Cheatography

# Gr. 12 Chemical Systems and Equilibrium Cheat Sheet by nescafeabusive32

### 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 K``eq``) = 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 K``eq`` formula

### Equili­brium Constant (Keq)

 K``eq`` formula: If a[A] + b[B] ⇌ c[C] + d[D]; then K``eq`` = ([C]c[D]d) ÷ ([A]a[B]b) Magnitude of K``eq``: states whether the equili­brium position favours produc­ts/­rea­ctants If K = 1 [products] = [react­ants] If K1 [products]  [react­ants] If K1 [products]  [react­ants] K``forward`` vs K``backward`` If a[A] ⇌ b[B], then K``forward`` = ([B]b) ÷ ([A]a) K``backward`` = ([A]a) ÷ ([B]b) So ∴ K``forward`` = 1/``Kbackward`` @ equili­brium Purpose of K``eq``: 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 K``eq`` If Q < K``eq`` [products] < [reactants]Reaction has not reached ⇌ yet; reaction needs to shift right If Q > K``eq`` [products] > [reactants]Reaction has not reached ⇌ yet; reaction needs to shift left If Q = K``eq`` [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[A``a``B``b``] ``(s)`` ⇌ a[Ab+] ``(aq)`` + b[Ba-] ``(aq)``, then K``sp`` = [Ab+]a[Ba-]b K``sp`` (Solub­ility product constant): the product of the [conc] of ions in a saturated solution If K``sp``1 Amount (mol) of aqueous ionsamount (mol) of solid substance If K``sp``1 Amount (mol) of aqueous ionsamount (mol) of solid substance Molar solubility and the K``sp`` 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 K``sp`` to predict precip­itate formation If Q < K``sp`` Not enough ions in solution (unsat­ura­ted); no precip­itate; reaction will shift to the right If Q > K``sp`` Lots of ions in solution (satur­ated); precip­itate can form; reaction will shift to the left If Q = K``sp`` 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[H``3``O+] [H``3``O+] = [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 K``w`` (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 (K``a`` and K``b``) K``a``/K``b`` indicate the strength of an acid/base If K``a``/K``b``1 Strong acid/base (complete ioniza­tio­n/d­isa­sso­cia­tion) If K``a``/K``b``1 Weak acid/base (partial ioniza­tio­n/d­isa­sso­cia­tion) If one K is known, the other can be determined using K``w`` through the formula: K``w`` = K``a`` (of acid) ⋅ K``b`` (of conjugate base) K``w`` = K``b`` (of base) ⋅ K``a`` (of conjugate acid)

### Autoio­niz­ation of Water and Water Constant (Kw)

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

* Value of K
``w``
is always 1.0⋅10-14