Le Chatelier’s PrincipleLe Chatelier’s Principle

reactions continue in both directions reactions continue in both directions during equilibrium during equilibrium

any factor which changes the rate of any factor which changes the rate of one or both reactions will alter the one or both reactions will alter the relative amounts of reactants and relative amounts of reactants and products. products.


we can predict how an equilibrium we can predict how an equilibrium system will respond to a given system will respond to a given change by using change by using Le Chatelier’s Le Chatelier’s Principle:Principle:

““When a system at equilibrium is When a system at equilibrium is disturbed by application of a stress, disturbed by application of a stress, it attains a new equilibrium position it attains a new equilibrium position that minimizes the stress.”that minimizes the stress.”


in other words, in other words, for every action there for every action there is an equal and opposite reactionis an equal and opposite reaction; ; whatever you do to a reaction whatever you do to a reaction system, the system will respond in system, the system will respond in the the oppositeopposite direction. direction.

Effect of Changing Concentraton Effect of Changing Concentraton on a Chemical Systemon a Chemical System

Increasing the concentration of any Increasing the concentration of any component will cause the reaction component will cause the reaction system to try to decrease it, by system to try to decrease it, by favouring the opposite side. favouring the opposite side.

Effect of Changing Concentraton Effect of Changing Concentraton on a Chemical Systemon a Chemical System

3 CH3 CH4 (g)4 (g) + 6 H + 6 H22OO(g)(g) + 4 N + 4 N2 (g)2 (g) 8 NH 8 NH3 (g)3 (g) + 3 CO + 3 CO2 (g)2 (g)

Increase [CHIncrease [CH44]; system responds by trying ]; system responds by trying decreasedecrease the [CH the [CH44] by making more product. ] by making more product. The concentration of reactants will go down The concentration of reactants will go down while the concentration of products goes up.while the concentration of products goes up.

In other words, increasing [CHIn other words, increasing [CH44] in this ] in this reaction favours the creation of product. reaction favours the creation of product.

Effect of Changing Temperature on Effect of Changing Temperature on a Chemical Systema Chemical System

Increasing the temperature of a Increasing the temperature of a system will favour the endothermic system will favour the endothermic direction.direction.

Decreasing the temperature will Decreasing the temperature will favour the exothermic direction.favour the exothermic direction.

Effect of Changing Temperature on Effect of Changing Temperature on a Chemical Systema Chemical System

3 H3 H2(g)2(g) + N + N2 (g)2 (g) 2 NH 2 NH3 (g)3 (g) + 92 kJ + 92 kJ

adding energy causes the system to try to adding energy causes the system to try to absorb energy; the endothermic direction absorb energy; the endothermic direction is favoured. More reactant is made and is favoured. More reactant is made and the amount of product decreases.the amount of product decreases.

taking energy away favours the taking energy away favours the exothermic reaction; in this case products exothermic reaction; in this case products are favoured.are favoured.

Effect of Changing Pressure on a Effect of Changing Pressure on a Chemical SystemChemical System

equal numbers of moles at the same equal numbers of moles at the same temperature and pressure have the same temperature and pressure have the same volume.volume.

increasing pressure has the effect of increasing pressure has the effect of favouring the side of the equation with favouring the side of the equation with fewer particles.fewer particles.

Effect of Changing Pressure on a Effect of Changing Pressure on a Chemical SystemChemical System

3 CH3 CH4 (g)4 (g) + 6 H + 6 H22OO(g)(g) + 4 N + 4 N2 (g)2 (g) 8 NH 8 NH3 (g)3 (g) + 3 CO + 3 CO2 (g)2 (g)

in this reaction the reactant side has in this reaction the reactant side has 13 particles; the product side has 11.13 particles; the product side has 11.

increasing the pressure will favour increasing the pressure will favour the products.the products.

decreasing the pressure will favour decreasing the pressure will favour the reactants.the reactants.

Effect of Changing Volume on a Effect of Changing Volume on a Chemical SystemChemical System

in a closed system reducing the in a closed system reducing the volume has the effect of volume has the effect of increasing increasing the pressurethe pressure..

increasing the volume reduces the increasing the volume reduces the pressure.pressure.

Catalysts and InhibitorsCatalysts and Inhibitors

catalysts reduce activation energy in catalysts reduce activation energy in both directionsboth directions; they increase the ; they increase the rate of reaction in both directions but rate of reaction in both directions but does not change the equilibrium.does not change the equilibrium.

inhibitors have the opposite effect, inhibitors have the opposite effect, but also do not affect the but also do not affect the equilibrium.equilibrium.

Common Ion EffectCommon Ion Effect

is related to is related to solubilitysolubility equilibria. equilibria. adding an ionic substance with an ion adding an ionic substance with an ion

in common with an aqueous in common with an aqueous equilibrium will affect the equilibrium equilibrium will affect the equilibrium in the same way as changing in the same way as changing concentrationconcentration



(aq)(aq)

if NaNOif NaNO3(aq)3(aq) is added to this system, it is added to this system, it breaks into Nabreaks into Na1+1+

(aq)(aq) and NO and NO331-1-

(aq)(aq). . It is just like adding NaIt is just like adding Na1+1+ ions to the ions to the

NaCl system. That system responds NaCl system. That system responds by favouring the formation of solid by favouring the formation of solid NaCl; the concentration of ClNaCl; the concentration of Cl1-1- ions ions decreases.decreases.



(aq)(aq)

add AgNOadd AgNO33

AgAg1+1+ precipitates chloride(Cl precipitates chloride(Cl1-1-) ) removing it from solution.removing it from solution.

system responds by producing more system responds by producing more chloride, also increases sodium ions chloride, also increases sodium ions and reduces the NaCl solidand reduces the NaCl solid

Equilibrium ConstantEquilibrium Constant

For the reaction:For the reaction:

aAaA(g) (g) + bB + bB(g)(g) cC cC(g)(g) + dD + dD(g) (g)

the lower-case letters represent the lower-case letters represent number of moles (the numbers used number of moles (the numbers used to balance the equation)to balance the equation)

the upper-case letters represent the the upper-case letters represent the chemical species. chemical species.


aAaA(g) (g) + bB + bB(g)(g) cC cC(g)(g) + dD + dD(g) (g)

At equilibrium a mathematical relationship At equilibrium a mathematical relationship exists where a exists where a constantconstant results from the results from the equation:equation:

Keq = Keq = [Products] [Products] = = [C][C]cc[D][D]dd

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

where [ ] represents the concentration. where [ ] represents the concentration.


this a useful mathematical relationship.this a useful mathematical relationship.

no matter what concentrations of reactant no matter what concentrations of reactant or product you start with we know that the or product you start with we know that the final ratio of products over reactants will final ratio of products over reactants will equal the value of the Kequal the value of the Keqeq..

This applies only if conditions of This applies only if conditions of temperature and pressure stay constant.temperature and pressure stay constant.

Writing Equilibrium ExpressionsWriting Equilibrium Expressions

Consider the reaction:Consider the reaction:

NN2 (g)2 (g) + 3 H + 3 H2 (g)2 (g) 2 NH 2 NH3 (g)3 (g) + 92 kJ + 92 kJ

KKeqeq = = [NH [NH33]]22

[N[N22] [H] [H22]]33

energy is not included in the equilibrium energy is not included in the equilibrium constant expressionconstant expression


2 CaCO2 CaCO3 (S) 3 (S) 2 Ca 2 Ca(s)(s) + 2 CO + 2 CO2 (g)2 (g) + O + O2 (g)2 (g)

KKeqeq = [CO = [CO22]]22[O[O22]]

solids and liquids are not included in solids and liquids are not included in the equilibrium expression.the equilibrium expression.


2 H2 H2 (g) 2 (g) + O+ O2 (g)2 (g) 2 H 2 H22OO (l) (l)

KKeqeq = = 1 1

[H[H22]]22[O[O22]]

since the liquid is not included in the since the liquid is not included in the equilibrium expression the equilibrium expression the numerator becomes a 1.numerator becomes a 1.

