EquilibriumEquilibrium

Reversible ReactionsReversible Reactions

Many chemical reactions exist in which the Many chemical reactions exist in which the reactants can be converted to products reactants can be converted to products and the products converted to reactants at and the products converted to reactants at the same time. There are two opposing the same time. There are two opposing processes happening.processes happening.

Many biologically important reactions are Many biologically important reactions are reversible reactions.reversible reactions.

CO + 3HCO + 3H22 CH CH44 + H + H22OO

The double arrow shows that a reaction is The double arrow shows that a reaction is reversiblereversible

If you put carbon monoxide and hydrogen If you put carbon monoxide and hydrogen into a container and let them react, they’ll into a container and let them react, they’ll form methane and water. There will be a mix form methane and water. There will be a mix of some products and some reactants.of some products and some reactants.

If you put methane and water into a container If you put methane and water into a container and let them react, they’ll form methane and and let them react, they’ll form methane and water. There will be a mix of some products water. There will be a mix of some products and some reactants.and some reactants.

CO + 3HCO + 3H22 CH CH44 + H + H22OO

In fact- you can put in the reactants only, In fact- you can put in the reactants only, the products only or some of each and the products only or some of each and within a few limits from stoichiometry you’ll within a few limits from stoichiometry you’ll end up with a mix of products and end up with a mix of products and reactants at the end. The same mix no reactants at the end. The same mix no matter where you start.matter where you start.

MOLES

Time

Once equilibrium is reached, There are no changes in the Concentrations of products orreactants

Time

RXN

RATE

EquilibriumEquilibrium

A state reached by a chemical system in A state reached by a chemical system in which the forward and reverse reaction which the forward and reverse reaction rates are equalrates are equal

This means that there is no net change in This means that there is no net change in the concentration of any chemicalthe concentration of any chemical

Equilibrium may take seconds to be Equilibrium may take seconds to be reached or billions of years (we can say reached or billions of years (we can say that catalysts get chemical systems to that catalysts get chemical systems to equilibrium faster).equilibrium faster).

CO + 3HCO + 3H22 CH CH44 + H + H22OO

Suppose you put 2.00 moles of CO and 4.50 Suppose you put 2.00 moles of CO and 4.50 moles of hydrogen into a 1.0 L flask. They moles of hydrogen into a 1.0 L flask. They react and after a while the concentration of react and after a while the concentration of methane is found to be constant. Equilibrium methane is found to be constant. Equilibrium has been reached and from now on, no has been reached and from now on, no matter how long you wait, the concentration matter how long you wait, the concentration of each chemical will not change. You find of each chemical will not change. You find that the concentration of methane is 0.65 M that the concentration of methane is 0.65 M ([CH([CH44] = 0.65M). Find the concentration of all ] = 0.65M). Find the concentration of all the chemicals at equilibrium based on this the chemicals at equilibrium based on this data.data.

Make an ICE chartMake an ICE chart

ICE stands for initial, change, equilibrium ICE stands for initial, change, equilibrium CO + 3HCO + 3H2 2 CH CH44 + +

H H22OO

Initial 2.0 M 4.5 M 0 0Initial 2.0 M 4.5 M 0 0

ChangeChange

Equil. 0.65MEquil. 0.65M

ICE stands for initial, change, equilibrium ICE stands for initial, change, equilibrium CO + 3HCO + 3H2 2 CH CH44 + H + H22OO

Initial 2.0 M 4.5 M 0 0Initial 2.0 M 4.5 M 0 0

Change -x -3x +x +xChange -x -3x +x +x

Equil. 0.65MEquil. 0.65M

The change in methane must be +0.65M (= x)The change in methane must be +0.65M (= x)

Use the stoichiometric ratios (in Use the stoichiometric ratios (in greengreen))

ICE stands for initial, change, equilibrium ICE stands for initial, change, equilibrium CO + 3HCO + 3H2 2 CH CH44 + H + H22OO

Initial Initial 2.0 M2.0 M 4.5 M4.5 M 00 00

Change Change -0.65-0.65 -3(0.65)-3(0.65) +0.65+0.65 +0.65+0.65

Equil. Equil. 1.35M1.35M 2.55M2.55M 0.65M 0.65M 0.65M0.65M

ICE tables are very useful ways to organize ICE tables are very useful ways to organize information I n this (and other) chapters.information I n this (and other) chapters.

