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Gr. 12 Chemical Systems and Equilibrium Cheat Sheet by

Introd­uction

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]c[D]d) ÷ ([A]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
Concen­tra­tio­n/T­emp­erature
[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+]a[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+]
[H3O+] = [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 [H3O+] = 1.0⋅10-7
pH + pOH = 14 pOH = 7
[OH¯] = 1.0⋅10-7
If [H3O+] = 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]H3O+(aq): Hydronium ion

* Value of Kw is always 1.0⋅10-14
                                                   
 

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