Chemical Equilibrium

Dynamic Equilibrium

definition: equilibrium state where forward and reverse reactions are happening simultaneously at equal rates

ZERO change to the concentrations of substances

reactants are consumed at the same rate they are produced

forward reaction rate = reverse reaction rate

Chemical Equilibrium

definition: state where the [reactant] and [product] are constant

ONLY applies to closed chemical systems

Different types of Equilibrium: Phase Equilibrium

H2O(l) ⇌ H2O(g)

system will eventually reach equilibrium

rate of forming water vapor = rate of forming liquid water

overall concentration of liquid water and water vapor is constant

Equilibrium Position

definition: the relative ratio between the amount of [reactant] and the amount of [product] at equilibrium

equilibrium position is unique for every system

for any closed chemical equilibrium system in constant environmental conditions, the same equilibrium concentration are reached regardless of reaction direction

Equilibrium Calculations

ICE table

Initial

Change

Equilibrium

purpose: helps organizes calculations involving equilibrium

concentration will always be measured in molarity (mol/L)

Example: H2(g) + F2(g) ⇌ 2HF(g)
The initial concentration of H2 and F2 are both 2.00 mol/L. There is no HF initially. What is the concentration of H2 and HF at equilibrium if the equilibrium concentration of F2 is 0.48 mol/L?

Answer:
H2(g) F2(g) ⇌ 2HF(g)
I 2.00 2.00 0
C -x -x 2x
E 2.00-x 0.48 2x

[F2]eq = 2.00 - x
0.48 = 2.00 - x
2.00 - 0.48 = x
x = 1.52 mol/L

[H2]eq = 2.00 - 1.52
= 0.48 mol/L

[HF]eq = 2(1.52)
= 3.04 mol/L

Equilibrium Law

definition: mathematical description of a chemical system at description

can be written for any balanced chemical equation

exponents in the equilibrium laws comes from the balanced equation

pure solids and pure liquids are ignored and only concentrations of gases or aqueous solutions are involved

Equilibrium Constant K

definition: a number that defines the equilibrium law for any system

units of K are not important, just the number value

equilibrium constant K is the NUMERICAL VERSIOn of the equilibrium position

the constant K is always the same value, regardless of the initial concentration used

K is specified for a specific temperature

Magnitude

K < 1

equilibrium position is far left (favours the reactants)

more reactant than product

the reaction occurs very little (99% stay as reactant)

K > 1

equilibrium position is far right (favours the products)

more product than reactant

the reaction goes to completion

K = 1

[reactant] = [product]

Homogenous/Heterogeneous Equilibria

Homogenous Equilibrium: equilibrium system where chemical substances are in the SAME state

Heterogenous Equilibrium: equilibrium system where chemical substances are in DIFFERENT states

Writing Equilibrium Laws and Calculating K

In a closed vessel at 500 C, nitrogen and hydrogen combine in a equilibrium system to form ammonia gas. [N2(g)]eq = 1.50 x 10^-5 mol/L [H2(g)]eq =3.45 x 10^-1 mol/L [NH3(g)]eq = 2.00 x 10^-4 mol/L

Write the Equilibrium Law and determine the value of K:

K = [NH3(g)]^2/[N2(g)][H2(g)]^3
= (2.00 x 10^-4)^2/(1.50 x 10^-5)(3.45 x 10^-1)^3
= 0.064939
K = 0.0649

Solubility Equilibrium

definition: dynamic equilibrium between the dissolution (dissolving) and precipitation of an ionic compound in closed system

Solubility Equilibrium Laws are the same as writing an equilibrium law (use the solubility product constant Ksp)

Ksp = [A(aq)]^a[B(aq)]^b

Solubility Product Constant Ksp

constant only used for solubility

product of the [dissolved ions] raised to their stoichiometric coefficients

specific for a single temperature just like the normal K (usually listed at SATP)

units are ignored

In an ice table: solubility is the equilibrium concentration for a dissolved ion in a saturated solution

