# 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*
`3`­O`(­aq)`: Hydr­onium ion

* Value of K`w` is always 1.0⋅10­-14 2 Pages
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