EquilibriumAP Chem

Mr. Nelson

The Concept of EquilibriumThe term reversible reaction is used to describe reactions that can go in either the forward or the reverse direction

Equilibrium is used to specify that a reversible reaction has reached an equal rate for both directions

The Concept of EquilibriumChemical equilibrium occurs when opposing reactions are proceeding at equal rates◦Concentrations of the reactants/products remains constant

The Concept of Equilibrium

Its important to note that forward and reverse reactions still occur, although outwardly no apparent change occurs

)(2)(3)( 322 gNHgHgN

The Concept of EquilibriumNote that the equilibrium concentrations are independent of initial concentrations

A mathematical relationship exists between concentration of the reactants and products once equilibrium has been reached

The Concept of EquilibriumFish Tank Example

Connection between Equilibrium & KineticsFor generalized reaction:

A⇋BForward reaction:

AB◦Forward Rate Law: Ratef = kf[A]

Reverse reaction:

BA◦Reverse Rate Law: Rater = kr[B]

Connection between Equilibrium & KineticsBy definition, at equilibrium

forward and reverse rates must be equal, so:

Ratef = Rater

kf[A] = kr[B]Also, concentrations remain

constant,so: eqK constant A

B

r

f

k

k

The Equilibrium ConstantThe equilibrium constant (Keq)

can be determined at constant temperature for any reaction at equilibrium:

aA+ bB cC + dDKeq =

[C]c [D]d

[A]a [B]b

Keq is unitless

The Equilibrium ConstantNote that only equilibrium

concentrations can be placed into an equilibrium constant expression!

Also, no solids or liquids are included in an equilibrium constant expression

The reason for this is their concentrations remain fairly constant in equilibrium

Note that they are still required for the reaction to proceed however

The Equilibrium Constant

The Many Types of

Equilibrium Constants, Keq

Kc

Kp

KspKa/Kb

Kw

Concentration vs. Pressure EquilibriaEquilibrium expressions can be

written in terms of concentration (Kc) measured in molarity or pressure (Kp) measured in atm◦Example:

)(2)( 242 gNOgON

][

][

42

22

ON

NOKc

42

2

2

ON

NO

p P

PK or

Concentration vs. Pressure EquilibriaIn general, Kc is not equal to Kp

The following equation converts between the two:

n = moles of gas products – moles of gas reactants◦Note: When no change in moles,

Kp=Kc

ncp RTKK

Homogeneous vs. Heterogeneous Equilibria

The term homogeneous equilibrium is used to describe reactions where all the species are in the same phase◦ Example:

The term heterogeneous equilibrium is used to describe reactions where species are in different phases◦ Example:

H2CO3 (aq) + H2O (l) HCO3- (aq) + H3O+

(aq)

)(2)( 242 gNOgON

Multiple EquilibriaWhen the products of one

reaction are the reactants of another

Example: A + B C + D

C + D E + F

K1

A + B E + F

Kc

K2

Multiple EquilibriaUsing their equilibrium

expressions, we can derive a mathematical relationship:

]][[

]][[

]][[

]][[

]][[

]][[

DC

FEx

BA

DC

BA

FEKc

21 KKKc

Manipulating Equilibria Expressions

If an equilibrium is written in the reverse, the following relationship exists:

]][[

][

23

5

ClPCl

PClKc

][

]][[

5

23

PCl

ClPClKc

cc KK /1

PCl3 + Cl2 PCl5

PCl5 PCl3 + Cl2

Manipulating Equilibria Expressions

If an equilibrium is multiplied by a factor, the following relationship exists:

]][[

][

23

5

ClPCl

PClKc PCl3 + Cl2 PCl5

2PCl3 + 2Cl2 2PCl5 22

23

25

][][

][

ClPCl

PClKc

2cc KK

The Magnititude of Equilibrium Constants

What does Keq tell us?

