CHEMICAL EQUILIBRIUM

Up to this point we have mostly been considering reactions “to completion”, where all the reactants change into product.

However, most reactions are reversible = occurs in both the forward and the reverse directions.

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

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

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

orN2(g) + 3H2(g) ↔ 2NH3 (g)

Combined in one equation using double arrows

Chemical Equilibrium

A state in which the forward and reverse reactions balance each other and when the forward reaction proceeds at the same rate as the reverse reaction.

Concentrations of reactants and products are constant at equilibrium.

(Constant ≠ equal)

Rate vs. Time

Question

Can we change the equilibrium position thereby increasing the amount of products in a reaction?

Yes – by adding stress to a system in equilibrium.

LE CHÂTELIER’S PRINCIPLE

Le Châtelier’s Principle

If a stress is applied to a system at equilibrium, the system shifts in the direction that relieves the stress.

Stress is anything that upsets equilibrium – changes in concentration, pressure, or temperature.

Le Châtelier’s Principle

If a stress is applied to a system at equilibrium, the system shifts in the direction that relieves the stress. Concentration Pressure TemperatureNote: Only temperature

affects K. The larger the value of K, the more product at equilibrium.

Concentration

Measure of molarity (moles/L) If you ↑concentration of a

reactant, equilibrium will shift toward the products.

If you ↓concentration of a reactant, equilibrium will shift toward the reactants.

CO(g) + 3H2(g) ↔ CH4(g) + H2O(g)

CO(g) + 3H2(g) ↔ CH4(g) + H2O(g)

CO(g) + 3H2(g) ↔ CH4(g) + H2O(g)

CO(g) + 3H2(g) ↔ CH4(g) + H2O(g)

CO(g)

H2O(g)

Changes in Concentration

A + B <=> C + D

Increasing the concentration of “A” will shift the reaction to the right we need to get rid of excess “A”

Decreasing the concentration of “C” will shift the reaction to the right there is a deficit, so more “C” needs to be

made

Temperature Increasing the temperature shifts

the reaction away from the side that contains the “heat”

Decreasing the temperature shifts the reaction toward the side that contains the “heat”

heat

CO(g) + 3H2(g) ↔ CH4(g) + H2O(g) + heat

CO(g) + 3H2(g) ↔ CH4(g) + H2O(g) + heat

Sample Reaction

heat + NH4Cl (s) <=> NH3 (g) + HCl (g)

In the above endothermic reaction, increasing the temperature will drive the reaction to the right (in other words, forward)

Pressure

Ideal Gas Law: PV=nRT If ↑P then ↑n, which means

more number of atoms. If ↑P, then the equilibrium will

shift toward the side with fewer moles of gas.

CO(g) + 3H2(g) ↔ CH4(g) + H2O(g)

4 moles of gas 2 moles of gas

Note: If moles of gaseous reactant = moles of gaseous product, then no shift in equilibrium will occur from a change in pressure

Equilibrium constant

K is called the equilibrium constant. It is a ratio of the concentrations of products to the concentration of reactants.

Equilibrium Constant Expression

aA + bB ↔ cC + dD

Keq= [C]c[D]d

[A]a[B]b

A & B = molar [ ] of reactantsC & D = molar [ ] of productsExponents a, b, c, and d = coefficients in the balanced equation.

Equilibrium Constant

If Keq > 1: products are favored at equilibrium

If Keq < 1: reactants are favored at equilibrium

Important!

Only substances that are gases and aqueous get factored into the equilibrium expression

Pure liquids and solids do not appear in the expression.

Example #1

Write the equilibrium expression for the following reaction:

2 CO (g) + O2 (g) ↔ 2 CO2 (g)

Answer

Keq = [CO2]2 / ([CO]2[O2])

Example #2

Write the equilibrium expression for the following reaction:

CO (g) + 3 H2 (g) ↔ CH4 (g) + H2O (g)

Answer

Keq = [CH4][H2O] / ([CO][H2]3)

Drill #14 May 19 & 20, 2014 What are the 3 types of stress that can

affect the equilibrium of a system? Which states of matter get factored

into an equilibrium expression? If Keq > 1: are favored at

equilibrium If Keq < 1: are favored at

equilibrium

Drill #16 May 27 & 28, 2014 An equilibrium mixture of N2, O2, and NO

gases at 1500 K is determined to consits of 6.4 x 10-3 mole/L of N2, 1.7 x 10-3 mol/L of O2 and 1.1 x 10-5 mo/L of NO. What is the equilibrium constant for the system at this temperature?

Given: the molarity of the three gases Unknown: K Write the balanced equation Substitute the given values for the [ ]s

into the equilibrium expression.

Answer to Drill

N2 + O2 2NO

K = [NO]2

[N2][O2]

K = (1.1 x 10-5 mo/L)2 (6.4 x 10-3 mole/L )(1.7 x 10-3

mol/L )

K = 1.1 x 10-5 The value of K is < 1 therefore reactants are favored at equilibrium.