Example 2Example 2

One More ExampleOne More Example

4 NH4 NH33 + 5 O + 5 O22 4 NO + 6 H 4 NO + 6 H22OOTwo moles of ammonia and oxygen and Two moles of ammonia and oxygen and

three moles of water are placed in a 1L three moles of water are placed in a 1L flask at room temperature. Pressure is flask at room temperature. Pressure is monitored and observed to reach a monitored and observed to reach a constant P. At this point, there are 1.6 constant P. At this point, there are 1.6 moles of ammonia. Calculate the moles of ammonia. Calculate the concentrations of all substances.concentrations of all substances.

Another ExampleAnother Example

4 NH4 NH33 + 5 O + 5 O22 4 NO + 6 H 4 NO + 6 H22OO

Initial 2.00 2.00 0.00 3.00Initial 2.00 2.00 0.00 3.00

ChangeChange

Equil 1.6Equil 1.6

This means that [NHThis means that [NH33] decreased by 0.4 moles. So ] decreased by 0.4 moles. So

by coefficients, what must have happened to the by coefficients, what must have happened to the other quantities?other quantities?

4 NH4 NH33 + 5 O + 5 O22 4 NO + 6 H 4 NO + 6 H22OO

Initial 2.00 2.00 0.00 3.00Initial 2.00 2.00 0.00 3.00

Change –0.40 -0.50 + 0.40 +0.60Change –0.40 -0.50 + 0.40 +0.60

Equil 1.60 1.50 0.40 3.60Equil 1.60 1.50 0.40 3.60

Things reacted in ratios according to the Things reacted in ratios according to the coefficients of the balanced rxn. Make coefficients of the balanced rxn. Make sure you understand + and - and why sure you understand + and - and why each substance has a different #each substance has a different #

Equilibrium ConstantEquilibrium Constant

A mathematical ratio that relates amounts A mathematical ratio that relates amounts of products and reactants to a number (the of products and reactants to a number (the size of which tells about relative amounts size of which tells about relative amounts of products and reactants (and of products and reactants (and spontaneity). spontaneity).

May be expressed in terms of May be expressed in terms of concentrations (Kc or sometimes Keq) or concentrations (Kc or sometimes Keq) or Pressures (Kp)Pressures (Kp)

Generic ReactionGeneric Reaction

aA = bB aA = bB cC + dD cC + dD

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

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

Products on top, coefficients become Products on top, coefficients become exponentsexponents

Solids and liquids are left out Solids and liquids are left out

ExamplesExamples

3 H3 H2(g)2(g) + N + N2(g)2(g) 22NHNH3(g)3(g)

Keq =Keq = [NH [NH33]]22

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

COCO(g)(g) + 3 H + 3 H2(g)2(g) CH CH4(g)4(g) + H + H22OO(g)(g)

CaClCaCl2(s)2(s) Ca Ca+2+2(aq)(aq) + 2Cl + 2Cl--(aq)(aq)

CaCOCaCO3(s)3(s) CaO CaO(s)(s) + CO + CO2(g)2(g)

Keq = [CHKeq = [CH44][H][H22O]O]

[H[H22]]33[CO][CO]

Keq = [CaKeq = [Ca+2+2][Cl][Cl-1-1]]22

Keq = [COKeq = [CO22]]

Quantitative Approaches to Quantitative Approaches to Equilibrium ConstantsEquilibrium Constants

For this example we will deal with the For this example we will deal with the chemical systemchemical system

NN22OO4(g)4(g) 2NO 2NO2(g)2(g)

(A different example is used in your book to (A different example is used in your book to make the same point).make the same point).

