Chemical Equilibrium

Chapter 13

Chemical EquilibriumThe rate of the forward reaction equal the rate of the reverse reaction.

The state where the concentrations of all reactants and products remain constant with time.

On the molecular level, there is frantic activity. Equilibrium is not static, but is a highly dynamic situation.

Equilibrium & Kinetics

At equilibrium, Rf = Rr

but kf is not necessarily equal to kr.

K= the equilibrium constant

The Law of Mass ActionFor

jA + kB lC + mD

The law of mass action is represented by the equilibrium expression:

Kl m

j k C DA B

Equilibrium Expression

4NH3(g) + 7O2(g) 4NO2(g) + 6H2O(g)

K NO H O

NH O2

2

24 6

34 7

2 SO3(g) 2 SO2(g) + O2(g)

Fe3+(aq) + SCN-(aq) FeSCN2+(aq)

Equilibrium ExpressionsN2(g) + 3H2(g) ---> 2NH3(g)

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

K´ = 1/K do the math!

322

23

HN

NHK

2

3

322'

NH

HNK

If there are two steps for a reaction

K1 X K2 = Overall reaction K

Magnitude of KA K value larger than 1 means that the equilibrium

system contains mostly products -- equilibrium lies far to the right.

A very small K value means the system contains mostly reactants -- equilibrium lies far to the left.

The size of K and the time required to reach equilibrium are not directly related!

Reactions That Appear to Run to Completion

1. Formation of a precipitate.

2. Formation of a gas.

3. Formation of a molecular substance such as water.

These reactions appear to run to completion, but actually the equilibrium lies very far to the right. All reactions in closed vessels reach equilibrium.

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

[N2]= 2.0M

[H2] =3.0

[NH3]=.5

Equilibrium PositionFor a given reaction at a given temperature,

there is only one equilibrium constant (K), but there are an infinite number of equilibrium positions.

Where the equilibrium position lies is determined by the initial concentrations of the reactants and products. The initial concentrations do not affect the equilibrium constant.

Three equilibrium positions but only one equilibrium constant (K).

Heterogeneous Equilibria

. . . are equilibria that involve more than one phase.

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

K = [CO2]

The position of a heterogeneous equilibrium does not depend on the amounts of pure solids or liquids present.

Write the equilibrium expression for the following reactions .

Ba(OH)2(s) Ba+2 (aq) + 2OH- (aq)

HCl(g) + NH3(g) NH4Cl (s)

Kp

K in terms of equilibrium partial pressures:

.

3

2

22

3

HN

NHp PP

PK

K versus Kp

For

jA + kB lC + mD

Kp = K(RT)n

n = sum of coefficients of gaseous products minus sum of coefficients of gaseous reactants.

Calculate the value of K at 25°C if the value for Kp is 1.9x103 for the reaction

2NO(g) + Cl2(g) 2NOCl(g)

K versus Kp

When n = 0 and Kp = K(RT)o, then

Kp = K

K and Kp are equal numerically but do not have the same units.

KC[SO2] = 1.5M [O2] = 1.25M

[SO3] = 3.5M

2SO2(g) + O2 2SO3

KpSO2 = 1.5 atm

O2 = 2 Atm

SO3 = 1.3 Atm

Review book

p306 1,2,5-8,13,14 and20,22

Reaction Quotient

Reaction Quotient is used for concentrations not at equilibrium . It helps to determine the direction of the move toward equilibrium.

The law of mass action is applied with initial concentrations.

Reaction Quotient

H2(g) + F2(g) 2HF(g)

Q HF

H F2 2

02

0 0

Q versus K

Q = K System is at equilibrium.

Q > K System will shift to the left.

Q < K System will shift to the right.

.

Kp= .133 atm at a particular temp for the reaction below

N2O4(g) 2NO 2 (g)

Calculate the reaction Quotient if pN2O4 =. 048 atm and pNO2= .058

What will the system do to reach equilibrium ?

Solving Equilibrium Problems

1. Balance the equation.

2. Write the equilibrium expression.(law of mass action)

3. ICE IT. Initial , Change, Equilibrium

4. Substitute and run the math

4. Check 5% rule

Systems With Small K’sWhen K is small, the change (x) is small

compared to the initial concentration and a simplification can be made for the calculation.

This approximation 5 % rule must be checked to see if it is valid..

ICE

The reaction between nitrogen and oxygen to form nitric oxide has a value for the equilibrium constant at 2000 K of K = 4.1 x 10-4.

If 0.50 moles of N2 and 0.86 mole of O2 are put into a 2.0 L container at 2000 K, what would the equilibrium concentrations of all species be?

N2(g) + O2(g) 2NO(g)

Rbook302

X=3.3x10-3

At 35º C K=1.6x10-5 for the reaction

2NOCl(g) 2NO(g) + Cl2(g)

Calculate the concentrations of all species at equilibrium for the following mixtures

A. 2.0 mol pure NOCl in a 2.0 l flask

Tbp648 sg330 # 51

1.6x10-2

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

Kp= 6.8x10-9 at 100º C

The initial pressure for COCl2 is 1 atm calculate the equilibrium pressures for all species

Tbp648 #54 sq332

X = 8.2x10-5

.125 moles of oxygen gas is added to carbon in a .250 container. The mixture is at equilibrium at 500k Calculate the equilibrium concentration of Carbon monoxide, knowing that K=.086 at 500k

2C(s) + O2(g) 2CO(g)

[.207]=CO

Hypobromous acid, HOBr dissociates in water.

Ka = 2.06 x 10 -9

Calculate [H+] of a solution with originally

1.25M HOBr

5.07x10-5

Le Châtelier’s Principle

. . . if a change is imposed on a system at equilibrium, the position of the equilibrium will shift in a direction that tends to reduce that stress (change).

Le Chatelier’s Principle

If a reactant or product is added to a system at equilibrium, the system will shift away from the added component.

If a reactant or product is removed, the system will shift toward the removed component.

.

Effects of Changes on the System

1. Concentration: The system will shift away from the added component.

2. **** adding a solid or liquid has no effect

2. Temperature: the system will change depending upon the temperature change (treat heat as a reactant or product).

Effects of Changes on the System (continued)

3. Pressure: only with gas

a. Increasing the volume(decrease in pressure) shifts the equilibrium toward the side with more gaseous moles.

b. Decreasing the volume(increasing the pressure) shifts the equilibrium toward the side with fewer gaseous moles.

Addition of inert gas does not affect the equilibrium position.

13_318

Key:

N2

H2

NH3

The system of N2, H2, and NH3 are initially at equilibrium. When the volume is decreased, thesystem shifts to the right -- toward fewer molecules.

Equilibrium Constant, K

For an exothermic reaction, if the temperature increases, K decreases. (Indirect)

For an endothermic reaction, if the temperature increases, K increases.(direct)