Chemical Equilibrium

Reactant and Product Concentrations at Equilibrium

Reactions are reversible

A + B C + D ( forward)

C + D A + B (reverse)

Initially there is only A and B so only the forward reaction is possible

As C and D build up, the reverse reaction speeds up while the forward reaction slows down.

Eventually the rates are equal

React

ion

Rate

Time

Forward Reaction

Reverse reaction

Equilibrium

What is equal at Equilibrium?

Rates are equal

Concentrations are not.

Rates are determined by concentrations and activation energy.

The concentrations do not change at equilibrium.

or if the reaction is verrrry slooooow.

The units for K

Are determined by the various powers and units of concentrations.

They depend on the reaction.

K is CONSTANT

At any temperature.

Temperature affects rate.

The equilibrium concentrations don’t have to be the same only K.

Equilibrium position is a set of concentrations at equilibrium.

There are an unlimited number.

Calculate K

N2 + 3H2 3NH3

Initial At Equilibrium

[N2]0 =1.000 M [N2] = 0.921M

[H2]0 =1.000 M [H2] = 0.763M

[NH3]0 =0 M [NH3] = 0.157M

Equilibrium and Pressure

Some reactions are gaseous

PV = nRT

P = (n/V)RT

P = CRT

C is a concentration in moles/Liter

C = P/RT

Equilibrium and Pressure

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

Kp = (PSO3)2

(PSO2)2 (PO2)

K = [SO3]2

[SO2]2 [O2]

The Reaction Quotient

Tells you the directing the reaction will go to reach equilibrium

Calculated the same as the equilibrium constant, but for a system not at equilibrium

Q = [Products]coefficient

[Reactants] coefficient

Compare value to equilibrium constant

What Q tells us

If Q<K Not enough products Shift to right

If Q>K Too many products Shift to left

If Q=K system is at equilibrium

Checking the assumption

The rule of thumb is that if the value of X is less than 5% of all the other concentrations, our assumption was valid.

If not we would have had to use the quadratic equation

More on this later.

Our assumption was valid.

Using KC to determine [products] and [reactants] at equilibrium

Only known KC and initial concentrations of reactants

Changes in concentrations (ΔC) of chemical compounds related to STOICHIOMETRIC rations in chemical equation.

Steps for Complex Problems

1) Make an ICE Chart

2) Set up the equilibrium expression

3) Solve for ΔC

Ice Chart

“ICE” Reactants Products

Initial Concentrations (I):

Change in Concentration (ΔC or x):

Equilibrium Concentrations (E):

Example 3:

H2O(g) is present in a rigid container at 25°C with an initial partial pressure of 0.784 atm. What are the partial pressures of H2(g) and O2(g) at equilibrium? (KP = 2.0 x 10 -42)

Example 3: continued

If Keq < 1x10-4, remove the “x or ΔC” value in denominator. ΔC is very small compared to initial

concentration so subtraction would not be a huge difference.

Only works when adding or subtracting ΔC If concentrations or partial pressures are

very small where their magnitude is approximately equal to Keq, CANNOT discount ΔC value.

Example 4:

Sulfur trioxide decomposes to form sulfur dioxide and oxygen at 300°C°. Calculate the concentrations of all chemical compounds at equilibrium with an initial SO3 concentration of 0.100M and KC = 1.6 x 10-10.

Example 5:

0.194 mol of COCl2 comes to equilibrium in a 5.8L container at 25°C (KC = 7.27 x 10-38). Find the equilibrium concentrations of all chemical compounds in the following equation. (Hint: first find the initial [COCl2] )

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

% Ionization

Le’Chatlier’s Principle

Le Chatelier’s Principle

If a stress is applied to a system at equilibrium, the position of the equilibrium will shift to reduce the stress.

3 Types of stress

Change amounts of reactants and/or products

Adding product makes Q>K

Removing reactant makes Q>K

Adding reactant makes Q<K

Removing product makes Q<K

Determine the effect on Q, will tell you the direction of shift

Variables disrupting equilibrium

1) Addition or removal of chemical compounds from the reaction.

2) Pressure

3) Temperature/Heat

Basic Concept

Main concept in chemical equilibrium

Change in a variable that alters the equilibrium of a system (chemical reaction) products a shift in the OPPOSITE direction Reaction shifts to counteract the variable’s influence

Reaction tries to get back to equilibrium state-----SO reaction shifts

Basic Concept (cont.)

What happens to a chemical reaction when equilibrium shifts

When one side of a chemical reaction is stressed, the reaction shifts to the side of LEAST stress !

Using KC to determine [products] and [reactants] at equilibrium

Only known KC and initial concentrations of reactants

Changes in concentrations (ΔC) of chemical compounds related to STOICHIOMETRIC rations in chemical equation.

Change Pressure

By changing volume

System will move in the direction that has the least moles of gas.

