Equilibrium Study Guide

Equilibrium A system where two opposing processes have equal rates ( aA + bB  cC + dD ) At equilibrium, the rate forward = rate backward, but kf ≠ kr Homogenous Equilibria – involve all same phase (all g or all aq) Heterogeneous Equilibria – involve more than one phase (omit s and l phases) Equilibrium Constant, Keq [C ]c [D] d at equilibrium, no units K c= [ A ]a [ B]b If Kc < 1  more reactants than products at equilibrium If Kc > 1  more products than reactants at equilibrium (C )c ( D)d for gaseous reactions that indicated pressure. Exponents in K p= ( A)a (B)b expression again relate to stoichiometry **Pure solids and liquids are not included in Keq – they don’t change [ ] K p=K c ( RT ) Δn o R = 0.08206 L atm/mol K, Δn = moles of gas product – moles gas reactant Can have a mechanism that has an equilibrium (use K) and a reaction that goes to completion (use k in rate law). Can solve for variables and make substitutions Calculating the Value of Keq [C ]c [D] d K c= Forward Reaction: use [ ] or partial pressures (atm) to plug into [ A ]a [ B]b 1 ' Reverse Reaction: K reverse= K forward '' n Reaction of different stoichiometry: K stoich=K original o If n is the factor by which the stoichiometry changed ''' Adding Reaction Mechanism K’s together: K overal =(K 1)(K 2) Reaction Quotient, Q Apply the law of mass action using initial concentrations instead of equilibrium concentrations c d [C ] [D] not at equilibrium, initially, still no units Q= a b [ A ] [B] Tells you which way the reaction will go o Q = K  at equilibrium, no shift o Q < K  system shift right. Consume reactants, form products until equil o Q > K  system shift left. Consume products, form reactants until equil ICE Tables Write balanced equation on top I = initial, write in what you are starting with C= change, write in the expected change (x) with attention to stoichiometry E=equilibrium, initial minus/plus the change depending on reaction shift Can write Keq and solve for unknown variable (solver function on calc) If reaction starts with a solid or liquid, want to know final mass of solid/liquid o Still right out ICE table o Solid or liquid, not included in Kp o Can solve for x still (how the gas pressures changed) o Add gas pressures together for total pressure o Use PV=nRT to find moles of solid/liquid used up o Convert to mass and subtract that from the original amount LeChatlier’s Principle Equilibrium can be disturbed by three things: (1) changes in concentration (2) changes in volume for gas (3) changes in temperature o LeChatlier predicts which way the system shifts to restore equil o Catalysts do not shift equilibrium, just affects Ea and mechanism, not ΔH *LeChatlier can affect solid and liquid amounts due to shift, but adding more solid or liquid does not cause shifts 1. Changes in Concentration o Kp and Kc do not change o Add something  increase concentration  shift away from that side o Take something away  decrease concentration  shift toward that side 2. Changes in Volume o With gases only o Kp and Kc do not change o Decrease volume  pressure increases  to try to maintain constant pressure, system shifts to side with less moles of gas (get from balanced eq) o Increase volume  pressure decreases  to maintain constant pressure, system shifts to side with more moles of gas (get from balanced eq) o If add an inert gas (not in the equation) no shift because no change in partial pressure of the system 3. Changes in Temperature o Know shift based on enthalpy (ΔH: endo or exo)  If ΔH = positive, then heat is a reactant. Add heatshift right. Less heat shift left  If ΔH = negative, then heat is a product. Add heatshift left. Less heat shift right o Kp and Kc do change: van Hoff Equation K − ΔH rxn 1 1  ln 2 = units???? − K1 R T2 T1 ( ) ( ) Acid Nomenclature If name of anion ends in –ide: o The acid is named with prefix hydro- and ends with suffix –ic If anion contains oxygen: o If anion name ends in –ate, the suffix –ic is added to the root name o If anion name ends in –ite, the suffix –ous is added to the root name Arrhenius Acids and Bases Arrhenius acid: any substance that produces an H+ in aqueous solution o Ex) HCl (aq)  H+(aq) + Cl- (aq) or HCl (aq) + H2O(l)  H3O+(aq) + Cl-(aq) Arrhenius base: any substance that produces OH- in aqueous solution and has an OH- in its formula o Ex) NaOH(aq)  Na+(aq) + OH-(aq) Strong vs. Weak Acids and Bases Strong acids – completely break into ions (K >>>1) o HCl, HBr, HI, HNO3, HClO4, H2SO4 Weak acids – equilibrium, not all transfer “incomplete dissociation” (K<<<1) o Ex) CH3COOH(aq) + H2O(l)  CH3COO-(aq) + H3O+(aq) Same thing with bases (strong=complete ionization, weak=equil, incomplete) o