IntroductionEquilibrium = 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 EquilibriaLe 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 CompoundsMolar 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 pOHpH/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 BasesArrhenius 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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