EQUILIBRIUM

Quantitative vs. Reversible

• So far we have studied quantitative reactions, reactions which proceed until the limiting reactant is consumed

• However, many reactions are not quantitative, they are reversible meaning they can proceed in both forward and reverse directions

• A + B C + D

)(2)(2

)(3)(2)(2 23

gl

ggg

OHOH

NHHN

Dynamic Equilibrium• When the rate of the forward reaction and the rate of the reverse reaction are equal, the system is in dynamic equilibrium

• This does NOT mean that nothing is happening!

Properties of Equilibrium Systems

• 1. Macroscopic properties (e.g. colour, pressure, concentration, pH) are constant• I.e. It appears as if nothing is changing

• 2. Can only be reached in a closed system• 3. Forward rate = Reverse rate• 4. Can be established from either direction

At equilibrium, the concentrations of all reactants and products will remain constant

Types of Equilibrium• Solubility Equilibrium: A dynamic

equilibrium between a solute and solvent

in a saturated solution in a closed system

• Phase Equilibrium: a dynamic equilibrium between different physical states of a pure substance in a closed system• Ex: water/liquid in a sealed container

• Chemical Reaction equilibrium: a dynamic equilibrium between reactants and products of a chemical reaction in a closed system

PracticeSketch a graph of Concentration of Reactants and products vs. time for Experiment 1 and 2

More Practice

1. Consider the reaction: • Hydrogen gas and hydrogen iodide gas are both clear

colourless, while iodine gas is purple.• Explain why when hydrogen gas and iodine gas are

added to a closed flask and allowed to react, the colour never fades to colourless.

2. Sketch a graph to illustrate the changes in concentration of each component over time as equilibrium is reached for Experiment 1

• 3. If experiment 1 did not reach equilibrium, what would the theoretical yield of HI be?

• 4. What is the actual yield of HI in experiment 1?

Solving Equilibrium Problems• Use ICE Tables• I: Initial Concentration (mol/L)• C: Change in concentration

• +x if increasing in concentration• -x if decreasing in concentration• Multiply x by the # of moles

• E: Equilibrium Concentration (mol/L)

Practice #1• In a 4.00L container, 2.50 mol of carbon dioxide gas is

decomposed. At equilibrium, [CO2(g)]eq = 0.125 mol/L. Use an ICE table to find [O2(g)]eq and [CO(g)]eq

Practice #2

• 0.500 mol of NOCl(g) is decomposed in a closed 2.00 L container. [NO(g)]eq = 0.040 mol/L. Find [Cl2(g)]eq and [NOCl(g)]eq.

Practice #3• When 2.00 mol of ethene gas and 1.50 mol of bromine

vapour come to equilibrium in a closed 1.00 L container, the equilibrium concentration of bromine vapour is measured at 0.150 mol/L. Find [C2H4(g)]eq and [C2H4Br2(g)]eq.

Practice #3 Continued• Graph the equilibrium reaction for the addition of bromine

to ethene.

At equilibrium, the rate of the forward and reverse reactions are equal

Eq

uilib

rium

For the reaction:

aA + bB cC + dD

At equilibrium:

rfwd = rrev

kfwd[A]a[B]b = krev[C]c[D]d

kfwd = [C]c[D]d

krev [A]a[B]b

Keq = [C]c[D]d

[A]a[B]b

The Equilibrium Constant, Keq

Sub in rate law equation

Rearrange

Sub in Keq for

kfwd/krevEquilibrium constant

Keq Conditions

• Note: the equilibrium constant expression only works with reactions that occur in a single step

• Keq will remain the same as long as the temperature is kept constant (changing the temperature changes the forward and reverse reactions by different amounts and therefore change the Keq)

Examples

)(3)(2)(2

)(42)(2

23)

2)

ggg

gg

NHHNb

ONNOa

Write the equilibrium constant expression for the following two reactions

)(3)(2)(2

)(42)(2

23)

2)

ggg

gg

NHHNb

ONNOa

Heterogeneous Equilibria• Equilibrium systems can involve all states of matter• However, the concentration of a pure solid or liquid cannot change

• Therefore, equilibrium constant expressions will not include solids and liquids

)()(3)(4

)(2)(2

)

