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Week 9 CHEM 1310 - Sections L and M 1

Exam #2 Results

Class Average = 77 (Great Job!)

Week 9 CHEM 1310 - Sections L and M 2

Exam #2 Results

4 Perfect Scores!

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Week 9 CHEM 1310 - Sections L and M 3

Ch 6: Chemical Equilibrium

What is Equilibrium? Equilibrium Constant, K Equilibrium Expressions Involving Pressures Activity Heterogeneous Equilibria Applications of Equilibrium Constant Solving Equilibrium Problems Le Chatelier’s Principle - very important Equilibria Involving Real Gases

Week 9 CHEM 1310 - Sections L and M 4

What is Equilibrium?

Equilibrium is the phenomenon that occurs whenthe rate of the forward reaction equals

the rate of the reverse reaction.

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Week 9 CHEM 1310 - Sections L and M 5

What is Equilibrium?

Example

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

Forward Rxn

Reverse Rxn

Forward Rxn: Product = HIReverse Rxn: Products = H2 and I2

At equilibrium, the concentrations of all reactants and products remain constant with time.

[HI] = constant; [H2] = constant; [I2] = constant

Week 9 CHEM 1310 - Sections L and M 6

What is Equilibrium?

Notice how the concentrations of products for the forward and reverse reactions are not necessarily equal

at equilibrium!

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Week 9 CHEM 1310 - Sections L and M 7

What does equilibrium look like in a chemical system?

N2O4 2 NO2colorless brown

Equilibrium

Closed system reaches the same equilibrium concentrationswhether the reaction starts with the N2O4 or the NO2!

Equilibrium Characteristics

Week 9 CHEM 1310 - Sections L and M 8

The Equilibrium Constant

aA + bB cC + dD

EquilibriumConstant

K = [C]c x [D]d [A]a x [B]b

Characteristics

Exponents are coefficients from balanced chemical equation.

Units for K will vary depending upon coefficients.

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Week 9 CHEM 1310 - Sections L and M 9

The Equilibrium Constant

cC + dD aA + bB

EquilibriumConstant K =

[C]c x [D]d [A]a x [B]b

Characteristics

Reversing the reactants and products inverts the equilibriumexpression. Thus,

Kforward = 1Kreverse

Week 9 CHEM 1310 - Sections L and M 10

Law of Mass Action

K is constant despite different initial and equilibriumconcentrations of reactants and products!

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Week 9 CHEM 1310 - Sections L and M 11

Equilibrium equations can be reversed, scaled orcombined.

aA + bB cC + dD

Forward:K1 =

[C]c x [D]d

[A]a x [B]b

K2 = [C]c x [D]d

[A]a x [B]b

K1

K2

Reverse:

cC + dD aA + bB

By defn: K1 x K2 = 1

Manipulation of Equilibrium Eqns

Week 9 CHEM 1310 - Sections L and M 12

Equilibrium equations can be reversed, scaled orcombined.

PCl3 + Cl2 PCl5K1 =

[PCl5][PCl3] x [Cl2]

K2 =

K

Scaled:

Example

K2 = (K1)2

2 PCl3 + 2 Cl2 2 PCl5K

[PCl5]2

[PCl3]2 x [Cl2]

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Manipulation of Equilibrium Eqns

When stoichiometry is scaled, the resulting K is raised to the power of

the scale factor

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Week 9 CHEM 1310 - Sections L and M 13

Equilibrium equations can be reversed, scaled orcombined.

K1

K3 =

Equation #1

Equation #2-

Equation #3

K2

K1

K2

K1

K3 =

Equation #1

Equation #2+

Equation #3

K2

K1 x K2

Subtraction Addition

Subtraction of Equilibrium Eqns

Week 9 CHEM 1310 - Sections L and M 14

What is the relationship between Q and K?

aA + bB cC + dD

Reaction Quotient vs. Equilibrium Constant

Q = [C]c x [D]d

[A]a x [B]b

Holds whether at equilibrium or not!

K = [C]c x [D]d

[A]a x [B]b

Holds at equilibrium only!