◦If Keq is very large, products will be favored This implies the numerator is large in the Keq

expression compared to denominator

◦If Keq is very small, reactants will be favored This implies the denominator is large in the

Keq expression compared to numerator

◦ If Keq is close to 1, then roughly equal amounts of reactants and products are present at equilibrium

The Reaction Quotient (Q)

Recall: In electrochemistry, we utilized “Q” which is similar in all respects to “Keq” except in definition◦Q is utilized when given initial concentrations, not concentrations at equilibrium

Predicting the Direction of Reversible ReactionsIf given an initial set of conditions for a reversible reaction, utilizing Q and Keq, a prediction of the direction of the reaction (towards reactants/towards products) can be made

Comparing Kc and Qc

Qc<Kc

• Ratio of prds to rcts is too small• To reach equilibrium, rcts must go

towards prds• Shifts to the right

Qc=Kc•Initial concentrations are at equilibrium concentrations•System is at equilibrium

Qc>Kc

• Ratio of prds to rcts is too large• To reach equilibrium, prds must go

towards rcts• Shifts to the left

Comparison of Kc and Qc

Reactants → Products

Equilibrium: no net change

Reactants ← Products

Qc

Kc KcKc

Qc

Qc

Calculating Equilibrium ConcentrationsTabulate all known initial and

equilibrium concentrations◦When both are known, calculate the

changeUsing coefficients, calculate

changes in concentration for all participants in a rxn

Calculate equilibrium concentrations

The ICE MethodThis is the process used for

determining equilibrium concentrations

Initial molarityChange in molarityEquilibrium molarity

Example #1: Calculating Equilibrium Concentrations

A mixture of 0.500 mol H2 and 0.500 mol I2 was placed in a 1.00 L stainless-steel flask at 430 °C. The equilibrium constant Kc for the reaction:

H2 (g) + I2 (g) 2 HI (g)

is 54.3 at this temp. Calculate the concentrations of H2, I2, and HI at equilibrium.

Initial (M)0.500 0.500 0

Change (M) -x -x +2x

Equilibrium (M) 0.500 – x 0.500 – x 2x

H2 (g) + I2 (g) 2 HI (g)

]][[

][3.54

22

2

IH

HIKc

Substituting, we get:

Taking the square root of both sides:

)500.0)(500.0(

)2(3.54

2

xx

x

)500.0(

237.7

x

x

x = 0.393 M

So, the concentrations at equilibrium would be the following:◦[H2] = (0.500 – 0.393) M = 0.107 M

◦[I2] = (0.500 – 0.393) M = 0.107 M

◦[HI] = 2 x 0.393 M = 0.786 M

Example #2: Calculating Equilibrium ConcentrationsWhen 3.0 mol of I2 and 4.0 mol of

Br2 are placed in a 2.0 L reactor at 150oC, the following reaction occurs until equilibrium is reached:

I2(g) + Br2(g) 2IBr(g)

Chemical analysis then shows that the reactor contains 3.2 mol of IBr. What is the value of the equilibrium constant Kc for the reaction?

Initial (M)1.5 2 0

Change (M) -x -x +2x

Equilibrium (M) 1.5 – x 2 – x 1.6

I2 (g) + Br2 (g) 2 IBr (g)

]][[

][

22

2

BrI

IBrKc

When initial and final concentrations are known, calculate the change:◦0 + 2x = 1.6◦x = 0.8

Then determine all equilibrium concentrations:◦[I2] = 1.5 – 0.8 = 0.7 M

◦[Br2] = 2 – 0.8 = 1.2 M

◦[IBr] = 1.6 M

Plug equilibrium concentrations into equation and solve:

]][[

][

22

2

BrI

IBrKc

05.3]2.1][7.0[

]6.1[ 2

cK

Calculating Equilibrium Concentrations when Keq is small

When solving equilibrium concentrations and Keq is small, assume the effect of the change (“x”) is negligible (for rcts only)

Example: Calculating Equilibrium Concentrations when Keq is small

At 100 °C the equilibrium constant (Kc) for the reaction:

COCl2 (g) 2 CO (g) + Cl2 (g)

is 2.19 x 10-10. If the initial concentration of [COCl2] = 1.5 M, calculate the concentrations of all species at equilibrium.

