Gr.12 - Chemistry Unit 4 PDF

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Chemical Systems & Equilibrium: Dynamic Equilibrium - a balance between forward and reverse processes occurring at the same time Chemical Equilibrium - a chemical equilibrium between reactants and products of the chemical reaction in a closed system Phase Equilibrium - a dynamic equilibrium between different physical states of a pure substance in a closed system Homogenous Equilibrium - one in which all of the reactants and products are present in a single solution Heterogenous Equilibrium - a system whose reactants, products, or both are in more than one phase Ionization - is the process by which an atom or a molecule acquires a negative or positive charge by gaining or losing electrons to form ions Dissociation - the splitting of a molecule into smaller molecules, atoms, or ions, especially by a reversible process Concentrated - high molarity Dilute - low molarity Strong - 100% ionization/dissociation (NO equilibrium) Weak - less than 50% ionization/disassociation (equilibrium form) Equilibrium Constant - a number that expresses the relationship between the amounts of products and reactants present at equilibrium Acid - proton donor Base - proton acceptor Conjugate Acid - contains one more + charge than the base that formed it Conjugate Base - contains one more - charge than the acid that formed it Auto-ionization - the reaction between two water molecules producing a hydronium ion and a hydroxide ion Hydronium Ion - the ion H3O+

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Chemical Equilibrium: Reference Video: Crash Course Chemistry #28 (up to 4:25)

Throughout this unit when we speak of equilibrium reactions, we are usually referring to a closed system: the system can exchange energy, but not matter (no atoms can escape). Equilibrium Overview: ! Up to this point, you have generally been told that reactions occur in one direction only: reactants turn into product or products turn into reactants. This is actually not the case. In many natural chemical reactions, forward and reverse processes are both occurring simultaneously. ! ! ! ! Forward reaction: left-to-right reaction → ! ! ! ! Reverse reaction: right-to-left reaction ← ! The rates of these forward and reverses reactions are dependant on several factors, but mostly depend on the concentration of the substances. Typically when a reaction begins, the forward and reverse reaction rates are not equal. (If you start with only reactants, then the rate of forming the products will be fast due to the high amount of reactants particles available to collide and react. As the concentration of these reactant particles decreases, and the concentration of the product particles increases, the rate of the forward reactions flows, and threat of the reverse reaction speeds up.) ! Eventually a balance is formed between the forward and reverse reactions and the rate of the forward reaction matches the rate of the reverse reaction. When this happens, we say that the system is at equilibrium. Qualitatively, at equilibrium the system will look stable. We will se no observable changes (ie. colour change, gas bubbles forming etc). This is because the number of particles changing into products will equal the number o particles changing into reactants. The systems as a whole will look constant. ! It is important to note that equilibrium does not mean there are equal concentrations of reactants and products (we very rarely see this). Dynamic Equilibrium: ! ! ! ! ! Solubility Equilibrium: ! ! ! ! !

! ! ! !

A balance between forward and reverse processes ! ! occurring at the same rate A dynamic equilibrium between a solute and a solvent ! in a saturated solution in a closed system

! !

dissolving and crystallizing taking place at the same rate

Phase Equilibrium:! ! ! ! !

! !

A dynamic equilibrium between different physical states of !! a pure substance in a closed system evaporation and condensation at the same rates

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Equilibrium Stoichiometry: ! ICE tables (math strategy) I → initial concentrations of reactants and products [2.0mol/L] (square brackets imply concentration) C → change in concentrations E → equilibrium concentrations of reactants and products example 1; ! ! !

H

the reaction begins with 1.00mol/L of H2(g) and F2(g), calculate the concentrations of the H2(g) and HF(g) at equilibrium if the concentration of F2(g) at equilibrium is 0.24mol/L

mol/L

H2(g)

F2(g)

2HI(g)

I

1.00

1.00

0

C

-x

-x

+2x

E

1.00-x 1.00-x

F2(g) 0.24 = 1.00 - x x = 0.76 mol/L

2x

H2(g) 1.00 - x = [H2(g)] 1.00 - 0.76 = [H2(g)] 0.24 mol/L = [H2(g)]

2HF(g) 2 (0.76) = [HF(g))] 1.52 mol/L = [HF(g))]

example 2;

when 4.0mol of NH3(g) is introduced into a 2.0L container and heated to a particular temperature, the amount of ammonia changes to 2.0mol. Determine the equilibrium concentrations of the two entities

mol/L

2NH3(g)

3H2(g)

N2(g)

I

2.0

0

0

C

-2x

+3x

+x

E

2.0-2x

3x

x

[NH3(g) initial] = 4.0mol ÷ 2.0L ! ! = 2.0mol/L [NH3(g) equil] = 2.0mol ÷ 2.0L ! ! = 1.0mol/L