The equilibrium constant expression is:The equilibrium constant expression is:

Keq = [NOKeq = [NO22]]22/[N/[N22O]O]44

NN22OO4(g)4(g) 2NO 2NO2(g)2(g)

Suppose we do a series of experiments in Suppose we do a series of experiments in which we sometimes start with all NOwhich we sometimes start with all NO22, ,

sometimes start with all Nsometimes start with all N22OO44 and and

sometimes have some of each. If we let sometimes have some of each. If we let the system reach equilibrium (what is true the system reach equilibrium (what is true at that time?), we would see something at that time?), we would see something like this (based on real data)like this (based on real data)

NN22OO4(g)4(g) 2NO 2NO2(g)2(g)

INITIALINITIAL

[N[N22OO44] ]

(M)(M)

INITIALINITIAL

[NO[NO22]]

(M)(M)

EquilibriumEquilibrium

[N[N22OO44] (M)] (M)

EquilibriumEquilibrium

[NO[NO22]]

(M)(M)

[NO[NO22]]22

[N[N22OO44]]

0.1000.100 0.0000.000 0.04910.0491 0.10180.1018 0.2110.211

0.0000.000 0.1000.100 0.01850.0185 0.06270.0627 0.2120.212

0.05000.0500 0.05000.0500 0.03320.0332 0.08370.0837 0.2110.211

0.07500.0750 0.02500.0250 0.04110.0411 0.09300.0930 0.2100.210

0.02500.0250 0.07500.0750 0.02570.0257 0.07360.0736 0.2110.211

NoticeNotice

At equilibrium, the ratio described by the At equilibrium, the ratio described by the Keq (or Kc) expression is the same. Keq (or Kc) expression is the same.

Equilibrium will be reached whether the Equilibrium will be reached whether the system starts will all products, all reactants system starts will all products, all reactants or some mixor some mix

Finding Keq Using DataFinding Keq Using Data

2HI2HI(g)(g) H H2(g)2(g) + I + I2(g)2(g)

3.50 moles of HI are placed in a 5.00 L 3.50 moles of HI are placed in a 5.00 L flask. At equilibrium, the flask contained flask. At equilibrium, the flask contained 0.360 moles of iodine. Find the equilibrium 0.360 moles of iodine. Find the equilibrium concentration of each substance and the concentration of each substance and the value of the equilibrium constant.value of the equilibrium constant.

First get amounts into Molarity, then use First get amounts into Molarity, then use an ICE chartan ICE chart

[HI][HI]00 = 3.50moles/5.00L = 0.70 M = 3.50moles/5.00L = 0.70 M

[I[I22]]eqeq = 0.36moles/5.00L = 0.072 M = 0.36moles/5.00L = 0.072 M

2HI2HI(g)(g) H H2(g)2(g) + I + I2(g)2(g)

Initial 0.70 0 0Initial 0.70 0 0

Change Change -0.144-0.144 +0.072+0.072 +0.072+0.072

Equilibrium 0.556 0.072 Equilibrium 0.556 0.072 0.0720.072

Keq = (0.072)(0.072) = 0.0168Keq = (0.072)(0.072) = 0.0168

(0.556)(0.556)22

Another ExampleAnother Example

In a study of possible synthetic fuels the In a study of possible synthetic fuels the following reaction is studied:following reaction is studied:

CHCH4(g)4(g) + H + H22OO(g)(g) CO CO(g)(g) + 3H + 3H2(g)2(g)

The value of Keq = 0.26. At equilibrium, the The value of Keq = 0.26. At equilibrium, the following data are obtained:following data are obtained:

[CH[CH44] = 0.145M, [H] = 0.145M, [H22] = 0.291M, [CO] = 0.88M] = 0.291M, [CO] = 0.88MFind the equilibrium concentration of water Find the equilibrium concentration of water

vapor.vapor.

Plan and SolutionPlan and Solution

Write out Keq expressionWrite out Keq expressionPlug in and solvePlug in and solveKeq = [CO][HKeq = [CO][H22]]33

[CH[CH44][H][H22O]O]

0.26 = (0.88)(0.291)0.26 = (0.88)(0.291)33

(0.145)(X)(0.145)(X)

Cross-multiply and solve to get x = 0.575MCross-multiply and solve to get x = 0.575M

Still One More ExampleStill One More Example

CO +HCO +H22O O CO CO22 + H + H22

0.250 moles of carbon monoxide and 0.250 moles of carbon monoxide and water are placed in a 125 ml flask. The water are placed in a 125 ml flask. The equilibrium constant for the reaction is equilibrium constant for the reaction is 1.56. What will be the composition of the 1.56. What will be the composition of the mixture at equilibrium?mixture at equilibrium?