Because partial pressures (and concentrations) change a new equilibrium must be reached.

System tries to minimize the moles of gas.

Change in Pressure

By adding an inert gas

Partial pressures of reactants and product are not changed

No effect on equilibrium position

Change in Temperature

Affects the rates of both the forward and reverse reactions.

Doesn’t just change the equilibrium position, changes the equilibrium constant.

The direction of the shift depends on whether it is exo- or endothermic

Example 5:

Predict the influence of certain changes to the following chemical reaction. 2SO3(g) 2SO2(g) + O2(g) ΔH°= 197.84 kJ/mol

A) increasing reaction’s temperature

B) increasing reaction’s pressure

C) adding more O2 at equilibrium

D) Removing O2 at equilibrium

Free Energy and Equilibrium

ΔG° ΔG

Concentration and partial pressures can change for reactants and products.

ΔG = ΔG° + RTlnQ

What happens when we leave the standard state conditons?

At equilibrium, ΔG = 0, so reaction quotient (Q) = equilibrium constant (K)

At equilibrium ΔG° = - RTlnK

Enables the reaction’s equilibrium constant (K) to be calculated from the change in free energy (ΔG°)

ΔG = ΔG° + RTlnQ

The magnitude of ΔG° indicates how far the chemical reaction in its standard state is from equilibrium. ΔG° = 0 , equilibrium ΔG°= large value, far from equilibrium ΔG° = small value, close to equilibrium

The sign (+, - ) indicates which direction the reaction needs to shift to achieve equilibrium Positive (+) -------- shift to left, no reaction Negative (-) -------- shift to right, reaction goes to completion

What is the relationship between free energy(ΔG) and K?

Calculate the equilibrium constant for the following reaction given ΔG° = -33 kJ/mol.

N2 + 3H2 2NH3

Example 1

Weak Acid/Base Calculations

Reaction between H2O and Acid

HA + H2O H3O+ + A-

Write the equilibrium expression

Acid Dissociation Constant (Ka)

Quantifies how much an ACID dissociates in water

Describes the acid strength

Ka = [H3O+] [A-]

[HA]

Acid Dissociation Constant

(Ka) Strong Acids > 1

Weak Acids <<<< 1

Examples: HCl Ka = 1x106

CH3CO2H Ka = 1.8 x 10-5

Base Ionization Constant (Kb)

Quantifies how much a BASE dissociates in water

Describes base strength

B + H2O BH+ + OH-

Kb = [BH+] [OH-]

[B]

Other helpful equations

pKa = -logKa

pKb = -logKb

Weak Acid pH Calculations

Weak acids only partially dissociate

Calculating: Use ICE method Compare reaction quotient (Qa) to equilibrium

constant (Ka)----- % Ionization When can we ignore change in initial acid

concentration (ΔC) ?

When can we ignore change in initial acid concentration (ΔC)?

If ΔC < 5% of initial acid concentration

Ka < 1x10-4

THEN, ΔC can be disregarded for equilibrium acid concentration

Example 1:

Determine the H3O+ ion concentration and pH of a 1.0M acetic acid solution with a Ka = 1.8 x 10-5

Example 2:

Calculate the pH of a 0.10M NH3 solution with a

pKb of 4.74

Ka and Kb

MUST determine Kb from Ka value

Use relationship between an acid and its conjugate base

KaKb = Kw

Example 5:

Find the pH of a 3.5M NH3 solution with a Kb value of 1.8 x 10-5

Acid/Base Mixtures

NaOAc

HOAc

Now we would have 2 sources for OAc - ion 1) Acetic Acid Dissociation

2) Salt Hydrolysis/Dissociation

**Acetate ion is COMMON to both reactions**

What would happen to the pH if we added….

Common ion Ion present in both acid/base dissociation and

salt ionization

Common ion effect Seen with weak acids and bases Hinders the ionization of a weak acid or base due

to the presence of a common ion between the acid/base and a salt

Common Ion Effect

HF + H2O(l) H3O+ + F-

Addition of KF

KF K+ + F-

Result is less [H3O+] so pH ----more BASIC

Example 1:

NH3 + H2O(l) OH- + NH4+

Addition of NH4Cl

NH4Cl NH4+ + Cl-

Result is less [OH-] so pH ----more ACIDIC

Example 2:

450ml of 1.0M HCO2H is mixed with 250ml of 0.40M NaHCO2 at 25°C. Find the pH of the solution (Ka = 1.8 x 10-4).

Example 3:

Salt Hydrolysis

Salts

Ionic compound made up of CATION and ANION

Has acidic and basic properties Based on ions produced when salts dissociate No acid/base properties—group I/II cations

(ex. Na+, Li+, K+, Ca+2) No basic properties—conjugate bases from monoprotic

acids (ex. Cl-, Br-, NO3-)

Ex. NaCl, CaBr2

1. Salt Formation from Strong Base and Weak

Acid Salt forms a BASIC solution.