Equilibrium CalculationsEquilibrium Calculations

the equilibrium expression is a the equilibrium expression is a mathematical relationship of several mathematical relationship of several variables.variables.

given the values of all but one of the given the values of all but one of the variables allows you to calculate the variables allows you to calculate the unknown, whether it be the Kunknown, whether it be the Keqeq or or any of the concentrations.any of the concentrations.


consider the following case:consider the following case:

NN2 (g)2 (g) + 3 H + 3 H2 (g)2 (g) 2 NH 2 NH3 (g)3 (g) + 92 kJ + 92 kJ

given this equation we can write the given this equation we can write the equilibrium expression:equilibrium expression:


[N[N22] [H] [H22]]33


at equilibrium the concentration of at equilibrium the concentration of

NN22 = 0.142 mol/L, the = 0.142 mol/L, the

[H[H22] = 0.341 mol/L and the ] = 0.341 mol/L and the

[NH[NH33] = 0.220 mol/L. ] = 0.220 mol/L. calculate the value of the Kcalculate the value of the Keqeq


[N[N22] = 0.142 mol/L ] = 0.142 mol/L

[H[H22] = 0.341 mol/L ] = 0.341 mol/L

[NH[NH33] = 0.220 mol/L] = 0.220 mol/L

KKeqeq = = [NH [NH33]]22 = = (0.220 mol/L)(0.220 mol/L)22

[N[N22] [H] [H22]]33 (0.142 mol/L)(0.341 mol/L) (0.142 mol/L)(0.341 mol/L)33

KKeqeq = 8.60 = 8.60

This is slightly larger than 1; products are This is slightly larger than 1; products are favoured over reactantsfavoured over reactants

ICE box problemsICE box problems

This type of problem gives you This type of problem gives you information about the information about the initialinitial situation, before equilibrium is situation, before equilibrium is established. established.

Then you get 1 or more pieces of Then you get 1 or more pieces of information concerning the situation information concerning the situation at at equilibriumequilibrium. .

You are required to fill in the blanks You are required to fill in the blanks by figuring out the by figuring out the changechange..

ICE box problemsICE box problems

2 A2 A(g)(g) + B + B(g)(g) C C(g)(g) + 3 D + 3 D(g)(g)

0.100 mol/L each of A and B are 0.100 mol/L each of A and B are placed in a flask and allowed to come placed in a flask and allowed to come to equilibrium. to equilibrium.

At equilibrium 0.022 mol/L of C is At equilibrium 0.022 mol/L of C is present. present.

Calculate the value of the KCalculate the value of the Keqeq. .

2 A2 A(g)(g) + B + B(g)(g) C C(g)(g) + 3 + 3 DD(g)(g)

[[IInitial] nitial] 0.100 0.100 MM 0.100 0.100 MM 0.00.0 0.0 0.0

[[CChange]hange]

[[EEquilibrium]quilibrium] 0.022 0.022 MM

this is what you know. this is what you know. Nothing is said in the problem about the amount of Nothing is said in the problem about the amount of

C and D initially, so we assume they start at C and D initially, so we assume they start at zerozero..

2 A2 A(g)(g) + B + B(g)(g) C C(g)(g) + 3 + 3 DD(g)(g)

[[II] 0.100 M 0.100 M 0.0] 0.100 M 0.100 M 0.0 0.0 0.0

[[CC] ] +0.022 M+0.022 M

[[EE]] 0.022 0.022 MM

since C started with nothing and ended with 0.022 since C started with nothing and ended with 0.022 mol/L we can see that it changed by +0.022 mol/L. mol/L we can see that it changed by +0.022 mol/L.


[[II] 0.100 M 0.100 M 0.0] 0.100 M 0.100 M 0.0 0.0 0.0

[[CC] ] +0.022 M +0.022 M 3/1(+0.022 M)3/1(+0.022 M)

[[EE]] 0.022 0.022 MM

if C goes up, so does D. if C goes up, so does D. the stoichiometry is 1 C : 3 D. For every 1 that C goes the stoichiometry is 1 C : 3 D. For every 1 that C goes

up, D goes up 3.up, D goes up 3.


[[II] 0.100 M 0.100 M 0.0] 0.100 M 0.100 M 0.0 0.0 0.0

[[CC] ] 2/1(-0.022M) 1/1(-0.022M)2/1(-0.022M) 1/1(-0.022M) +0.022 M +0.022 M 3/1(+0.022 3/1(+0.022 M)M)

[[EE]] 0.022 0.022 MM

the change to A and B are negative (teeter-totter)the change to A and B are negative (teeter-totter) always take the stoichiometry into account.always take the stoichiometry into account.