Plan- write out a Keq expression, make an Plan- write out a Keq expression, make an ICE chart and solve (remember to use M ICE chart and solve (remember to use M and not moles)and not moles)

Keq (or Kc) = [COKeq (or Kc) = [CO22][H][H22]]

[CO][H[CO][H22O]O]

Initial [CO]=[HInitial [CO]=[H22O] = 0.250 mole/.125L = 2.00MO] = 0.250 mole/.125L = 2.00M

CO + HCO + H22O O CO CO22 + H + H22

Initial 2.00 2.00 0 0Initial 2.00 2.00 0 0

Change -x -x +x +xChange -x -x +x +x

Equilibrium 2.00-x 2.00-x x xEquilibrium 2.00-x 2.00-x x x

1.56 = (x)(x)1.56 = (x)(x)

(2.0-x)(2.0-x)(2.0-x)(2.0-x)

This may look like you’ll need the quadratic This may look like you’ll need the quadratic but the alert student (that’s you) will notice but the alert student (that’s you) will notice that the right side is a perfect squarethat the right side is a perfect square

So, take the square root of both sides and So, take the square root of both sides and getget

√√1.56 = 1.25 = (x)/2-x) and x = 1.11 so1.56 = 1.25 = (x)/2-x) and x = 1.11 so

Plug back into the ICE chart and getPlug back into the ICE chart and get

[CO] = [H[CO] = [H22O] = 2-x = 0.89MO] = 2-x = 0.89M

[CO[CO22] = [H] = [H22] = x = 1.11 M] = x = 1.11 M

Go back and plug into the original Go back and plug into the original expression as a check:expression as a check:

(1.11)(1.11)(1.11)(1.11)

(0.89)(0.89) (0.89)(0.89)

= 1.56 (!)= 1.56 (!)

So What Does anSo What Does anEquilibrium Constant Tell Us?Equilibrium Constant Tell Us?

It is, roughly speaking a ratio of amounts It is, roughly speaking a ratio of amounts of products vs reactants at equilibriumof products vs reactants at equilibrium

The larger the value of Keq, the more the The larger the value of Keq, the more the products predominate at equilibriumproducts predominate at equilibrium

If Keq = 1 then there would be equal If Keq = 1 then there would be equal amounts of products and reactants amounts of products and reactants (depending on exponents in the (depending on exponents in the expression).expression).

For ExampleFor Example

Keq = 3 x 10Keq = 3 x 10-6-6 suggests primarily reactants suggests primarily reactants at equilibriumat equilibrium

Keq = 6 x 10Keq = 6 x 1077 suggests primarily products suggests primarily products are equilibriumare equilibrium

Keq = 1.2 suggests significant amounts of Keq = 1.2 suggests significant amounts of both products and reactants at equilibriumboth products and reactants at equilibrium

(this is usually said to be the case if Keq is (this is usually said to be the case if Keq is between 0.1 and 10)between 0.1 and 10)

PressurePressure

The equilibrium constant can also be The equilibrium constant can also be written in terms of the partial pressures of written in terms of the partial pressures of each gas in a mixture.each gas in a mixture.

The general rules still applyThe general rules still applyThe Kc and Kp expressions can be related The Kc and Kp expressions can be related

by the equation:by the equation:Kp =Kc(RT)Kp =Kc(RT)ΔΔnn, where , where ΔΔn = change in # n = change in #

moles of gas as the reaction goes from moles of gas as the reaction goes from reactants to productsreactants to products

Predicting The Direction of a Predicting The Direction of a ReactionReaction

You can also use the equilibrium constant You can also use the equilibrium constant idea to determine if a system is at idea to determine if a system is at equilibrium, and if not whether it will tend equilibrium, and if not whether it will tend to reach equilibrium via the forward or to reach equilibrium via the forward or reverse reaction.reverse reaction.

