AP Chapter 15Equilibrium

*Chapters 15, 16 and 17 are all EQUILIBRIUM chapters*

HW: 4 13 15 17 21 27 29 31 37 41 55 69

15.1 - The Concept of Equilibrium

Chemical equilibrium occurs when a reaction and its reverse reaction proceed at the same rate.

N2O4 2 NO2

Reactants and Products CANNOT escape the container!

Depicting Equilibrium

In a system at equilibrium, both the forward and reverse reactions are being carried out; as a result, we write its equation with a double arrow

N2O4 (g) 2 NO2 (g)

At equilibrium the forward and reverse directions of the reaction are occurring at equal rate.

Ratef = Rater

The Concept of Equilibrium• As a system approaches

equilibrium, both the forward and reverse reactions are occurring.

• At equilibrium, the forward and reverse reactions are proceeding at the same rate.

A System at Equilibrium

Once equilibrium is achieved, the amount of each reactant and product remains constant.

The amounts are not equal!!

K = The Equilibrium Constant

15.2 -

The Equilibrium Constant• Consider the reaction:

• The equilibrium expression for this reaction would be

Kc = [C]c[D]d

[A]a[B]b

aA + bB cC + dD

Multiplication of Products to their “coefficient powers”

Multiplication of Reactants to their “coefficient powers”

concentration

“reactants” “products”

The Equilibrium Constant

Kc = [C]c[D]d

[A]a[B]b

aA + bB cC + dD

K values are usually UNITLESS!(even if they don’t truly cancel)

Examples:

Kc = [NO2]2

[N2O4]N2O4 (g) 2 NO2 (g)

Kc = [N2O4][NO2]2

N2O4 (g)2 NO2 (g)

Example:

Kc = [NO2]2

[N2O4]N2O4 (g) 2 NO2 (g)

Kc = = ??[NO2]2

[N2O4]

15.2 – Writing Equilibrium Constant Expressions

• Homogeneous Equilibrium – Equilibrium when the species are in the same phase

N2O4 (g) 2 NO2 (g)

Kc = [NO2]2

[N2O4]Kp = PNO2

2

PN2O4

Because pressure is proportional to concentration for gases in a closed system, the equilibrium expression can also be written based on pressures.

Relationship between Kc and Kp

• From the ideal gas law we know that

• Rearranging it, we get

PV = nRT

P = RTnV

• Using the Kp expression, Kp becomes:

Kp = RT

nB

Vb

RTnA

Va

For the reaction: aA(g) bB(g)

Kp = PBb

PAa

Relationship between Kc and Kp

n/V is concentration and the RT terms can be pulled out to give:

Where

Kp = Kc (RT)n

n = b-a = moles of gaseous product − moles of gaseous reactant

Kp = b

a(RT)Dn

[A]

[B]

R = 0.08206

Relationship between Kc and Kp

• Example:H2(g) + Br2(g) = 2HBr (g)

Dn = 2 – 2 = 0, Kc = Kp Example:

Dn = 2 – 1 = 1

Kp = Kc (RT)n

N2O4 (g) 2 NO2 (g)

Relationship between Kc and Kp

N2(g) + 3H2(g) = 2 NH3(g)

If Kc = 9.60 at 300oC, what is Kp?

(Kp = 4.34 x 10-3)

15.3 – Interpreting the Equilibrium Constant

• If K >> 1, the reaction is product-favored; product predominates at equilibrium.

• If K << 1, the reaction is reactant-favored; reactant predominates at equilibrium.

15. 3 - Form of K-How you write the reaction determines the value of K

The equilibrium constant of a reaction in the reverse reaction is the reciprocal of the equilibrium constant of the forward reaction.

10.212=

Kc = = 0.212 at 100C[NO2]2

[N2O4]N2O4 (g) 2 NO2 (g)

Kc = = 4.72 at 100C

[N2O4][NO2]2N2O4 (g)2 NO2 (g)

Form of K-K values depend on the way the reaction is balanced.-The equilibrium constant of a reaction that has been multiplied by a number is the equilibrium constant raised to a power that is equal to that number.

Kc = = 0.212 at 100C[NO2]2

[N2O4]N2O4 (g) 2 NO2 (g)

Kc = = (0.212)2 at 100C[NO2]4

[N2O4]22 N2O4 (g) 4 NO2 (g)

Multiple Equilibria

C + D E + F

K = K =

Conclusion: The overall reaction is the product of the equilibrium constants of the individual reactions!