COCl2 (g) 2 CO (g) + Cl2 (g)Initial (M)

1.5 0 0

Change (M) -x +2x +x

Equilibrium (M) 1.5 – x +2x +x

][

][][

2

22

COCl

ClCOKc

)5.1(

)()2( 2

x

xxKc

)5.1(

)()2( 2 xx

5.1

41019.2

310 x

x

4

4

4

10285.3 310 xx

3 33 11102125.8 xx

Le Chatelier’s Principle

If a stress is applied to a system in equilibrium, the equilibrium shifts to relieve that stress

Forms of “stress”:◦Concentration◦Pressure◦Temperature

Change in Reactant or Product ConcentrationAdding a substance shifts a system at equilibrium away from the added substance◦System must consume added item

Removing a substance from a system at equilibrium shifts the reaction toward the removed substance◦System must replace removed item

Changes in Pressure

Increase in pressure forces an equilibrium to shift in the direction that reduces the number of moles of gas in the system◦Example:◦An increase in pressure causes a shift to the left

N2O4 (g) 2 NO2 (g)

Changes in PressureRecall: Changes in volume can also

affect changes in pressure◦Increase volume = decrease pressure◦Decrease volume = increase pressure

Also the addition of an inert gas (one not in the reaction) can cause an increase in pressure as well

Changes in TemperatureEnthalpy of the reaction plays a major roll in how heat effects the equilibrium◦Recall: Endothermic (+∆H) means heat is a reactant, while exothermic (- ∆H) means it is a product

Treat changes in heat in the same manner that you treat changes in concentration

Effects of Catalysts on EquilibriumRecall: Catalysts lower energy

barrier between reactants and products◦Ea for both forward and reverse are

lowered to the same extentCatalysts increase the rate at

which equilibrium is reachedHowever, it does NOT change the

composition of the equilibrium mixture.

Solubility Equilibria

Saturated solutions of salts are another type of chemical equilibria

Slightly soluble salts establish a dynamic equilibrium with the hydrated cations and anions in solution

The Solubility Product, Ksp

The equilibrium constant, the Ksp, is no more than the product of the ions in solution (Recall: solids do not appear)

For a saturated solution of AgCl, the equation would be:

The solubility product expression would be:

)()()( aqClaqAgsAgCl

]][[ ClAgK sp

Sequence of Steps for Solubility Equilibria

Solubility

of compoun

d

Molar solubility of

compound

Concentrations of cations and anion

s

Ksp of

compoundKsp

of compound

Concentrations of cations and anion

s

Molar solubility of

compound

Solubility

of compoun

d

SolubilitySolubility can be expressed two ways:◦Molar solubility is the moles of solute per 1L of saturated solution (mol/L)

◦Solubility is the number of grams of solute per 1 L of saturated solution (g/L)

Ksp expressions require molar solubility

Determining Ksp by Experimental MeasurementsUse ICE table, but with “s” for change (instead of “x”)

“s” is not only the change, but the molar solubility

Determining Ksp: Example #1

The solubility of CaSO4 is found to be 0.67 g/L. Calculate the value of Ksp for calcium sulfate.

Solubility of CaSO4 in g/L

Molar solubility of CaSO

4

[Ca2

+] and [SO4

2-]

Ksp of

CaSO4

Determining Ksp: Example #1

)()()( 24

24 aqSOaqCasCaSO

Initial (M)0 0

Change (M) -s +s +s

Equilibrium (M) s s

224

2 ]][[ sSOCaK sp

Determining Ksp: Example #1

As CaSO4 dissolves, we get 1 mole Ca2+, and 1 mole SO4

2-, so at equilibrium:

sMxCaSOg

CaSOmol

solutionL

CaSOg 3

4

44 109.42.136

1

1

67.0

MxSOCa 324

2 109.4][][ 533 104.2)109.4)(109.4( xxxK sp

Determining Ksp: Example #2

Lead (II) chloride dissolves to a slight extent in water according to the equation:

Calculate the Ksp if the solubility of lead (II) chloride has been found to be 1.62 x 10-2 M

)(2)()( 22 aqClaqPbsPbCl

Determining Ksp: Example #2Molar

solubility of PbCl2

[Pb2+

] and [Cl-]

Ksp of PbCl2

)(2)()( 22 aqClaqPbsPbCl

Initial (M)0 0

Change (M) -s +s +2s

Equilibrium (M) s 2s

Determining Ksp: Example #2

Mxs 5108.1 35 )108.1(4 xK sp

141033.2 xK sp

3222 4)2)((]][[ sssClPbK sp

Calculating Solubility from Ksp: Example #1

The Ksp for CaCO3 is 3.8 x 10-9 at 25 °C. Calculate the solubility (in g/L) of calcium carbonate in pure water.