The term Keq or Kc means we know that a The term Keq or Kc means we know that a system is at equilibrium (no net change in system is at equilibrium (no net change in [reactants] or [products], forward and [reactants] or [products], forward and reverse rate are =reverse rate are =

Q - the reaction quotientQ - the reaction quotient

We define a term Q, the reaction quotient We define a term Q, the reaction quotient to be the same ratio as the equilibrium to be the same ratio as the equilibrium constant, just not known to be at constant, just not known to be at equilibriumequilibrium

We can compare Q to K and one of three We can compare Q to K and one of three things must be truethings must be true

1.1. Q > K (not at equilibrium goes to left)Q > K (not at equilibrium goes to left)2.2. Q < K (not at equilibrium goes to right)Q < K (not at equilibrium goes to right)3.3. Q = K (at equilibrium)Q = K (at equilibrium)

ExampleExample

Ammonia is made by the following reactionAmmonia is made by the following reaction 3 H3 H2(g) 2(g) + N+ N2(g)2(g) 2 NH 2 NH3(g) 3(g) , which as an , which as an

equilibrium constant value of 0.95 at 300Kequilibrium constant value of 0.95 at 300K A mixture in a 5.0 L flask has the following A mixture in a 5.0 L flask has the following

composition (at 300K): composition (at 300K): 2.0 moles hydrogen, 4.0 moles nitrogen and 1.0 2.0 moles hydrogen, 4.0 moles nitrogen and 1.0

moles ammonia.moles ammonia. Is the system at equilibrium and if not will the Is the system at equilibrium and if not will the

forward reaction (shift right) or reverse reaction forward reaction (shift right) or reverse reaction (shift left) happen to reach equilibrium?(shift left) happen to reach equilibrium?

Q vs KQ vs K

Write out QWrite out QPlug in and solve for QPlug in and solve for QCompare Q vs KCompare Q vs KExplainExplain

Q = [NHQ = [NH33]]2 2 = (0.2) = (0.2)22 = .78 = .78

[H[H22]]33[N[N22] (0.4)] (0.4)33(0.8)(0.8)

Q vs KQ vs K

Q is less than K so the reaction mix has Q is less than K so the reaction mix has too few products (or too many reactants) too few products (or too many reactants) to be at equilibriumto be at equilibrium

The forward rxn will happen until The forward rxn will happen until equilibrium is reached (shift right)equilibrium is reached (shift right)

This does not tell us how fast this will This does not tell us how fast this will happenhappen

Example:  A mixture consisting initially of 3.00 moles NH3,

2.00 moles of N2, and 5.00 moles of H2, in a 5.00 L container was

heated to 900 K, and allowed to reach equilibrium.  Determine the equilibrium concentration for each species present in the equilibrium mixture. 2 NH3(g)        N2(g) + 3 H2(g)    Kc = 0.0076 @ 900 K

•Convert the initial quantities to molarities as shown for NH3.

                                                                      •Create a chart as illustrated below and enter in the known quantities.   NH3 N2 H2

Initial Concentration (M)

0.600 0.400 1.00

Change in Conc. (M)

     

Equilibrium Conc. (M)

     

•Calculate Qc and compare to Kc to determine the direction the reaction •will proceed.

                                                                                                                            

Le Chatelier’s PrincipleLe Chatelier’s Principle

Equilibrium systems have an ability to Equilibrium systems have an ability to restore equilibrium if some change restore equilibrium if some change happens that makes the system not at happens that makes the system not at equilibrium (a chemical is added or equilibrium (a chemical is added or removed, the temperature or pressure is removed, the temperature or pressure is changed).changed).

The formal statement of this is called The formal statement of this is called LeChatelier’s principleLeChatelier’s principle

LeChatelierLeChatelier

A system at equilibrium, when A system at equilibrium, when stressedstressed, , will reacheive equilibrium by having either will reacheive equilibrium by having either the forward or reverse reaction increase in the forward or reverse reaction increase in rate, until equilibrium is restoredrate, until equilibrium is restored

Adding A ChemicalAdding A Chemical

Adding a reactant will mean that the ratio Adding a reactant will mean that the ratio of [products]/]reactants] is too low (excess of [products]/]reactants] is too low (excess of reactants or denominator)of reactants or denominator)

How does this get solved?How does this get solved?You need to “use” up the excess reactants You need to “use” up the excess reactants

and this is done by having the forward and this is done by having the forward reaction take place, until equilibrium is reaction take place, until equilibrium is restored. We say the system “shifts right.”restored. We say the system “shifts right.”