NH3(g) 1.0 = 2.0 - 2x x = 2.0 - 1.0 x = 0.5mol/L

H2(g) [H2(g)] = 3x [H2(g)] = 3 (0.5) [H2(g)] = 1.5mol/L

N2(g) [N2(g)] = x [N2(g)] = 0.5mol/L

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Equilibrium Constant (K): ! When chemical reactions take place in a closed system, forward and reverse reactions occur continuously and the system always contain products and reactants. In previous studies, you used stoichiometry to calculate amounts of reactants and products. In equilibrium systems, equilibrium law is used: ! ! For the general chemical reaction

K = [C]c[D]d ———— [A]a[B]b

A,B,C,D are: chemical entities in gas/aqueous phases (concentration) a,b,c,d are: coefficients in the balanced equations

(products) (reactants)

K is constant called the equilibrium constant. Each chemical system has a unique K value that is dependant on the temperature. ! ! ! ! ! ! ! H Exp. #

Initial concentrations

Equilibrium Concentrations

[H2(g)]

[I2(g)]

[HI(g)]

[H2(g)]

[I2(g)]

[HI(g)]

1

2.000

2.000

0

0.442

0.442

3.116

2

0

0

2.000

0.221

0.221

1.560

3

0

0.010

0.350

0.035

0.045

0.280

Experiment Ratio of Equilibrium Concentrations K = 1 K = 2 K = 3

Value of K

[HI(g)]2 ————— [H2(g)][HI(g)]

(3.116)2 ————— = 49.7 (0.442)(0.442)

49.7

[HI(g)]2 ————— [H2(g)][HI(g)]

(1.560)2 ————— = 49.8 (0.221)(0.221)

49.8

[HI(g)]2 ————— [H2(g)][HI(g)]

(0.280)2 ————— = 49.7 (0.035)(0.045)

49.8

Note the following characteristics of equilibrium law:

- the molar concentrations of the products are always multiplied by one another and written in the

-

numerator, and the molar concentrations of the reactants are always multiplied by each other and written in the denominator the coefficients in the balanced chemical equation are equal to the exponents of the equilibrium expression the concentrations in the equilibrium law expression are the molar concentrations of the entities at equilibrium (mol/L - for gases, this means the moles of gas per litre of space occupied)

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Example 1: ! write the equilibrium law expression for the reaction of nitrogen gas with hydrogen gas to produce ammonia gas in a closed system. !

!

K =

N

[NH3(g)]2 ————— [N2(g)][H2(g)]3

Example 2: ! calculate the value (K) of the equilibrium constant for the reaction of nitrogen gas with hydrogen gas to produce ammonia gas if the equilibrium concentrations are 1.50 x 10-5 mol/L, 3.45 x 10-1 mol/L, and 2.00 x 10-4 mol/L respectively. ! ! ! ! N K =

[NH3(g)]2 ————— [N2(g)][H2(g)]3

** favours

(2.00 x 10-4)2 ——————————— (1.50 x 10-5)(3.45 x 10-1)3

K = 0.065 or 6.49 x 10-2

reactants

Forward & Reverse Reactions: - K represents the equilibrium constant for the forward reaction - Kʹ represents the equilibrium constant for the reverse reaction Forward Reaction Example;

[NH3(g)]2 K = ————— [N2(g)][H2(g)]3

= 6.49 x 10-2

[N2(g)][H2(g)]3 Kʹ = ————— [NH3(g)]2

= 15.4

K = 1! 1 — = —! = 6.49 x 10-2 Kʹ 15.4

! The equilibrium constant of a forward reaction and the equilibrium constant of the reverse reaction are reciprocal quantities

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Equilibrium Law in Chemical Reactions: Limitations of Equilibrium Constants and Precent Reaction Values: - the position of the equilibrium is a measure of the extent to which reactants become products in a closed system - the value of the equilibrium constant, K, depends on the temperature example: ! percent reaction values are dependant on temperature and concentration

K = 4.26 x 108 at 25℃ K = 1.02 x 10-5 at 300℃ K = 8.00 x 10-7 at 400℃

Homogenous Equilibria: - the equilibria in which all entities are in the same phase (all gases or all aqueous solution) - however in some systems reactants and produce are in different phases, this is called heterogenous equilibria

-

Heterogeneous Equilibria: - equilibrium systems can involve (s), (l), (g), (aq), most will be homogenous solutions - equilibria in which reactants and products are in more than one phase example;