Ksp of CaCO3

Concentrations of [Ca2+]

and [CO32-]

Molar solubility of

CaCO3

Solubility of CaCO3

Calculating Solubility from Ksp: Example #1

223

2 ]][[ sCOCaK sp

)()()( 23

23 aqCOaqCasCaCO

Initial (M)0 0

Change (M) -s +s +s

Equilibrium (M) s s

Calculating Solubility from Ksp: Example #1

3

335

1

100

1

1016.6

CaCOmol

CaCOg

solutionL

CaCOmolx

29108.3 sx

Mxs 51016.6

Lgx /1016.6 3

Predicting Precipitation Reactions using QUsing the reaction quotient, Q, we can

◦Decide whether a ppt will form◦Calculate ion concentration required

to begin the ppt. of an insoluble salt

Q < Ksp

• System has too much reactant

• No ppt

Q = Ksp

• System at equilibrium

• No ppt

Q > Ksp

• System has too much product

• Forms ppt!

Predicting Precipitation ReactionsSteps

1. Determine insoluble product using solubility rules

2. Calculate cation and anion concentration of insoluble product (if mixed)

3. Calculate Q4. Compare to Ksp

Example: Predicting Precipitation

Exactly 200 mL of 0.0040 M BaCl2 are mixed with exactly 600 mL of 0.0080 M K2SO4. Will a precipitate form? (Ksp of insoluble salt is 1.1 x 10-10)

The Common Ion EffectLe Chatelier’s Principle can describe the effect of an added common ion◦For example, if NaCl were added to a saturated solution of PbCl2, what effect would occur?

◦Increase product concentration shifts equilibrium towards reactants, forming more ppt.

Solubility and the Common Ion EffectThe presence

of a common ion will cause the equilibrium to shift so that even less of the substance (with smaller Ksp value) will dissolve

The Common Ion EffectBe aware: pH can affect solubility

◦The addition of an acid or base could have a side reaction, decreasing concentration of an ion

◦If an acid were added to the following, what effect would be observed?

H+ combines with OH-, so [OH-] is decreased, forcing equilibrium to the right

)(2)()()( 22 aqOHaqMgsOHMg

Example: Common Ion Effect

Calculate the solubility of silver chloride (in g/L) in a 6.5 x 10-3 M silver nitrate solution. (Ksp = 1.6 x 10-10)

◦Since silver nitrate is a soluble, strong electrolyte: )()()( 33

2 aqNOaqAgsAgNO OH

6.5 x 10-3 M

6.5 x 10-3 M

Example: Common Ion Effect

)()()( aqClaqAgsAgCl Initial (M)

6.5 x 10-3 0.00

Change (M) -s +s +s

Equilibrium (M)

(6.5 x 10-3 + s)

s

sxxK sp )105.6(106.1 310

Because Ksp is so small, s must be very small in comparison to 6.5 x 10-3, so we approximate that 6.5 x 10-3 + s 6.5 x 10-3

Mxs 8105.2

Selective PrecipitationIn certain problems, you will have

to determine which of two possible precipitates will form◦The salt with the lower Ksp will

precipitate firstTo determine the amount of an ion

needed to precipitate, recall that Ksp concentrations are saturated, so anything greater than this will form a ppt.

Example: Selective Precipitation

A solution contains 1.0 x 10-4 M Cu+ and 2.0 x 10-3 M Pb2+. If a source of I- is added gradually to this solution, will PbI2 (Ksp = 1.4 x 10-8) or CuI (Ksp = 5.3 x 10-12) precipitate first? Specify the concentration of I- necessary to begin precipitation of each salt