[[II] 0.100 M 0.100 M 0.0] 0.100 M 0.100 M 0.0 0.0 0.0

[[CC] 2/1(-0.022M) 1/1(-0.022M) +0.022 M 3/1(+0.022 M)] 2/1(-0.022M) 1/1(-0.022M) +0.022 M 3/1(+0.022 M)

= -0.044 M = -0.022 M= -0.044 M = -0.022 M = +0.066 M= +0.066 M

[[EE] ] 0.056 M 0.078 M0.056 M 0.078 M 0.022 M 0.066 M 0.022 M 0.066 M

adding or subtracting the changes from the initial adding or subtracting the changes from the initial concentration yields the equilibrium concentration of concentration yields the equilibrium concentration of each species.each species.

all that remains is to write the equilibrium expression all that remains is to write the equilibrium expression and solve for Kand solve for Keqeq

KKeqeq = = [C][D] [C][D]33 [A] [A]2 2 [B][B]

= = (0.022 mol/L)(0.066 mol/L)(0.022 mol/L)(0.066 mol/L)33

(0.056 mol/L)(0.056 mol/L)22(( 0.078 mol/L)0.078 mol/L)

KKeqeq = 0.026 = 0.026

Another ICE boxAnother ICE box

NN2 (g)2 (g) + 3 H + 3 H2 (g)2 (g) 2 NH 2 NH3 (g)3 (g)

If 1.65 mol of ammonia is placed in a If 1.65 mol of ammonia is placed in a 3.00 L flask at 142 3.00 L flask at 142 °C and allowed to °C and allowed to come to equilibrium the come to equilibrium the concentration of hydrogen gas is concentration of hydrogen gas is 0.771 mol/L0.771 mol/L

Calculate the value of the KCalculate the value of the Keqeq

NN2 (g)2 (g) 3 H3 H2 (g)2 (g) 2 NH2 NH3 (g)3 (g)

[I][I] 0.00 mol/L0.00 mol/L 0.00 mol/L0.00 mol/L 1.65 mol/3.00L1.65 mol/3.00L

= 0.550 mol/L= 0.550 mol/L

[C][C] 1/3 1/3 (+(+0.771 0.771 M)M)

= 0.257 = 0.257 mol/Lmol/L

++0.771 0.771 mol/Lmol/L

2/3 2/3 ((--0.771 0.771 mol/L)mol/L)

= -0.514 mol/L= -0.514 mol/L

[E][E] 0.257 mol/L0.257 mol/L 0.771 mol/L0.771 mol/L 0.036 mol/L0.036 mol/L

NN2 (g)2 (g) + 3 H + 3 H2 (g)2 (g) 2 NH 2 NH3 (g)3 (g)


[N[N22] [H] [H22]]33

= = ( (0.036 mol/L0.036 mol/L))22

((0.257 mol/L0.257 mol/L)()(0.771 mol/L0.771 mol/L))33

= 0.011= 0.011

AA(g)(g) + B + B(g)(g) 2 C 2 C(g)(g)

If the initial concentration of C is If the initial concentration of C is 0.350 mol/L, calculate the 0.350 mol/L, calculate the equilibrium concentrations of all equilibrium concentrations of all three species if the value of the Keq three species if the value of the Keq is 145.is 145.

AA(g)(g) + B + B(g)(g) 2 C 2 C(g)(g)

[[II] ] 0.0 0.0 0.0 0.0 0.350 mol/L 0.350 mol/L

[[CC]]

[[EE]]

this is what we know, other than the Kthis is what we know, other than the Keqeq value. value. where do we go from here ?where do we go from here ?

AA(g)(g) + B + B(g)(g) 2 C 2 C(g)(g)

[[II] ] 0.0 0.0 0.0 0.0 0.350 mol/L 0.350 mol/L

[[CC] ] + x + x - 2x+ x + x - 2x

[[EE]]

We don’t know the changes in concentration. We don’t know the changes in concentration. When we don’t know something in math we call it When we don’t know something in math we call it xx. .

AA(g)(g) + B + B(g)(g) 2 C 2 C(g)(g)

[[II] ] 0.0 0.0 0.0 0.0 0.350 mol/L 0.350 mol/L

[[CC] ] + x + x - 2x+ x + x - 2x

[[EE]] x x 0.350 mol/L x x 0.350 mol/L - 2x- 2x

The equilibrium concentrations are expressed in The equilibrium concentrations are expressed in terms of x.terms of x.