Equilibrium Constant

If Keq > 1: products are favored at equilibrium

If Keq < 1: reactants are favored at equilibrium

Temperature• Think of heat as a reactant or a product.

CO(g) + 3H2(g) ↔ CH4(g) + H2O(g) + heat

Is this Exothermic or Endothermic?

Reversible reactions are exothermic (energy as heat shown on product side) in one direction and endothermic (energy as heat shown on reactant side) in the other.

Drill #17 May 28 & 29, 2014

1. Concentration Changed 2A (g) + 3B (g) 2C (g) + D (g)Action Effect?

Increase [A] Shift to the ? Increase [C] Shift to the ? Decrease [B] Shift to the ? Decrease [C] Shift to the ?

Drill #17 May 28 & 29, 2014

1. Concentration Changed 2A (g) + 3B (g) 2C (g) + D (g)Action Effect?

Increase [A] Shift to the

right Increase [C] Shift to the

left Decrease [B] Shift to the

left Decrease [C] Shift to the

right

Drill #17 May 28 & 29, 2014

2. Volume Changed 2A (g) + 3B (g) 2C (g) + D (g)Action Effect?

Volume decreased Shifts to the ? Volume increased Shifts to the ?

Drill #17 May 28 & 29, 2014

2. Volume Changed 2A (g) + 3B (g) 2C (g) + D (g)Action Effect?

Volume decreased Shifts to right (side with

fewest moles) Volume increased Shifts to left (side with

most moles)

Drill #17 May 28 & 29, 2014

3. Temp. Changed for Endothermic Rxn.

A (g) + B (g) C (g) + D (g)Action Effect?

Increase temp Shift to

the ? Decrease temp Shift to

the ?

Drill #17 May 28 & 29, 2014

3. Temp. Changed for Endothermic Rxn.

A (g) + B (g) C (g) + D (g)Action Effect?

Increase temp Shifts to right

(endothermic rxn.) Decrease temp Shifts to left

(exothermic rxn.)

Agenda

Quiz Finish Notes on Ice Table Finish Worksheet and new worksheet on

Ice Table Chemical Energy notes

INTRO TO ICE BOXES

Equilibrium Problems

The equilibrium constant can also be used to calculate equilibrium concentrations from initial concentrations

Use an ICE box! To start with, we will use a generic

equation A(g) + B(g) 2 C(g)

[A] = .100 M and [B] = .100 M

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

A B 2 C

Initial .100 M .100 M 0

Change -x -x +2x

Equilibrium

.100 – x .100 – x 2x

A and B are assigned –x values because they are being reacted, while C gets a +x because products are being

formed. The 2x comes from the coefficient in front of the C

ICE BOX PRACTICE PROBLEMSWe will complete in class examples and then you will try on your own!

Drill #18 June 2 & 3, 2014 What is energy? What are the two different types of

energy?

What is Energy?The ability to do work or produce heat.

Potential Energy – stored energy. Kinetic Energy – energy of motion.

Agenda

Equilibrium Constant for Acids and Bases Worksheets Chemical Energy Diagrams Final Exam Review Project

Strong Acids/Bases vs. Weak Acids/Bases

STRONG Ionize almost

100% in water (react to completion)

WEAK Do not ionize

completely, achieve an equilibrium

HOW CAN I TELL THE DIFFERENCE BETWEEN STRONG AND WEAK ACIDS?

Two different ways:

Study the list of strong acids and bases Look at the arrow in an acid/base

balanced equation If arrow is one direction it is STRONG If arrow is two directions it is WEAK

Introduction

Strong Acid Example HNO3(aq) + H2O(l) H3O+

(aq) + NO3-(aq)

Weak Acid Example CH3COOH(aq) + H2O(l) H3O+

(aq) + CH3COO-(aq)

Strong Base Example: NaOH(aq) + H2O(l) Na+

(aq) + OH-(aq)

Weak Base Example: NH3(aq) + H2O(l) NH4

+(aq) + OH-

(aq)

INSTEAD OF USING KEQ WE USE A DIFFERENT CONSTANT FOR WEAK ACIDS AND BASES

Ionization Constant of a Weak Acid Ka – acid ionization constant Ka varies at different temperatures Generic Example:

Weak Acid Example

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

(aq)

Weak acids have a Ka < 1 Leads to small [H3O+] pH of weak acids is calculated by solving

the expression for [H3O+] at equilibrium and then taking the –log[H3O+]

–3 3

a3

[H O ][CH COO ]K

[CH COOH]

Ionization Constant of a Weak Base Kb – base ionization constant Weak bases = Kb < 1 Leads to a small [OH-] To find pOH take -log[OH-]

Weak Acid Example

HA(aq) + H2O(l) H3O+(aq) + A-

(aq)

Initially, you have 1.00 M HA. Calculate the equilibrium concentrations of HA, H3O+, A-, and calculate the pH. Ka = 1.80 x 10-5

Use an ICE table!