[H2(g)]2[O2(g)] K = ————— [H2O(l)]2

= 6.49 x 10-2

- liquid water written in the denominator is a problem because the concentration of -

liquid cannot change; it is fixed and equal to the substances density 1L of liquid water has a mass of 1.00kg = 55.5mols ∴ the [H2O(l)] = 55.5mol/L adding.removing water doesn’t change the concentration adding/ removing H2(g) or O2(g) does change the concentration the [H2O] is a constant , its value is incorporated into the K value shown

example; K[H2O(g)] = [H2(g)]2[O2(g)] K 55.52 = [H2(g)]2[O2(g)] K = [H2(g)]2[O2(g)]

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- the concentrations of entities in a condenses state (s) or (l) are not included as -

variables in the equilibrium law expression, but rather incorporated into the value of the equilibrium constant water vapour is a gas just like H2 or O2, its concentration varies equilibrium law expressions are always written from the net ionic form, balanced with simplest numbers

Magnitude of K: - the magnitude of the equilibrium constant provides a measure of the extent to which the reaction has gone to completion when equilibrium is reached

Equilibrium Constant: Limitations of Equilibrium Constant

- the position of equilibrium is a measure of the extent to which reactants become products in a closed system

- it is important to note that the value of the equilibrium constant, K, depends on temperature !

!

N

!

!

!

K = 4.26 x 108 at 25℃

!

!

!

K = 1.02 x 10-5 at 300℃

!

!

!

K = 8.00 x 108 at 400℃

Generally, any calculation involving an equilibrium law equation must specify a temperature.

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Heterogeneous Equilibria ! equilibrium systems can involve solids, liquids, gases, and aqueous solutions ! ! Homogenous Equilibria - all entities are in the same phase ! ! Heterogeneous Equilibria - reactants and/or products are in more than one phase ! ! most of the systems we will study will be homogenous equilibria. however, in some systems, the reactants and products are in different phases. the concentration of solids, and pure liquids (ie. water) cannot change, their concentrations are fixed and equal to their densities. because these substances have fixed concentrations, their values become part of the value K, and not written in the equilibrium law expression. only substances in the gas or aqueous phases are included, as their concentrations can change. example;

K = [Zn2+(aq)] ———— [Cu2+(aq)]

Magnitude

Explanation

K >> 1

[product] much greater than [reactants]

K=1

[product] and [reactants] are equal

K K

too much product so the system shifts left to produce more reactant

Q K too much product ∴the system will shift left to produce more reactants

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Equilibrium Calculations: example 2; ! carbon monoxide reacts with water vapour to produce carbon dioxide and hydrogen. at 900℃, K is 4.200. calculate the concentration of all entities at equilibrium if 4.000 mol of each entity is initially placed in a 1.000-L closed container. Step 1: write a balanced chemical equation and list known values

V = 1.000L!

K = 4.200!

nall entities = 4.000mol

Step 2: calculate concentrations if necessary !

[all] = 4.000mol ÷ 1.000L [all] = 4.000mol/L

Step 3: calculate Q and determine the direction of shift Q = [CO2(g)][H2(g)]! ! Q = (4.000)(4.000) ——————! ! —————— [CO(g)][H2O(g)] (4.000)(4.000) ! ! ! ! Q = 1.000 ! ! ! ! K = 4.000 K > Q !too much reactant ∴ the system will shift right to produce more product

Step 4: ICE table CO(g)

H2O(g)

CO2(g)

H2(g)

I

4.000

4.000

4.000

4.000

C

-x

-x

+x

+x

E

4.000 - x

4.000 - x

4.000 + x

4.000 + x

Step 5: sub into equilibrium law expression and find x 4.200 = (4.000 + x)(4.000 + x)! ! ————————— (4.000 - x)(4.000 - x)! ! → 8.198 - √4.200 - x = 4.000 + x → 4.198 = 3.049x →→ x ≐ 1.38

√4.200 = √(4.000 + x)2! √4.200 = 4.000 + x →!! ————— → ——— ———— ! √(4.000 - x)2! 1 4.000 - x

Step 6: use x value to solve for the equilibrium concentrations

[CO(g)] = [H2O(g)] = 4.000 - x ! ! ! ! = 4.000 - 1.38! ! = 2 623 mol/L

! !

[CO2(g)] = [H2(g)] = 4.000 + x ! ! = 4.000 + 1.38 = 5 377 mol/L

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The Solubility Product:

! this is a special case of equilibrium where excess solute is in equilibrium with its aqueous solution.

Solubility: the concentration of a saturated solution of a solute in a particular solvent at a particular temperature; a specifi c maximum concentration Ksp: the solubility product constant the value obtained from the equilibrium law applied to a saturated solution copper (I) chloride is dissolved in water. (has a low solubility, so even a small amount results in a saturated solution) ! Balanced Equation: ! !

Solubility Equilibrium Law Expression: ! Ksp = [Cu(aq)+] [Cl(aq)-]

Calculating Solubility Using Ksp Values: example; calculate the Ksp for magnesium fluoride at 20℃, given a solubility of 0.00172 g/100mL !

!

MgF

1. n = m÷M !! ! ! = 0.00172g ÷ 62.31g/mol! = 2.76 x 10-5 mol ! !

! ! !

! ! !

2. 100mL x 1L ! ——! ! ! 1000mL

3. C = 2.76 x 10-5 mol ÷ 0.100mL = 2.76 x 10-4 mol/L ! ! ! ! ! ! ! ! ! ! ! ! 4.

! !

! !

MgF2(g)

Mg2+(aq)

2F-(aq)

I

—

0

0

C

—

+x

+ 2x

E

—

x

x

= 0.1L

5. Ksp = (2.76 x 10-4)(5.52 x 10-4)2 = 8.3 x 10 -17

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Q>K • supersaturated • ∴ precipitate formed

Predicting Precipitation: find Q (trial in product) and compare to K ! example 1; ! ! !

! ! !

! ! !

Q=K • saturated solution • ∴ at equilibrium

Ksp = [Cu+ (aq)][Cl-(aq)] Q = (____)(____) = _____

Q Ksp ∴ supersaturated

Common Ion Effect: - when equilibrium exists in a solution, the equilibrium can be shifted by dissolving into the solution a compound that adds a common ion example; - saturated solution of sodium chloride - in equilibrium with undissolved sodium

- if a few drops of HCL is added, additional crystals of sodium chloride will form - use Le Chatelier’s principal to explain the HCL releases large numbers of !

!

chloride ions into the solution HCl

- those additional ions increase the concentration of chalice ions, shifting the NaCl equilibrium to the left

- causing NaCl to precipitate out - a similar result would happen if a solution containing Na+(aq) ions were added -

instead in this case, the common ion would be the sodium ions

Common Ion Effect - a reduction in the solubility of a salt caused by the presence of another salt having a common ion

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Acids & Bases: - Acids and bases are electrolytes that for aqueous solutions with unique properties: Properties of Acids & Bases

! Acids

Bases

Property

sour

bitter

taste

7 (greater)

Auto-ionization of Water: base ! ! ! ! acid ! ! -14 ! Kw = 1.0 x 10 ! ! ! + ! Kw = [H3O (aq)][OH (aq)] 1.0 x 10-14 = [H+(aq)][OH-(aq)]

conjugate acid ! conjugate base [product ÷ reactant] !

example; ! if we had pure water, what would be the concentration of H+ and OH-? ! ! !

! ! !

1.0 x 10-14 ! = [ x ][ x ] √1.0 x 10-14 ! = √x2 1.0 x 10-7 ! = x

[H+(aq)] = 5.3 x 10 -6! [H+(aq)] = 6.2 x 10 -3! [H+(aq)] = 7.8 x 10 -6!

pH = -log [H+(aq)] ! ! !

pH = 5.28 pH = 2.21 pH = 5.11

Dissolving an Acid - dissolving an acid increases the concentrations Dissolving a Base - dissolving a base increase the concentration of OH-(aq)

- in neutral solutions [H+(aq)] = [OH-(aq)] - in acidic solutions [H+(aq)] > [OH-(aq)] - in basic solutions [H+(aq)] < [OH-(aq)] - since Kw = [H+(aq)][OH-(aq)], if you know one of the concentrations, you can find the other example; ! a 0.15 mol/L solution of HCl is at SATP, calculate the concentration of hydroxide ions. ! !

C = 0.15 mol/L! ! [H +(aq)] = 1.0 x 10-7 mol/L + ! ! Kw = [H (aq)][OH (aq)] ! 1.0 x 10-7 (0.15)x ! ———— = ——— ! ! 6.67 x 10-7 mol/L = x ! 0.15 ! 0.15!! ** strong acid ∴ 100% ionization !

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pH: pH = -log [H+(aq)]

[H+(aq)] = 10-pH

0 - 2 → strong acid ;!3 - 6 → weak acid ;! 8 - 11 → weak base !; 12 - 14 → strong base example; ! a sample of sodium hydrozide of 1.0 mol/L, determine the concentration of H+ ions ! ! NaOH! →! Na+(aq) ! +! OH-(aq) ! c = 1.0 mol/L ! ! ! ! c = 1.0 mol/L ! ! ! Kw = [H+(aq)][OH-(aq)]! ! 1.0 x 10-14 = [H+]1.0 ! ! ! ! ! ! ! ! ———— ———— ! ! ! ! ! ! ! ! [H+] = 1.0 x 10-14 Terms: • strong - 100% ionization/dissociation (NO equilibrium) • weak -...