Now we write the KNow we write the Keqeq expression expression

KKeqeq = = [C] [C]22

[A] [B][A] [B] 145 = 145 = (0.350 mol/L – 2x) (0.350 mol/L – 2x)22

((x)(x)(x)x)

145 = 145 = (0.350 mol/L – 2x) (0.350 mol/L – 2x)22

((x)x)22

√ √145 = 145 = 0.350 mol/L – 2x 0.350 mol/L – 2x xx

x = 0.0249 mol/Lx = 0.0249 mol/L

AA(g)(g) + B + B(g)(g) 2 C 2 C(g)(g)

[[II] ] 0.0 0.0 0.0 0.0 0.350 mol/L 0.350 mol/L

[[CC] ] + x + x - 2x+ x + x - 2x

[[EE]] x x 0.350 mol/L - x x 0.350 mol/L - 2x2x

[A] = [B] = x = [A] = [B] = x = 0.0249 mol/L 0.0249 mol/L [C] = 0.350 M – 2x = 0.350 M – 2(0.0249 M)[C] = 0.350 M – 2x = 0.350 M – 2(0.0249 M)

= 0.300 mol/L= 0.300 mol/L

AA(g)(g) + B + B(g)(g) C C(g)(g)

If the initial concentration of C is If the initial concentration of C is 0.350 mol/L, calculate the 0.350 mol/L, calculate the equilibrium concentrations of all equilibrium concentrations of all three species if the value of the Keq three species if the value of the Keq is 145.is 145.

AA(g)(g) + B + B(g)(g) C C(g)(g)

[[II] ] 0.0 0.0 0.0 0.0 0.350 mol/L 0.350 mol/L

[[CC] ] + x + x - x+ x + x - x

[[EE]] x x 0.350 mol/L x x 0.350 mol/L - x- x

KKeqeq = = [C] [C] [A] [B][A] [B] 145 = 145 = (0.350 mol/L – x) (0.350 mol/L – x) ((x)(x)(x)x)

145 = 145 = 0.350 mol/L – x 0.350 mol/L – x xx22

145x145x22 = 0.350 mol/L – x = 0.350 mol/L – x

145x145x22 + x - 0.350 mol/L = 0 + x - 0.350 mol/L = 0

now we apply the quadratic equation:now we apply the quadratic equation:

145x145x22 + x - 0.350 mol/L = 0 + x - 0.350 mol/L = 0

a b ca b c

x = x = -b ± √ b-b ± √ b22 – 4ac – 4ac

2a 2a

x = 0.0458 mol/L x = 0.0458 mol/L

AA(g)(g) + B + B(g)(g) C C(g)(g)

[[II] ] 0.0 0.0 0.0 0.0 0.350 mol/L 0.350 mol/L

[[CC] ] + x + x - x+ x + x - x

[[EE]] x x 0.350 mol/L - x x 0.350 mol/L - xx

[A] = [B] = 0.0458[A] = [B] = 0.0458 mol/L mol/L [C] = 0.350 M – x = 0.350 M – [C] = 0.350 M – x = 0.350 M – 0.04580.0458 M M

= 0.304 mol/L= 0.304 mol/L

Calculations using the KCalculations using the Kspsp A saturated solution of CuBr has a concentration of 2.0 x A saturated solution of CuBr has a concentration of 2.0 x

1010-4-4 mol/L. Calculate the K mol/L. Calculate the Kspsp.. Write the dissociation equation and determine the Write the dissociation equation and determine the

concentration of each ion in solution:concentration of each ion in solution:

CuBrCuBr(s)(s) Cu Cu1+1+ (aq)(aq) + Br + Br1-1- (aq)(aq)

2.0 x 102.0 x 10-4-4 mol/L 1(2.0 x 10 mol/L 1(2.0 x 10-4-4 mol/L) 1(2.0 x 10 mol/L) 1(2.0 x 10-4-4 mol/L) mol/L)

= 2.0 x 10= 2.0 x 10-4-4 mol/L = 2.0 x 10 mol/L = 2.0 x 10-4-4 mol/L mol/L Write the KWrite the Kspsp expression, insert the numbers and solve: expression, insert the numbers and solve:

KKspsp = [Cu= [Cu1+1+][Br][Br1-1-]]= (2.0 x 10= (2.0 x 10-4-4 mol/L)( 2.0 x 10 mol/L)( 2.0 x 10-4-4 mol/L) mol/L)= 4.0 x 10= 4.0 x 10-8-8

Calculate the Ksp for Bi2S3, which has a solubility of 1.36 x 10-15 mol/L at 25°C.

Bi2S3 (s) 2 Bi3+(aq) + 3 S 2- (aq)(aq)

1.36 x 10-15 mol/L 2(1.36 x 10-15 mol/L) 3(1.36 x 10-15 mol/L)= 2.72 x 10-15 mol/L = 4.08 x 10-15 mol/L

Ksp = [Bi3+]2[S2-]3

= (2.72 x 10-15 mol/L)2(4.08 x 10-15 mol/L)3

= 5.02 x 10-73

The Ksp value for Cu(IO3)2 is 1.4 x 10-7 at 25°C. Calculate its solubility at this temperature.

Like the ICE box, express concentrations in terms of x:Like the ICE box, express concentrations in terms of x:

Cu(IO3)2 (s) Cu2+(aq) + 2 IO3

1-(aq)

x 1(x) 2(x)

Ksp = [Cu2+][IO31-]2

= (x)(2x)2

1.4 x 10-7 = 4x3

x = 3√1.4 x 10-7 = 3.3 x 10-3 mol/L 4

Predicting a PrecipitatePredicting a Precipitate

KKspsp values can also be used to predict values can also be used to predict if mixing solutions will produce a if mixing solutions will produce a precipitate.precipitate.

e.g. e.g. • A solution is prepared by adding 750.0 A solution is prepared by adding 750.0

mL of 4.00 x 10mL of 4.00 x 10-3-3 mol/L Ce(NO mol/L Ce(NO33))33 to to 300.0 mL of 2.00 x 10300.0 mL of 2.00 x 10-2-2 mol/L KIO mol/L KIO33..

• The KThe Kspsp of Ce(IO of Ce(IO33))33 is 1.9 x 10 is 1.9 x 10-10-10. . • Will Ce(IOWill Ce(IO33))33 precipitate from this precipitate from this

solution ? solution ?

Step 1. Write a dissociation equation for each substance and determine the concentration of each ion in solution.

Ce(NO3)3 (aq) Ce3+(aq) + 3 NO3

1-(aq)

4.00 x 10-3 mol/L 1(4.00 x 10-3 mol/L) 3(4.00 x 10-3 mol/L)= 4.00 x 10-3 mol/L = 1.20 x 10-2

mol/L

KIO3 (aq) K1+(aq) + IO3

1-(aq)

2.00 x 10-2 mol/L 1(2.00 x 10-2 mol/L) 1(2.00 x 10-2 mol/L)= 2.00 x 10-2 mol/L = 2.00 x 10-2

mol/L

Step 2. Use the dilution equation to determine the concentration of each of the significant species. In this case they are Ce3+

(aq) and IO31-

(aq)

[Ce3+] = csvs = (4.00 x 10-3 mol/L)(750.0 mL)

vd 750.0 mL + 300.0 mL

= 2.86 x 10-3 mol/L

[IO31-] = csvs = (2.00 x 10-2 mol/L)(300.0 mL)

vd 750.0 mL + 300.0 mL

= 5.71 x 10-3 mol/L

Step 3. Write the Ksp expression for the substance you are testing and solve for the Ksp using the numbers you just calculated. This is a trial Ksp

Substance: Ce(IO3)3

Ce(IO3)3 (s) Ce3+(aq) + 3 IO3

1-(aq)

Ksp = [Ce3+][IO31-]3

Trial Ksp = (2.86 x 10-3 mol/L)( 5.71 x 10-3 mol/L)3

= 5.52 x 10-10

Step 4. Compare the trial Ksp with the actual Ksp from the question:

if the trial Ksp is larger than the actual Ksp, there is more solute than can dissolve; there will be a precipitate.

if the trial Ksp is smaller than the actual Ksp, there is less solute than can dissolve; there will not be a precipitate.

In this case 5.52 x 10-10 is larger than 1.9 x 10-10

there will be a precipitate

Equilibrium PhaseSolutionChemical. Reversible Reactions a number of chemical reactions have a ΔH...

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Transcript of Equilibrium PhaseSolutionChemical. Reversible Reactions a number of chemical reactions have a ΔH...