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Gr. 12 Chemical Systems and Equilibrium Cheat Sheet by


Equil­ibrium = the point in a chemical reaction where the reactants and the products are formed and broken at the same rate
Dynamic equili­brium = a balance between the forward and backward rates that are occu­rring simult­ane­ously
Equil­ibrium law = math­ema­tical descri­ption of a chemical system at equili­brium
Equil­ibrium constant (K or Keq) = the numerical value defining the equili­brium law for a system at a given temper­ature (changes with temper­ature)
Heter­oge­neous equili­brium = products and reactants are in at least 2 different states; pure solids­/li­quids are not included in Keq formula

Equili­brium Constant (Keq)

Keq formula:
If a[A] + b[B] ⇌ c[C] + d[D]; then
Keq = ([C][D­]d) ÷ ([A][B­]b)
Magnitude of Keq: states whether the equi­librium position favours produc­ts/­rea­cta­nts
If K = 1
[products] = [react­ants]
If K1
[products] [react­ants]
If K1
[products] [react­ants]
Kforward vs Kback­ward
If a[A] ⇌ b[B], then
Kfo­rward = ([B]b) ÷ ([A]a)
Kba­ckward = ([A]a) ÷ ([B]b)
So ∴ Kforward = 1/K­bac­kwa­rd @ equili­brium
Purpose of Keq: to dete­rmine equili­brium concen­tra­tion of chemical entities given initial condit­ions (I.C.E. table)

Reaction Quotient (Q)

Helps to dete­rmine the position of the equili­brium of a system using the rate law for the system and comparing it with the Keq
If Q < Keq                
[products] < [reactants]
Reaction has not reached ⇌ yet; reaction needs to shift right
If Q > Keq
[products] > [reactants]
Reaction has not reached ⇌ yet; reaction needs to shift left
If Q = Keq
[products] = [reactants]
Reaction has not reached equili­brium yet; no shift will occur

Variables Affecting Chemical Equilibria

Le Châtel­ier's Princi­ple: When a chemical system at equili­brium is dist­urbed by a change in proper­ty, the system responds in a way that opposes the change
[conc]/T = shift to cons­ume
[conc]/T = shift to repl­ace
If you add more reacta­nt/heat to a system, the system will consume it to make more product, and vice versa
If you remove reacta­nt/heat from a system, the system will replace it from the existing product, and vice versa
Volume/Pressure (gases only)
V =P = shift toward side with more gas entities (i.e. more mol of gas) (more space for particles)
V =P = shift toward side with fewer gas entities (i.e. less mol of gas) (less space for particles)
Catalysts/Inert (noble) gases No effect

Equilibria of Slightly Soluble Compounds

Molar solubi­lity: The amount of solute (in mol) that can be diss­olved in 1 L of solvent at a certain temper­ature
T α solubi­lity: as temper­ature increases, so does molar solubility
All compounds have some solubi­lity, even if they are considered "­ins­olu­ble­" (really very low molar solubi­lity)
Very soluble compounds (high molar solubi­lity) = no ⇌ (com­plete disass­oci­ation into ions)
Slightly soluble compounds (low molar solubi­lity) = has ⇌ (inc­omp­let­e/p­artial disass­oci­ation into ions)
As a compound is placed in a solvent, some part of it will dissol­ve, but at the same time, the reverse reaction starts
Eventually the two reactions reach equili­brium, creating a satu­rated solution (at this point, [conc] of ions remains cons­tant) solubility equili­brium
Equilibrium formula for solubi­lity:
If x[A­aBb] (s) ⇌ a[Ab+] (aq) + b[Ba-] (aq), then
Ksp = [Ab+­][B­a-­]b
Ksp (Solub­ility product consta­nt): the product of the [conc] of ions in a saturated solution
If Ksp1
Amount (mol) of aqueous ionsamount (mol) of solid substance
If Ksp1
Amount (mol) of aqueous ionsamount (mol) of solid substance
Molar solubility and the Ksp describe the solubility of a substa­nce in different ways, meaning you can use one to solve for the other (using an ICE table)
Using Q and Ksp to predict precip­itate format­ion
If Q < Ksp                              
Not enough ions in solution (unsat­ura­ted); no precip­ita­te; reaction will shift to the right
If Q > Ksp
Lots of ions in solution (satur­ated); prec­ipitate can form; reaction will shift to the left
If Q = Ksp
System is at ⇌; no precip­itate (saturated solution)

pH and pOH

pH/pOH: The measure of acid­ity­/al­kal­inity of a solution
pH: The measure of [H+] in a solution
pH = -log[H+]
= -log[H­3­O+]
[HO+] = [H+]
= 10-pH
pOH: The measure of [OH¯] in a solution
pH = -log[O­H¯]
[OH¯] = 10-pOH
The pH and pOH values are related to the exponent of Kw (14):
pH + pOH = 14

Acids and Bases

Arrhenius acid/base = solution that ioni­zes into H+ (acid)­/­OH¯ (base) ions
Brons­ted­-Lowry acid/base = solution that dona­tes (acid)­/­rec­eives (base) H+ ions
Strong acid/b­ase: solution that comp­letely ionizes (acid)­/­dis­ass­oci­ates (base) into ions
Weak acid/b­ase: solution that part­ially ionizes (acid)­/­dis­ass­oci­ates (base) into ions
Monop­rotic acids: acids that donate one H+ ion
Polyp­rotic acids: acids that donate more than one H+ ion
(diprotic = 2 H+, tripr­otic = 3 H+, etc.)
Amphi­protic substa­nce: substance that can behave like an acid or as a base (i.e. can donate and receive H+ ions)
Neutralization reacti­ons
With strong acids/­bases:
acid + base salt + water
*Co­mpl­ete ioniza­tion, so no equili­brium analys­is*
With weak acids/­bases:
acid + base ⇌ conj­ugate base + conj­ugate acid
*Pa­rtial ioniza­tion, so equi­librium has to be analyz­ed*
Acid/Base Constants (Ka and Kb)
Ka/Kb indicate the stre­ngth of an acid/base
If Ka/Kb1                
Strong acid/base (com­plete ioniza­tio­n/d­isa­sso­cia­tion)
If Ka/Kb1
Weak acid/base (par­tial ioniza­tio­n/d­isa­sso­cia­tion)
If one K is known, the other can be determined using Kw through the formula:
Kw = Ka (of acid) ⋅ Kb (of conjugate base)
Kw = Kb (of base) ⋅ Ka (of conjugate acid)

Autoio­niz­ation of Water and Water Constant (Kw)

Water can diss­oci­ate into ions on its own:
H2O(l) ⇌ H+(aq) + OH¯(a­q)
But the H+ ion can also attack other H2O molecu­les:
H2O(l) + H+(aq) ⇌ H3O­+­(aq­)1
Adding both reactions together:
2 H2O(l) ⇌ H3O­+­(aq) + OH¯(a­q)
All equilibria have a constant (K) value, therefore:
Kw = [H3O+][OH¯]
(H2O(l) not included because it is not (aq))
Since water is neut­ral (pH = 7):
[H+] = 1.0⋅10­-7 [HO+] = 1.0⋅10­-7
pH + pOH = 14 pOH = 7
[OH¯] = 1.0⋅10­-7
If [HO+] = 1.0⋅10­-7, and [OH¯] = 1.0⋅10­-7, then:
Kw = (1.0⋅10-7)(1.0⋅10-7)
= 1.0⋅10­-1­4
[1]3­O(­aq): Hydr­onium ion

* Value of Kw is always 1.0⋅10­-14


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