Ksp is a larger number

the product of the dissolved ion concentrations is very high

large amount of dissolved ions

high solubility

Ksp is a small number

the product of the dissolved ion concentrations is very low

small amount of dissolved ions

low solubility

Solubility Equilibrium Calculations

Example: Calculate the solubility product constant for a saturated solution of solid zinc hydroxide Zn(OH)2(s) at 25 C. The [Zn2+] is 2.7 x 10^-6 mol/L and [OH-] is 5.4 x 10^-6 mol/L

Answer: Ksp = [Zn2+][OH-]^2
= (2.7 x 10^-6)(5.4 x 10^-6)^2
= 7.9 x 10^-17

Predicting Precipitation

to calculate if a precipitate will form, use Q and compare it to Ksp

Q > Ksp

shift to the left

precipitate will form

Q = Ksp

solution is saturated

no precipitate will form

Q < Ksp

shift to the right

precipitate will not form

Predicting Precipitation Calculations

Example: Will a silver bromide precipitate form when 1.0 x 10^-3 mol/L silver nitrate is mixed with 5.0 x 10^-3 mol/L potassium bromide at 25 C? (Ksp AgBr = 5.1 x 10^-13)

Answer: AgBr(s) ⇌ Ag+(aq) + Br-(aq)

Q = [Ag+]{Br-]
= (1.0 x 10^-3)(5.0 x 10^-3)
= 5.0 x 10^-6 > Ksp

rxn move <- and PPT will form

Le Chatelier's Principle

definition: when a chemical system at equilibrium is disturbed by a change or stress, the system will behave in a way to oppose that change

this principle is used my chemical engineers to increase yield

reactants are continually added while products are continually removed

equilibrium will never be reached and the system will always favour the products

Types of Changes

1. Concentration

A(aq) ⇌ B(aq)

Increasing concentration of A
- adding A will shift equilibrium to produce more B (right)

Decreasing concentration of A
- removing A will shift equilibrium to produce more A (left)

2. Temperature

endothermic reactions absorb energy (energy is a reactant)

A(g) + B(g) + heat ⇌ C(g) + D(g)

Increase temp
- shift to right

Decrease temp
- shift to the left

exothermic reactions release energy (energy is a product)

A(g) + B(g) ⇌ C(g) + D(g) + heat

Increase temp
- shift to left

Decrease temp
- shift to right

3. Volume/Pressure

2A(g) + 3B(g) ⇌ 2C(g) + D(g)

Increasing Volume (decreasing pressure)
- shift to left (side w/ most gases)

Decreasing Volume (increasing pressure)
- shift to left (side w/ fewest gases)

Changes that DON'T affect Equilibrium

1. Adding Catalysts

reduce activation energy by introducing an alternative reaction pathway

only help a system reach equilibrium FASTER

2. Adding Inert gases

adding inert gases will increase the total pressure, but not individual partial pressures

even after colliding with other gases, they don't react

3. Adding Pure Solids/Liquids

The Haber-Bosch Process

N2(g) + 3H2(g) ⇌ 2NH3(g)

ammonia can be produced from N2 but the equilibrium heavily favors the reactants at SATP

equillibrium can shift if temp is raised 700 C

Fritz Haber discovered Iron (III) Oxide was an effective catalyst to speed up this reaction

Carl Bosch used this information to synthesize ammonia on an industrial scale

by adding reactant, removing product and using a catalyst (equilibrium is pushed towards the product)

Bosch was able to synthesize ammonia at more manageable conditions (500 C)

measures the relative amounts of products and reactants present during a reaction at a particular point in time

tells you if a chemical system is at equilibrium or not

calculating Q is the same as calculating K

How to use Q?

calculate Q and then compare the value to K

There are 3 possible cases:

