Exam #2 Results  Class Average = 77 (Great Job!)

Week 9

CHEM 1310 - Sections L and M

1

Exam #2 Results  4 Perfect Scores!

Week 9

CHEM 1310 - Sections L and M

2

1

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

3

What is Equilibrium?

Equilibrium is the phenomenon that occurs when the rate of the forward reaction equals the rate of the reverse reaction. Week 9

CHEM 1310 - Sections L and M

4

2

What is Equilibrium? At equilibrium, the concentrations of all reactants and products remain constant with time. Example Forward Rxn

H2(g) + I2(g)

2 HI(g) Reverse Rxn

Forward Rxn: Product = HI Reverse Rxn: Products = H2 and I2 [HI] = constant; [H2] = constant; [I2] = constant Week 9

CHEM 1310 - Sections L and M

5

What is Equilibrium?

Notice how the concentrations of products for the forward and reverse reactions are not necessarily equal at equilibrium! Week 9

CHEM 1310 - Sections L and M

6

3

Equilibrium Characteristics What does equilibrium look like in a chemical system? N2O4 colorless

2 NO2 brown Equilibrium

Closed system reaches the same equilibrium concentrations whether the reaction starts with the N2O4 or the NO2! Week 9

CHEM 1310 - Sections L and M

7

The Equilibrium Constant

aA + bB Equilibrium Constant

cC + dD 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.

Week 9

CHEM 1310 - Sections L and M

8

4

The Equilibrium Constant

cC + dD Equilibrium Constant

aA + bB K=

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

Characteristics Reversing the reactants and products inverts the equilibrium expression. Thus, Kforward = Week 9

1 Kreverse

CHEM 1310 - Sections L and M

9

Law of Mass Action K is constant despite different initial and equilibrium concentrations of reactants and products!

Week 9

CHEM 1310 - Sections L and M

10

5

Manipulation of Equilibrium Eqns Equilibrium equations can be reversed, scaled or combined. Forward: K1

aA + bB

cC + dD

K1 =

Reverse: K2

cC + dD

aA + bB

K2 =

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

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

By defn: K1 x K2 = 1 Week 9

CHEM 1310 - Sections L and M

11

Manipulation of Equilibrium Eqns Equilibrium equations can be reversed, scaled or combined.

PCl3 + Cl2

K

PCl5

K1 =

Scaled: 2 PCl3 + 2 Cl2

K

2 PCl5

When stoichiometry is scaled, the resulting K is raised to the power of the scale factor Week 9

CHEM 1310 - Sections L and M

K2 =

[PCl5] [PCl3] x [Cl2]

[PCl5]

2

2

[PCl3] x [Cl2]

2

Example K2 = (K1)2 12

6

Subtraction of Equilibrium Eqns Equilibrium equations can be reversed, scaled or combined. Equation #1

K1

Equation #1

K1

- Equation #2

K2

+ Equation #2

K2

Equation #3

K3 = K1 K2

Equation #3

Subtraction

Week 9

K3 = K1 x K2

Addition

CHEM 1310 - Sections L and M

13

K and the Reaction Quotient, Q What is the relationship between Q and K? aA + bB

cC + dD

Reaction Quotient Q=

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

vs.

Equilibrium Constant K=

Holds whether at equilibrium or not! Week 9

CHEM 1310 - Sections L and M

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

Holds at equilibrium only! 14

7

Q vs K 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

Week 9

CHEM 1310 - Sections L and M

15

Q vs K What is the relationship between Q and K? aA + bB

When Q = K =

cC + dD

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

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

Week 9

Q>> [A] and [B]

Week 9

Equilibrium occurs

Reverse rxn proceeds

Q>K

CHEM 1310 - Sections L and M

17

Q vs K 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

Thus, knowing K and calculating Q for any given state helps us predict which way a chemical reaction will proceed! Week 9

CHEM 1310 - Sections L and M

18

9

Equilibrium Equations for Gases Equilibrium equations for gaseous reactions can be written in terms of concentrations or partial pressures. Why?

Recall…

PV = nRT P=

n V

P=

M (RT)

RT

Pressure is proportional to molar concentration. Week 9

CHEM 1310 - Sections L and M

19

Equilibrium and Partial Pressure Equilibrium expressions can be written in terms of the partial pressures of the gases instead of their molar concentrations N2(g) + 3H2(g)

K=

2 PNH

3

P

N2

Week 9

2NH3(g)

x

PH3

2

In text, Kp denotes equilibrium constant expressed in terms of partial pressures

CHEM 1310 - Sections L and M

20

10

Heterogeneous Equilibria How can we express the equilibrium constant when the 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 Week 9

CHEM 1310 - Sections L and M

21

Heterogeneous Equilibria How can we express the equilibrium constant when the 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.

K=

x 13

P4

x 1

N2O O2

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

P2

Rule #4. Products multiplied in the numerator reactants multiplied in the denominator

P2

K=

N2O

P4

O2

Week 9

CHEM 1310 - Sections L and M

22

11

What To Study and Know… •

Know how to write equilibrium expressions



Know how to calculate K and mathematically manipulate K



Be able to calculate Q (via conc or partial pressures) and relate Q to K



Be able to calculate K for gases in equilibrium



Know how to express heterogeneous equilibria

Week 9

CHEM 1310 - Sections L and M

23

PRS Question 2 SO2 + O2

2SO3

Calculate K for SO3

K = 7.0 x 1025 SO2 + 0.5 O2

[1]

3.5 x 1025

[3]

1.2 x 10-13

[2]

7.0 x 10-25

[4]

1.4 x 10-26

Week 9

CHEM 1310 - Sections L and M

24

12

PRS Question 2 SO2 + O2

2SO3

Calculate K for SO3

K = 7.0 x 1025 SO2 + 0.5 O2

To solve this problem: 1st: Kreverse =

1

Kreverse = 1.4 x 10-26

Kforward 2nd: Molar ratio is half, so take the square root of Kreverse K = 1.2 x 10-13 Week 9

Answer = #3 CHEM 1310 - Sections L and M

25

PRS Question Which reaction will tend to proceed farthest toward completion? [1]

H2 + Br2

[2]

2NO

N2 + O2

K = 2.1 x 1030

[3]

2BrCl

Br2 + Cl2

K = 0.145

Week 9

2 HBr

CHEM 1310 - Sections L and M

K = 1.4 x 10-21

26

13

PRS Question Which reaction will tend to proceed farthest toward completion? [1]

H2 + Br2

[2]

2NO

N2 + O2

K = 2.1 x 1030

[3]

2BrCl

Br2 + Cl2

K = 0.145

K=

2 HBr

K = 1.4 x 10-21

[Products] [Reactants]

Week 9

CHEM 1310 - Sections L and M

27

14