Weak Acid Example

HA H3O+ A-

I 1.00 0 0

C -x +x +x

E 1.00 – x +x +x

Weak Acid Example

Write Ka expression:

Assume x is small because Ka is very small, therefore x can be ignored

Ka = 1.80 x 10-5 = [H3O+][A-] = x2

[HA] 1.00-x

Ka = 1.80 x 10-5 = x2

1.00

Weak Acid Example

x = [H3O+] = [A-] = 4.20 x 10-3 M pH = - log [H3O+] = 2.37

Ka = 1.80 x 10-5 = x2

1.00

Weak Base Example

NH3(aq) + H2O(l) NH4+

(aq) + OH-(aq)

Initially, you have a concentration of 0.010 M NH3. Find the concentrations at equilibrium. Calculate the pH. Kb = 1.80 x 10-5

Use an ICE table!

Weak Base Example

NH3 NH4+ OH-

I 0.010 0 0

C -x +x +x

E 0.010 – x +x +x

Weak Base Example

Write Kb expression:

Assume x is small because Kb is very small, therefore x can be ignored

Kb = 1.80 x 10-5 = [NH4][OH-] = x2

[NH3] 0.010-x

Kb = 1.80 x 10-5 = x2

0.010

Weak Base Example

x = [NH4+] = [OH-] = 4.20 x 10-4 M

pOH = - log [OH-] = 3.37 pH + pOH = 14 Therefore, 14 - 3.37 = 10.6 = pH

Kb = 1.80 x 10-5 = x2

0.010

DIFFERENT TYPES OF EQUILIBRIUM

Concentration Equilibrium Kc (or Keq)

nA + mB ↔ xC + yD

Kc = [C]x[D]y

[A]n[B]m

Remember…equilibrium is where the rates of forward and reverse reactions are the same. It means that the concentrations do not change, NOT that they are identical.

Because equilibrium expressions have to do with concentration (in molarity) we do not include items that are not in solution so NO LIQUID or SOLID states! They are in excess so can be ignored.

Acid Equilibrium

Acid + H2O ↔ H3O+ + Acid Ion

or HA + H2O ↔ H3O+ + A-

Ka = [H3O+][A-] Ka = [H3O+][A-]

[HA][H2O] [HA]

Because water is a solvent (liquid) and its conc. greatly exceeds the acid, we can assume that the conc. of water does not change.

Base Equilibrium

Base + H2O ↔ OH- + Base Ion

or B + H2O ↔ OH- + HB+

Kb = [OH-][HB+] Kb = [OH-][HB+]

[B][H2O] [B]

Acid Strength

A stronger acid will react completely to form ions (strong electrolytes) and hydronium ions (H3O+) in water (dissociation)

Hydrochloric acid (all HCL molecules are ionized into hydronium and chloride ions)

HCl + H2O H3O+ + Cl-

Acetic Acid (not all A.A. molecules are ionized into hydronium and ions)

CH3COOH + H2O H3O+ + CH3COO-

Examples of Strong & Weak Acids Strong Weak

HCl CH3COOH

HBr H2CO3

H2SO4 HClO

Base Strength

A strong base has the strongest affinity for H+ ions and dissociates entirely into metal ions and OH-.

For ex. Calcium oxide (CaO); oxygen strongly attracts H+ ions.

Strong bases: CaO, NaOH, KOH Weak base: ammonia (NH3)

Sample Problem

Write an equation to show the dissociation of hydrochloric acid in water.

Answer

HCl + H2O → H3O+ + Cl−

HCl is a STRONG ACID

Property Strong Acid Weak AcidKa value Ka is large Ka is small

Position of equilibrium

Far to the right (a lot of dissociation)

Far to the left (little dissociation)

[H+] compared to original [HA]

[H+]≈[HA]o [H+]<<[HA]o

Strength of conjugate base

A- is much weaker

A- is much stronger

Graphic Representation of the Behavior of Acids of

Different Strengths in Aqueous Solution

Solubility Equilibrium

Salt (s) ↔ Cation (aq) + Anion (aq)

Solids are not included in equilibrium equations! So…

Ksp = [Cation][Anion]

Solubility Equilibrium Example CaF2 (s) ↔ Ca+2 (aq) + 2F- (aq)

Solids are not included in equilibrium equations! So…

Ksp = [Ca+2][F-]2

Drill #15 May22 & 23, 2014

1. Write the equilibrium expression for the reaction:

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

2. How would the following shift the equilibrium in the equation (forward, reverse or no change):H+ (aq) + Cl- (aq) ↔ HCl (aq) + 10.3 kJ

a) Increasing temperatureb) Increasing pressurec) Adding Cl-

d) Removing HCl

Answers: May22 & 23, 2014

1. Write the equilibrium expression for the reaction:

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

Keq = [HI]2

[H2]

2. How would the following shift the equilibrium in the equation (forward, reverse or no change):

H+ (aq) + Cl- (aq) ↔ HCl (aq) + 10.3 kJa) Increasing temperature (reverse)b) Increasing pressure (forward)c) Adding Cl- (forward)d) Remove HCl (forward)