Q < K Not at equilibrium, there is too much reactant, reaction will shift RIGHT

Q = K The system is at equilibrium, nothing will happen

Q > K Not at equilibrium, there is too much product, reaction will shift LEFT

1. BE - Balanced Equation

2. GR - Givens + Required

3. Identify if system is at equilibrium or not / which direction reaction will it move
- If one substance is not present, system is NOT at equilibrium and will move in that direction
- If both are present, calculate Q and compare with K

4. ICE - Fill out Ice table using information from Q

5. K - Create Equilibrium Law

6. Solve for X

7. [EQ] Input X back into your equilibrium expressions to determine final concentration at equilibrium

1. Equilibrium law does not contain any x^2

2. Equilibrium law is a perfect square

Example: In a 250 mL sealed container at 150 C, 0.50 mol of iodine gas and bromine gas are mixed and allowed to react until they form equilibrium with IBr. K = 120. What are the equilibrium concentrations of iodine gas and bromine gas?

Answer:
I2(g) + Br2(g) ⇌ 2IBr(g)
I 2.0 2.0 0
C -x -x 2x
E 2.0-x 2.0-x 2x

[I2]ini = 0.50 mol/0.250 L
= 2.0 mol/L
[Br2]ini = 0.50 mol/0.250 L
= 2.0 mol/L
Not @ EQ, shift ->

K = [IBr]^2/[I2][Br]
120 = (2x)^2/(2.0-x)^2
10.95445 = 2x/2.0-x
21.90890 = 2x + 10.95445x
x = 1.691

[IBr]eq = 2x
= 2(1.691)
= 3.4 mol/L
[I2]eq = 2.0-x
= 2.0 - 1.691
= 0.31 mol/L
[Br2]eq = 2.0-x
= 0.31 mol/L

3. Equilibrium law is not a perfect square

100 Rule

NOT a perfect square:
Ex: 8.40 x 10^-6 = x^2/0.200-x
- since K is very small, we make a simplifying approximation
- if the equilibrium concentration is very small, the value of X is small as well
- K with a magnitude of 10^-6 means that the [reactant] is 1 000 000x greater than the [product]
-> so the approximation we can make is 0.200-x = 0.200 (x = 0.000001)
It makes it easier:
8.40 x 10^-6 = x^2/0.200
1.68 x 10^-6 = x^2
x = 1.2961 x 10^-3

BEFORE using the 100 rule:
smallest initial concentration/K
- > 100

AFTER using the 100 rule:
x/initial concentration x 100%
- < 5%

Example: Carbon monoxide gas is a primary starting material in the synthesis of many organic compounds, including methanol. At 2000 C, the K is 6.40 x 10^-7 for the decomposition of carbon dioxide gas, into carbon monoxide and oxygen gas. Calculate all [equilibrium] if 0.250 mol of CO2(g) is placed in a 1.000 L container.

Answer:
2CO2(g) ⇌ 2CO(g) + O2(g)
I 0.250 0 0
C -2x +2x +x
E 0.250-2x 2x x

[CO2]ini = 0.250 mol/1.000 L
= 0.250 mol/L

rxn will move ->

K = [CO]^2[O2]/[CO2]^2
6.40 x 10^-7 = (2x)^2(x)/(0.250-2x)^2
6.40 x 10^-7 = 4x^3/(0.250-2x)^2
6.40 x 10^-7 = 4x^3/(0.250)^2
x = 2.154 x 10^-3

Test:
[smallest]/K = 0.250/6.40 x 10^-7 = 390 000
390 000>100

Check:
x/[initial] x 100% = 2.154 x 10^-3/0.250 x 100% = 0.86%
0.86%<5%

[CO2]eq = 0.250-2(2.154 x 10^-3)
= 0.246 mol/L
[CO]eq = 2(2.154 x 10^-3)
= 4.31 x 10^-3 mol/L
[O2]eq = 2.15 x 10^-3 mol/L

4. Quadratic formula