Like ThisLike This

equilibrum excess of reactants equilibrium restored

Adding a ReactantAdding a Reactant

Causes the forward reaction to increase in Causes the forward reaction to increase in rate until the equilibrium ratio is restoredrate until the equilibrium ratio is restored

This uses up the excess reactants, This uses up the excess reactants, creating more productscreating more products

Adding A ProductAdding A Product

Removing a ReactantRemoving a Reactant

This creates a “shortage” of reactants (or This creates a “shortage” of reactants (or an excess of products)an excess of products)

How to get more reactants- reverse How to get more reactants- reverse reaction, uses up products and forms reaction, uses up products and forms reactantsreactants

Reactant/Product SummaryReactant/Product Summary

Add a ReactantAdd a ReactantShift rightShift right

Add a ProductAdd a ProductShift leftShift left

Remove a ReactantRemove a ReactantShift leftShift left

Remove a ProductRemove a ProductShift rightShift right

3H3H22 +N +N22 2 NH 2 NH33

Temperature EffectsTemperature Effects

This is not dealing with the rate of the reactionThis is not dealing with the rate of the reaction This is the effects of changing temperature on This is the effects of changing temperature on

the composition of an equilibrium mixturethe composition of an equilibrium mixture You need to know if the rxn is endothermic or You need to know if the rxn is endothermic or

exothermicexothermic Endothermic- heat goes in and is treated like a Endothermic- heat goes in and is treated like a

reactantreactant Exothermic- heat is released and is treated like a Exothermic- heat is released and is treated like a

productproduct

Example of Temp EffectsExample of Temp Effects

CO + 3 HCO + 3 H22 CH CH44 + H + H22O + heatO + heatThis is an exothermic rxn (This is an exothermic rxn (ΔH < 0)ΔH < 0) If the temperature is raised (heat is If the temperature is raised (heat is

added) then the rxn will shift to the leftadded) then the rxn will shift to the leftThis will form more carbon monoxide and This will form more carbon monoxide and

hydrogenhydrogenThis will use up (decrease) methane and This will use up (decrease) methane and

waterwater

Co(HCo(H22O)O)66+2+2 + 4 Cl + 4 Cl-- CoCl CoCl44-2-2 + 6 H + 6 H22OO

pink bluepink blue

This rxn is endothermicThis rxn is endothermic

So…So…Heat + Co(HHeat + Co(H22O)O)66

+2+2 + 4 Cl + 4 Cl-- CoCl CoCl44-2-2 + 6 H + 6 H22OO

pink bluepink blue

Heat SummaryHeat Summary

ExothermicExothermicAdd heat- shift leftAdd heat- shift leftRemove heat- shift rightRemove heat- shift right

EndothermicEndothermicAdd heat- shift rightAdd heat- shift rightRemove heat-shift leftRemove heat-shift left

Pressure EffectsPressure EffectsRaising the pressure of a system (usually accomplished by decreasing the volume) shifts the equilibrium to the side with fewer moles of gas.Adding more gas does change the pressure but that is treated as a concentration effect

ExamplesExamples

NN22OO4(g)4(g) 2 NO 2 NO2(g)2(g)

raising the pressure shifts leftraising the pressure shifts left

4 NH4 NH3(g)3(g) + 5 O + 5 O2(g)2(g) 6 H 6 H22OO(g)(g) + 4 NO + 4 NO(g)(g)

raising the pressure shifts leftraising the pressure shifts left

HH2(g)2(g) + I + I2(g)2(g) 2 HI 2 HI(g)(g)

pressure change has no effect pressure change has no effect (why?)(why?)

HH22COCO3(aq)3(aq) H H22OO(l)(l) + CO + CO2(g)2(g)

raising the pressure shifts leftraising the pressure shifts left

Adding an inert gas (Ne, Ar) has no effect b/c volume does not change and the partial pressure of any reacting gas does not change.

Watch the states of matter- only gases are affected by P. Don’t count total moles on each side

CatalystsCatalysts

Do not affect the equilibrium positionDo not affect the equilibrium positionDo affect the rate at which equilibrium is Do affect the rate at which equilibrium is

reached.reached. If a system is not at equilibrium, a catalyst If a system is not at equilibrium, a catalyst

will get it to equilibrium fasterwill get it to equilibrium faster If a system is at equilibrium, a catalyst will If a system is at equilibrium, a catalyst will

have no effecthave no effect

The Haber ProcessThe Haber Process

The reaction conditions by which ammonia The reaction conditions by which ammonia is made commercially.is made commercially.

Developed by Fritz Haber to increase Developed by Fritz Haber to increase Germany’s agricultural production Germany’s agricultural production (ammonia is a component of fertilizers as (ammonia is a component of fertilizers as a nitrogen source)a nitrogen source)

Used in WWII as a means to increase Used in WWII as a means to increase Germany’s weaponry Germany’s weaponry

3H3H22 +N +N22 2 NH 2 NH3 + 3 + heatheat

To maximize ammonia syntheis:To maximize ammonia syntheis:Add HAdd H22 and N and N22 (or keep them at high (or keep them at high

concentrations)concentrations)Remove ammonia as fast as it if formedRemove ammonia as fast as it if formedUse a high pressureUse a high pressureUse a lower temperature- can’t do too low Use a lower temperature- can’t do too low

as tis lowers the rate of the reactionas tis lowers the rate of the reactionKeep a catalyst in the system-Keep a catalyst in the system-

Ksp & The Common Ion EffectKsp & The Common Ion Effect

Application of LeChatelier’s PrincipleApplication of LeChatelier’s PrincipleWe did this same topic with acids and We did this same topic with acids and

bases- same results should applybases- same results should applyPbCrOPbCrO4(s)4(s) Pb Pb+2+2

(aq)(aq) + CrO + CrO44-2-2

(aq)(aq)

If the amount of chromate ion is increased, If the amount of chromate ion is increased, then the equilibrium will shift to the left and then the equilibrium will shift to the left and more solid will form. This means the more solid will form. This means the compound is less soluble.compound is less soluble.

One ExampleOne Example

What is the molar solubility of BaSO4 in a What is the molar solubility of BaSO4 in a solution of .10M sodium sulfate?solution of .10M sodium sulfate?Ksp = 1.1 x 10-10Ksp = 1.1 x 10-10

BaSOBaSO4(s)4(s) Ba Ba+2+2(aq)(aq) + SO + SO44

-2-2(aq)(aq)

x .10 + xx .10 + xKsp =[BaKsp =[Ba+2+2][SO4][SO4-2-2]]

Ksp =[BaKsp =[Ba+2+2][SO4][SO4-2-2]]1.1 x 101.1 x 10-10-10 = (x)(.1 +x) = (x)(.1 +x)Shortcut: is x is small, then .1 + x = .1Shortcut: is x is small, then .1 + x = .11.1 x 101.1 x 10-10-10 = (x)(.1) and x = = (x)(.1) and x =1.1 x 101.1 x 10-9-9

Molar solubility = 1.1 x 10Molar solubility = 1.1 x 10-9-9 M MHow would you turn this into solubility in How would you turn this into solubility in

grams/L?grams/L?Compare to solubility of BaSOCompare to solubility of BaSO44 in pure in pure

water which is 1.05 x 10water which is 1.05 x 10-5-5 M MPresence of a common ion decreases Presence of a common ion decreases

solubilitysolubility

Will a Precipitate Form?Will a Precipitate Form?

An application of theQ vs K problems fromequilibrium studies.

Will a Precipitate Form ?Will a Precipitate Form ?

The equilibrium condition in a saturated The equilibrium condition in a saturated solution is a dynamic equilibriumsolution is a dynamic equilibrium

The solid is dissolving as fast as ions are The solid is dissolving as fast as ions are combining to form solid (undissolving)combining to form solid (undissolving)

If the ion concentrations exceed Ksp then If the ion concentrations exceed Ksp then a precipitate forms (shift to the left which is a precipitate forms (shift to the left which is formation of solid)formation of solid)

Example 1Example 1

Will a precipitate form when equal Will a precipitate form when equal volumes of .2M Pb(NOvolumes of .2M Pb(NO33))22 and .2M KI are and .2M KI are

mixed? Ksp for lead (II) iodide = 1.4 x 10mixed? Ksp for lead (II) iodide = 1.4 x 10-8-8