)

ggs

gl

HClNHClNHb

OHOHa

)()(3)(4

)(2)(2

)

)

ggs

gl

HClNHClNHb

OHOHa

Magnitude of Keq

Keq >> 1 At equilibrium there is more products than reactants. The reaction is product favoured

Keq = 1 At equilibrium there is an equal amount of products and reactants

Keq << 1 At equilibrium there is more reactants than products. The reaction is reactant favoured

Keq = [C]c[D]d

[A]a[B]b

ExampleCalculate the value of Keq for the following system

)(2)()(2)(2 gggg OHCOHCO

At Equilibrium:

[CO2] = 0.0954 mol/L

[H2] = 0.0454 mol/L

[CO] = [H2O] = 0.00460 mol/L

Practice

For each reaction and their respective equilibrium constant, predict whether reactants or products are favoured.

• N2(g) + O2(g) 2NO(g) Keq = 4.7 x 10-31

• NO(g) + CO(g) 2N2(g) + CO2(g) Keq = 2.2 x 1059

Le Chatelier’s Principle

• When a chemical system at equilibrium is disturbed by a stress, the system adjusts (shifts) to oppose the change

• Stresses include:• Change in concentration• Change in pressure (or volume)• Change in temperature

Change in Concentration

• Increasing the concentration of the reactants OR• Decreasing the concentration of the products• Will favour the forward reaction, causing the equilibrium to

shift to the RIGHT

• Decreasing the concentration of the reactants OR• Increasing the concentration of the products• Will favour the reverse reaction, causing the equilibrium to

shift to the LEFT

• RECALL: Addition or removal of solid or liquids does not change the concentration. Therefore does not cause a shift. I.e. only applies to gases and aqueous solutions.

A(g) + 3B(g) 2C(g) + heat

Change in Concentration

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

Change in Pressure

• Increasing the volume of the container OR Decreasing the pressure

• Will cause a shift to the side with MORE gas molecules• In our example, it will shift left (4 molreactants > 2 molproducts)

• Decreasing the volume of the container OR Increasing the pressure

• Will cause a shift to the side with LESS gas molecules• In our example, it will shift right (4 molreactants > 2 molproducts)

A(g) + 3B(g) 2C(g) + heat

volume pressure volume pressure

Change in Temperature

In an exothermic reaction:

• Increasing the temperature will cause a shift to the LEFT

• Decreasing the temperature will cause a shift to the RIGHT

Change in Temperature

In an endothermic reaction:

• Increasing the temperature will cause a shift to the RIGHT

• Decreasing the temperature will cause a shift to the LEFT

Change in Temperature

Recall: Keq is temperature dependent. Therefore, changes in temperature will also affect Keq

Shift right = products, Keq

Shift left = reactants, Keq

DEMONSTRATION

Variables that do NOT Affect Equilibrium• Catalysts

• Increases reaction rate by lowering activation energy (of BOTH the forward and the reverse reactions equally)

• Decreases the time required to reach equilibrium but does not affect the final position of equilibrium

• Inert Gases• Increases the pressure, which will increase reaction rate• Increases the probability of successful collisions for BOTH

products and reactants equally • Decreases the time required to reach equilibrium but does

not affect the final position of equilibrium

Practice

The Reaction Quotient (Q)• If a chemical system begins with reactants only, it is

obvious that the reaction will shift right (to form products).• However, if BOTH reactants and products are present

initially, how can we tell which direction the reaction will proceed?

• Use a trial value called the reaction quotient, Q• When a reaction is NOT at equilibrium

• Q=Keq the system is at equilibrium• Q > Keq the system shifts towards reactants to reach equilibrium• Q < Keq the system shifts towards products to reach equilibrium

(p. 464) In a container at 450°C, N2 and H2 react to produce NH3. K = 0.064. When the system is analysed, [N2] = 4.0 mol/L, [H2] = 2.0 X 10-2 mol/L, and [NH3] = 2.2 X 10-4 mol/L. Is the system at equilibrium, if not, predict the direction in which the reaction will proceed.