Summary Pg 638

CH3COOH(aq) + H2O(l) H3O+(aq) + CH3COO-(aq)

K = ??? =

But – the [H2O] is VERY high (~55.5 M) so it is essentially constant, so K becomes:

K = ??? =

(Liquids are USUALLY not in K expressions)

15.4 – Heterogeneous Equilibrium

The Concentrations of Solids and Liquids Are Essentially Constant

d = g/LMM = g/mold/MM = = = mol/L

For solids and liquids, density and MM are constants at constant temperature concentration is a constant.

gramsLitergramsmole

grams molesx

Liter gram

The Concentrations of Solids and Liquids Are Essentially Constant

Therefore, the concentrations of solids and liquids do not appear in the equilibrium expression

Kc = [Pb2+] [Cl−]2

PbCl2 (s) Pb2+ (aq) + 2 Cl−

(aq)

Kc = [CO2] OR Kp = PCO2

CaCO3 (s) CO2 (g) + CaO(s)

15.5 – Calculating Equilibrium ConstantsExample 15.9 – Pg 643

• A closed system initially containing 1.000 x 10-3 M H2 and 2.000 x 10-3 M I2 at

448 C is allowed to reach equilibrium. Analysis of the equilibrium mixture shows that the concentration of HI is 1.87 x 10-3 M. Calculate Kc at 448 C for the reaction taking place, which is

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

What do we know:

[H2], M [I2], M [HI], M

Initially 1.000 x 10-3 2.000 x 10-3 0

Change -x -x +2x

At equilibrium 1.87 x 10-3

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

Now we know the change in [HI]

[H2], M [I2], M [HI], M

Initially 1.000 x 10-3 2.000 x 10-3 0

Change +1.87 x 10-3

At equilibrium 1.87 x 10-3

Can fill in the remainder of the changes based on stoichiometry…

[H2], M [I2], M [HI], M

Initially 1.000 x 10-3 2.000 x 10-3 0

Change -9.35 x 10-4 -9.35 x 10-4 +1.87 x 10-3

At equilibrium 1.87 x 10-3

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

Now fill in the Equilibrium Values…

[H2], M [I2], M [HI], M

Initially 1.000 x 10-3 2.000 x 10-3 0

Change -9.35 x 10-4 -9.35 x 10-4 +1.87 x 10-3

At equilibrium 6.5 x 10-5 1.065 x 10-3 1.87 x 10-3

And finally calculate K…

Kc =[HI]2

[H2] [I2]

= 51

=(1.87 x 10-3)2

(6.5 x 10-5)(1.065 x 10-3)

15.6 - Calculating Equilibrium

ConcentrationsThe ICE Method

ICE

InitialChangeEquilibrium

*Make a chart for these calculations*

Example:

Cis-stilbene trans-stilbeneKc = 24.0 at 200oC[cis]i = 0.850 M

[trans] i = 0 M

WORK:

More Examples:

• 15.11

• 15.12

% Yield

• % = Equilibrium [ ]of a product100

[ ]of a product if reaction was completex

On Page 13 of notes!!

15.6 – What does K tell us?• Predicting the direction of a reaction:• Calculate a “Reaction Quotient” (Qc). It has

the same FORM as Kc, but can be used at ANY time.

• If Q>K, then products are too high, and reaction will shift LEFT

• If Q<K, then products are too low, and reaction will shift RIGHT

• If Q = K, then the reaction is at equilibrium

Example – In what direction will the reaction proceed?

H2 + I2 2HIK = 54.3 at 430oCInitial concentrations in a 1 L container:H2 = 0.243 molI2 = 0.146 molHI = 1.98 mol

15.7 – Factors that Affect Chemical Equilibrium

• LeChatlier’s Principle – When an external stress is applied to a system at equilibrium, the system adjusts to offset the stress.

LeChatlier’s – Changes in Concentration

Does NOT change the value of KFeSCN+2 Fe+3 + SCN-1

-Add NaSCN

-Add Fe(NO3)3

-Add H2C2O4

Changes in Volume and Pressure• Only affects gases• Does not change the value of K• Lowering volume, creates an increase in concentration

(mol / L)• Lowering volume, increasing concentration, causes a

shift to the side of the reaction with FEWER MOLES OF GAS

• Increasing volume, decreasing concentration, causes a shift to the side of the reaction with MORE MOLES OF GAS

• If there is no change in the moles of gas, changing volume and pressure create NO SHIFT

Changes in Volume and Pressure

• Adding “an inert gas” does NOT effect equilibrium, because the partial pressures of all of the gases in the reaction do not change

Changes in Temperature

• ALTERS K – this is the only change that changes the Equilibrium Constant!

• If heat is added, the reaction shifts to the ENDOTHERMIC reaction

• If heat is removed, the reaction shifts to the EXOTHERMIC reaction

Catalyst

• Does not change the value of K• Lowers the activation energy (for both the

forward and the reverse reaction)• Makes the reaction reach equilibrium faster