Pb(NOPb(NO33))22 +2 KI +2 KI PbIPbI22 + 2KNO + 2KNO33

PbIPbI2(s)2(s) Pb Pb+2+2(aq)(aq) + 2 I + 2 I--

(aq)(aq)

Ksp = [PbKsp = [Pb+2+2][I][I--]]22

Treat as a Q vs K ProblemTreat as a Q vs K Problem

Q = [PbQ = [Pb+2+2][I][I--]]22

Since equal volumes of .2M PbSince equal volumes of .2M Pb+2+2 and .2 M and .2 M II-- ions are mixed, the concentration of ions are mixed, the concentration of each ion is .1 M (same # moles, double each ion is .1 M (same # moles, double volume = half concentration)volume = half concentration)

Q = (.1)(.1)Q = (.1)(.1)22 = .001 = 1 x 10 = .001 = 1 x 10-3-3

Q> K so it shifts left and forms a ppt.Q> K so it shifts left and forms a ppt.

One moreOne more

Does a precipitate form when 200 ml of 1 Does a precipitate form when 200 ml of 1 x 10x 10-4-4 M AgNO3 and 800 ml of 1 x 10 M AgNO3 and 800 ml of 1 x 10-6-6 M M KCl are mixed? Ksp for AgCl is 1.6 x 10-KCl are mixed? Ksp for AgCl is 1.6 x 10-10.10.

AgNOAgNO33 + KCl + KCl AgClAgCl + KNO + KNO33

AgClAgCl(s)(s) Ag Ag++(aq)(aq) + Cl + Cl--(aq)(aq)

Problem #2 cont’dProblem #2 cont’d

[Ag[Ag++] = 2 x 10] = 2 x 10-5-5 (why not the original 1 x 10 (why not the original 1 x 10--

44?)?) [Cl[Cl--] = 8 x 10] = 8 x 10-7-7 (same ?) (same ?)Q = (2 x 10Q = (2 x 10-5-5 )(8 x 10 )(8 x 10-7-7 ) = 1.6 x 10 ) = 1.6 x 10-11-11

Q < K so shift right (no precipitate forms).Q < K so shift right (no precipitate forms).

Q > KPpt forms

[ions] exceedsequilibrium

Q = K equilibrium

saturated sol’n

Q < K[ions] less

than equilibriummore could dissolve

More Help?More Help?

http://www.chemguide.co.uk/physical/equilhttp://www.chemguide.co.uk/physical/equilibria/kc.htmlibria/kc.html

http://dbhs.wvusd.k12.ca.us/webdocs/Equihttp://dbhs.wvusd.k12.ca.us/webdocs/Equilibrium/Equilibrium-Constant.htmllibrium/Equilibrium-Constant.html

http://www.chemguide.co.uk/physical/equilhttp://www.chemguide.co.uk/physical/equilibria/lechatelier.htmlibria/lechatelier.html

http://www.chm.davidson.edu/java/LeChathttp://www.chm.davidson.edu/java/LeChatelier/LeChatelier.htmlelier/LeChatelier.html

More Help 2?More Help 2?

http://www.chem.purdue.edu/gchelp/howtohttp://www.chem.purdue.edu/gchelp/howtosolveit/Equilibrium/Solubility_Products.htmsolveit/Equilibrium/Solubility_Products.htm

http://chemed.chem.purdue.edu/genchem/http://chemed.chem.purdue.edu/genchem/topicreview/bp/ch18/ksp.phptopicreview/bp/ch18/ksp.php

http://www.chem.sc.edu/goode/C112Web/http://www.chem.sc.edu/goode/C112Web/CH14NF/Ch14nf.htmCH14NF/Ch14nf.htm

http://www.shodor.org/unchem/advanced/http://www.shodor.org/unchem/advanced/equ/equ/

Wrap-UpWrap-Up

Make sure you understand definitions Make sure you understand definitions related to equilibriumrelated to equilibrium

Be able to do ICE tables and solve the Be able to do ICE tables and solve the algebraalgebra

Be able to explain how any changes can Be able to explain how any changes can affect equilibrium using Le Chatelier’s affect equilibrium using Le Chatelier’s Principle.Principle.