)(3)(2)(2 23 ggg NHHN

3

32

24

3)(2)(2

2)(3

105.1

)100.2)(0.4(

)102.2(

]][[

][

gg

g

HN

NHQ

rightshift willsystem The

064.00015.0

KQ

Practice #1

Practice #2

In a container, carbon monoxide and water vapour are producing carbon dioxide and hydrogen at 900oC.

CO(g) + H2O(g) H2(g) + CO2(g) Keq = 4.00 at 900oC

If the concentrations at one point in the reaction are: [CO(g)] = 4.00 mol/L, [H2O(g)] = 2.00 mol/L, [CO2(g)] = 4.00 mol/L, and [H2(g)] = 2.00 mol/L. Determine whether the reaction has reached equilibrium, and, if not, in which direction it will proceed to establish equilibrium.

Practice #2 Answer

products

Practice #3Calculating Equilibrium Concentrations from Initial Concentrations

• Carbon monoxide reacts with water vapour to produce carbon dioxide and hydrogen. At 900oC, Keq is 4.200. Calculate the concentrations of all entities at equilibrium if 4.000 mol of each entity are initially place in a 1.00L closed container.

Practice #4Calculating Equilibrium Concentrations Involving a Quadratic Equation

• If 0.50 mol of N2O4 is placed in a 1.0L closed container at 150oC, what will be the concentrations of N2O4 and NO2 at equilibrium? (Keq = 4.50)

Practice #5

Simplifying Assumption: 100 rule (for small K values)

If: [initial reactant] > 100, you can simplify the Keq expression

K

Ex: 2CO2(g) 2CO(g) + O2 (g)

If K = 6.40 x 10-7, determine the concentrations of all

substances at equilibrium if it starts with [CO2] = 0.250 mol/L

Solubility Equilibrium• Not all ionic compounds are equally soluble• Ionic compounds dissociate into individual ions in aqueous

solutions• This can be a reversible system• Example: CaCl2(s) Ca2+(aq) + 2Cl-(aq)

• Equilibrium can be reached between the solid substance and its dissolved ions (saturation point)• The solution is saturated at equilibrium (no more ions can dissociate)

• An equilibrium equation can be written for solubility reactions

• Ex: AgCl (s) Ag+ (aq) + Cl- (aq)

Solubility Product Constant (Ksp)

Recall: Since AgCl is a solid, the concentration is not changing, so it is “built in” to the K value:

• The new constant is the solubility product constant (Ksp)

Example• Eg: Lead (II) chloride has a molar solubility of 1.62x102

mol/L at 25oC. What is the Ksp of this salt?

PbCl2 Pb2+ + 2Cl-

Ksp = [Pb2+][Cl-]2

[Pb2+] = [PbCl2] = 1.62x10-2mol/L

[Cl-] = 2[PbCl2] = 2(1.62x10-2mol/L) = 3.24 x 10-2mol/L

Ksp = [1.62x10-2mol/L][3.24 x 10-2mol/L]2

= 1.7x10-5

Example 2• The Ksp of silver chloride at 25oC is 1.8x10-10. What is the molar

solubility of AgCl?

AgCl Ag+ + Cl-

AgCl Ag+ Cl-

Initial - 0 0

Change - + X + X

Equilibrium - X X

Ksp = [Ag+][Cl-]1.8x10-10 = [X][X]1.8x10-10 = X2

X = 1.34x10-5M

• The size of Ksp depends on the solubility of the salt. • Large Ksp: [ions] at equilibrium is high, salt is very soluble• Small Ksp: [ions] at equilibrium is low, salt has low solubility

• To determine whether a precipitate will form during a reaction, a trial solubility product constant can be determine which is denoted by the symbol Qsp.

Qsp < Ksp : Shifts right to equilibrium – all solid dissolving

Qsp > Ksp : Shifts left to equilibrium – precipitate forms

Qsp = Ksp : Equilibrium (saturated) – no precipitate

Example 1• Will a ppt form if a solution of CaSO4 is made such that

[Ca2+] = 0.0104 mol/L and [SO42-]=0.0082 mol/L?

• Ksp =7.1x10-5

• (Ans: Q = 8.5 X 10-5, ppt forms)

Example 2• Will a precipitate form when a solution is made by

dissolving PbCl2 into water. • [Pb2+]=0.0001mol/L. [Cl-]=0.05mol/L. Ksp = 1.2x10-5