K and the Reaction Quotient, Q

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Week 9 CHEM 1310 - Sections L and M 15

What is the relationship between Q and K?

aA + bB cC + dD

When Q = K = [C]c x [D]d

[A]a x [B]b

Equilibrium occurs

Q vs K

Week 9 CHEM 1310 - Sections L and M 16

What is the relationship between Q and K?

aA + bB cC + dD

When Q = K = [C]c x [D]d

[A]a x [B]b Equilibrium occurs

Q < KWHEN

[A] and [B] >>> [C] and [D]Forward rxn proceeds

Q vs K

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Week 9 CHEM 1310 - Sections L and M 17

What is the relationship between Q and K?

aA + bB cC + dD

When Q = K = [C]c x [D]d

[A]a x [B]b Equilibrium occurs

Q > KReverse rxn proceeds

WHEN [C] and [D] >>> [A] and [B]

Q vs K

Week 9 CHEM 1310 - Sections L and M 18

Thus, knowing K and calculating Q for any given statehelps us predict which way a chemical reaction

will proceed!

aA + bB cC + dD

When Q < K reaction proceeds to the right

When Q = K equilibrium occurs

When Q > K reaction proceeds to the left

Q vs K

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Week 9 CHEM 1310 - Sections L and M 19

Equilibrium equations for gaseous reactions can bewritten in terms of concentrations or

partial pressures.

Why? Recall… PV = nRT

P =

P = M (RT)

nV

RT

Pressure is proportional to molar concentration.

Equilibrium Equations for Gases

Week 9 CHEM 1310 - Sections L and M 20

Equilibrium and Partial PressureEquilibrium expressions can be written in terms of the partial pressures of the gases instead of their

molar concentrations

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

K = P2

NH3

PN2

P3H2

x

In text, Kp denotesequilibrium constant

expressed in termsof partial pressures

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Week 9 CHEM 1310 - Sections L and M 21

How can we express the equilibrium constant whenthe reactants and products are in different phases?

Si3N4(s) + 4 O2(g) 3 SiO2(s) + 2 N2O(g)

Rule #1. Express gases as partial pressures

Rule #2. Express solute in solution as molar conc.

Rule #3. Express pure solids/liquids as “1”.

Rule #4. Products multiplied in the numerator

reactants multiplied in the denominator

Heterogeneous Equilibria

Week 9 CHEM 1310 - Sections L and M 22

How can we express the equilibrium constant whenthe reactants and products are in different phases?

Si3N4(s) + 4 O2(g) 3 SiO2(s) + 2 N2O(g)

Rule #1. Express gases as partial pressures

Rule #2. Express solute in solution as molar conc.

Rule #3. Express pure solids/liquids as “1”.

Rule #4. Products multiplied in the numerator

reactants multiplied in the denominator

K =N2O

P2 x 13

O2P4 x 1

K =N2O

P2

O2P4

Heterogeneous Equilibria

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Week 9 CHEM 1310 - Sections L and M 23

• Know how to write equilibrium expressions

• Know how to calculate K and mathematicallymanipulate K

• Be able to calculate Q (via conc or partialpressures) and relate Q to K

• Be able to calculate K for gases in equilibrium

• Know how to express heterogeneous equilibria

What To Study and Know…

Week 9 CHEM 1310 - Sections L and M 24

[1] 3.5 x 1025

[2] 7.0 x 10-25

2 SO2 + O2 2SO3 K = 7.0 x 1025

Calculate K for SO3 SO2 + 0.5 O2

[3] 1.2 x 10-13

[4] 1.4 x 10-26

PRS Question

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Week 9 CHEM 1310 - Sections L and M 25

2 SO2 + O2 2SO3 K = 7.0 x 1025

Calculate K for SO3 SO2 + 0.5 O2

Kreverse = 1

Kforward

To solve this problem:

1st:

2nd: Molar ratio is half, so take the square root of Kreverse

Kreverse = 1.4 x 10-26

K = 1.2 x 10-13 Answer = #3

PRS Question

Week 9 CHEM 1310 - Sections L and M 26

Which reaction will tend to proceed farthest towardcompletion?

[1] H2 + Br2

[2] 2NO

[3] 2BrCl

2 HBr K = 1.4 x 10-21

N2 + O2 K = 2.1 x 1030

Br2 + Cl2 K = 0.145

PRS Question

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Week 9 CHEM 1310 - Sections L and M 27

Which reaction will tend to proceed farthest towardcompletion?

[1] H2 + Br2

[2] 2NO

[3] 2BrCl

2 HBr K = 1.4 x 10-21

N2 + O2 K = 2.1 x 1030

Br2 + Cl2 K = 0.145

K = [Products][Reactants]

PRS Question