Conjugate base ion reacts with water to give hydroxide (OH-) ions.

Ex. Potassium fluoride (KF) KF K+ + F-

F- + H2O HF + OH-

2. Salt Formation from a Strong Acid and Weak

Base Salt forms an ACIDIC solution

Conjugate acid reactions with water to give hydronium ion (H3O+)

Ex. Ammonium nitrate (NH4NO3)

3. Salt Formation from Strong Acid and Strong

Base Salt forms a NEUTRAL solution

Conjugate base resulting from salt dissociation is weak

Ex. Sodium chloride (NaCl)

Example 1:

Calculate the concentration of HOAc, OAc- and OH- at equilibrium in a 0.10M NaOAc solution (Ka for HOAc = 1.8 x 10-5).

Acid-Base Titrations

Titration

Lab technique commonly utilized to determine an UNKNOWN concentration of a chemical compound with a KNOWN concentration of another chemical compound.

Chemical compounds combine with exact stoichiometric proportions

Analyte— Chemical compound with unknown concentration

Titrant— Chemical compound with known concentration Measured with volume and concentration Added to chemical compound with unknown concentration

in titration

2 Types of Acid-Base Titrations

1) Strong Acid/Strong Base Titrations

2) Weak Acid/Strong Base Titrations

1. Strong Acid/Strong Base Titrations

Low initial pH value

Sharp increase in pH before equivalence point

Equivalence point is pH = 7

Rapid pH increase after equivalence point

**Indicators with pH range 4-10 helpful for these titrations

**Neutralization reactions

Strong Acid with Strong Base Titrant

Strong Base with Strong Acid Titrant

2. Weak Acid/Strong Base Titrations

High initial pH value

pH = pKa at half-neutralization [weak acid] = [conjugate base]

Ka = [H3O+] [A -] SO Ka = [A-]/[HA] is 1:1

[HA]

Ka = [H3O+], SO pH = pKa

2. Weak Acid/Strong Base Titrations

Equivalence point > 7 on pH scale

**Indicators with pH range > 7 helpful as pH equivalence point is basic

Example 1: Weak Acid/Strong Base Titration Calculations

30 ml of 0.5M HC2H3O2 titrated with 0.50M NaOH. Ka for acetic acid is 1.8x10 -5.

a) Find the initial pH of 0.5M acetic acid.

HC2H3O2 + OH- C2H3O2- + H2O (titration

view)

HC2H3O2 H+ + C2H3O2- (in detail)

Example 1: Weak Acid/Strong Base Titration Calculations

b) Find the pH after 15ml of NaOH were added

pH = pKa + log [C2H3O2-] / [HC2H3O2]

-used when dealing with buffer solutions

-buffer—mixture of weak acid/conjugate base

-more on this concept later

Example 1: Weak Acid/Strong Base Titration Calculations

c) Find the pH at the equivalence point.

C2H3O2- + H2O HC2H3O2 + OH-

Buffers

Weak acid/conjugate base mixtures OR weak base/conjugate acid mixtures

“buffers” or reduces the affect of a change in the pH of a solution Absorbs slight changes in pH resulting from the

addition of small acid/base amounts to water.

Buffers

HOAc + H2O(l) H3O+ + OAc –

What happens if an acid is added??? Reacts with OAc ion [HOAc] increases slight, [OAc] decreases

slightly, ratio mostly the same No pH change

Example 1: (cont.)

HOAc + H2O(l) H3O+ + OAc –

What happens if a base is added??? Reacts with HOAc ion More OAc ion formed, removes excess

OH- from solution No pH change

Example 1: (cont.)

1) Acidic Buffers Formed from mixing a weak acid and its conjugate

base pH < 7 Ex. HOAc and OAc –

2) Basic Buffers Formed from mixing a weak base and its conjugate

acid pH > 7 Ex. NH3 and NH4

+

Types of buffers

Compare Ka and Kb from acid-conjugate base pair

Ka > Kb, (generally > 1x10-7) -------- acidic buffer

Ka < Kb, (generally > 1x10-7) -------- basic buffer

How do we tell acidic vs. basic buffers?

1) Method applying “Common Ion Effect”

2) Henderson-Hasselbalch equation

Calculating the pH of buffers

Easier method

pH = pKa + log[A-]/[HA]

pOH = pKb + log[HB+]/[B] [B] = molarity of weak base [HB+] = molarity of conjugate acid

Assumption: weak acids and conjugate bases do NOT change concentration with equilibrium.

Henderson-Hasselbalch Equation

Find the pH of a buffered solution created by mixing 0.15mol NH4NO3 with 0.65L of a 0.25M NH3 solution. Assume that the volume change is negligible. (Kb = 1.8x10-5)

Example 1: