Chemical Equilibrium

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Properties Of Solution

Ved Siti Zahrah

The scientific approach

Ved Francesco Diamante

Acids and Alkali

Ved william yee

Eng 122 Research Topic

Ved Jeffrey Schmidt

Chemical Equilibrium

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4 Types of EQ Problems

4. Quadratic formula

3. Equilibrium law is not a perfect square

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

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

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

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

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

1. Equilibrium law does not contain any x^2

Strategy for Equilibrium Calculations

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

6. Solve for X

5. K - Create Equilibrium Law

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

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

2. GR - Givens + Required

1. BE - Balanced Equation

Reaction Quotient (Q)

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 product, reaction will shift LEFT

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

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

calculating Q is the same as calculating K

tells you if a chemical system is at equilibrium or not

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

Le Chatelier's Principle

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The Haber-Bosch Process

N2(g) + 3H2(g) ⇌ 2NH3(g)

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

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

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

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

equillibrium can shift if temp is raised 700 C

Changes that DON'T affect Equilibrium

3. Adding Pure Solids/Liquids

2. Adding Inert gases

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

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

1. Adding Catalysts

only help a system reach equilibrium FASTER

reduce activation energy by introducing an alternative reaction pathway

Types of Changes

3. Volume/Pressure

2A(g) + 3B(g) ⇌ 2C(g) + D(g)

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

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

2. Temperature

exothermic reactions release energy (energy is a product)

A(g) + B(g) ⇌ C(g) + D(g) + heat

Decrease temp - shift to right

Increase temp - shift to left

endothermic reactions absorb energy (energy is a reactant)

A(g) + B(g) + heat ⇌ C(g) + D(g)

Decrease temp - shift to the left

Increase temp - shift to right

1. Concentration

A(aq) ⇌ B(aq)

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

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

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

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

reactants are continually added while products are continually removed

this principle is used my chemical engineers to increase yield

Solubility Equilibrium

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

Predicting Precipitation

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

Q < Ksp

precipitate will not form

shift to the right

Q = Ksp

no precipitate will form

solution is saturated

Q > Ksp

precipitate will form

shift to the left

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

Solubility Product Constant Ksp

Ksp is a small number

low solubility

small amount of dissolved ions

the product of the dissolved ion concentrations is very low

Ksp is a larger number

high solubility

large amount of dissolved ions

the product of the dissolved ion concentrations is very high

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

units are ignored

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

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

constant only used for solubility

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

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

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

Equilibrium Law and Constant K

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

Homogenous/Heterogeneous Equilibria

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

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

Equilibrium Constant K

Magnitude

K = 1

[reactant] = [product]

K > 1

the reaction goes to completion

more product than reactant

equilibrium position is far right (favours the products)

K < 1

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

more reactant than product

equilibrium position is far left (favours the reactants)

K is specified for a specific temperature

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

equilibrium constant K is the NUMERICAL VERSIOn of the equilibrium position

units of K are not important, just the number value

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

Equilibrium Law

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

exponents in the equilibrium laws comes from the balanced equation

can be written for any balanced chemical equation

definition: mathematical description of a chemical system at description

Introduction to Equilibrium

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Equilibrium Calculations

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

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

ICE table

purpose: helps organizes calculations involving equilibrium

Equilibrium

Change

Initial

Equilibrium Position

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

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

Different types of Equilibrium: Phase Equilibrium

overall concentration of liquid water and water vapor is constant

rate of forming water vapor = rate of forming liquid water

H2O(l) ⇌ H2O(g)

system will eventually reach equilibrium

ONLY applies to closed chemical systems

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

Dynamic Equilibrium

forward reaction rate = reverse reaction rate

reactants are consumed at the same rate they are produced

ZERO change to the concentrations of substances

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

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