Strong Bases: all alkali metal hydroxides, Ca(OH)2, Sr(OH)2, Ba(OH)2 Bronsted-Lowry Definitions Focuses on the reaction between the acid and base, not just the aqueous forms Acid – proton donor Base – proton acceptor (must have a lone pair) o Ex) NH3 has a lone pair that it uses to accept a proton from water to form NH4+ (the conjugate acid) Writing Acid-Base chemistry Overall equation: write compounds together, no ionized Full Ionic equation: write out all ions that dissociate Net Ionic equation: cross off any spectator ions If writing a reaction with a SA and WB, write one way arrow (strong conquers weak) If WA and WB, write an equilibrium arrow  Weak Acids and Ka HA(aq) + H2O(l)  H3O+(aq) + A-(aq) H3 O ¿ +¿ A ¿ −¿ ¿¿ ¿ ¿ ¿ K a=¿ Larger Ka means a stronger weak acid Percent Dissociation concentrationofiondissociated x 100 % Dissociation = originalconcentrationof WA o Concentration of ion dissociated calculated from an ICE table Depends on: (1) initial concentration of an acid and (2) Ka – how strong the acid is Water Auto Ionization 2 H2O  H3O+ + OHWater is amphoteric acts as an weak acid or weak base Kw = [H3O+][OH-] = 1 x 10-14 at 25°C pH and pOH pH = -log[H3O+] pOH= -log[OH-] Take –log of both sides of Kw equation  pKw = pH + pOH 14 = pH + pOH Sig figs when calculating pH: # sig figs in [ ]  # sig figs past decimal point pH To find pH and pOH of strong acid/base  direct calculation of pH use given [ ] To find pH and pOH of weak acid/base  need to do ICE table to get [ ] Amphoteric Species (polyprotic acids and such) Can behave as an acid or base (donate or accept H+) Have an Ka for each possible reaction (H3PO4  H2PO4-  HPO42-  PO4-) Ka1 >>> Ka2 > Ka3 Weak Bases React WB with water and form an OH-. Use ICE table to calculate [OH-] and then pOH to finally get pH Can make an assumption in calculations: can ignore the –x in equilibrium expression for the reactant if % ionization is less than 5% and original concentration is > 0.1M Have Kb Relationship Between Acid Strength and Strength of Conjugate Base (Kb)(Ka) = 1 x 10-14 = Kw The stronger a WA, the weaker its weak conjugate base (and vice versa) Conjugate pairs of SB and SA, are considered non-acids/bases or acid/bases of no strength Acid-Base Reaction not just with Water The acid-base equilibrium reaction is favored in the direction of the weaker conjugate acid-base pair Ex) F- + HCN  HF + CN- Ka HCN = 6.2x10-10 Ka HF=6.6x10-4 So favors the reactants, because Ka HF is larger Ions as Acids and Bases Hydrolysis of salts – water reacts with cations or anions Only ions that conjugates of WA or WB will hydrolyze appreciably pH stay same if add ions of conjugates of SA and SB o KCl  K+ + ClIf add conj ion of WB and conj ion of SA o NH4Cl(s)  NH4+(aq) + Cl-(aq)  Then NH4+ + H2O  NH3 + H3O+  pH goes down <7 If add conj ion of WA and conj ion of SB o CH3COONa(s)  CH3COO-(aq) + Na+(aq)  Then CH3COO-(aq) + H2O(l)  CH3COOH + OH-  pH goes up >7 If add conj ion of WA and conj ion of WB  need to compare Ka and Kb to see which is stronger. Whichever is larger, then pH will go in the direction General Approach to Acid-Base Problem Solving 1. Identify major species in solution a. Major species – (1) present in a high conc (2) participates in acid-base chem 2. Write a reaction with major species 3. Choose the right arrow a.  if strong acid or strong base (use stoichiometry) b.  if both WB and WA (use equilibrium and ICE table) Common Ion Effect Lets say there was a weak acid equilibrium present in the solution. If you add ions of the conjugate weak base, that would be the common ion. Still use ICE table just change the initial concentration of the common ion to a number instead of zero Still assume x is small and approximate. Use Ka/b to find x and thus pH or pOH H ¿ Can rearrange of Ka to get +¿  ratio affects pH ¿ ¿ Buffers Mixture of a weak acid and its conjugate base, both in substantial concentrations o “Substantial concentrations” – means that the weak acid and its weak conjugate base must be present in concentrations within a factor of ten Can react with acids and bases Buffered solutions are simply solutions of weak acids or bases containing a common ion. pH calculations on buffered solutions require exact same procedures as before When a SA or SB is added to buffered solution, it is best to deal with the stoichiometry of the resulting reaction first. Be careful of limiting reagent. Then consider the equilibrium calculations. Example) o First use stoichiometry (take into account limiting reagent)  OH+ HA H2 O + Amoli molc molf o Then use molf for HA and A- in its equilibrium equation  HA + H2O H3 O+ + AI (use molf from above to calc [HA]i and [A-]i) C E o Then find the [H3O+] and thus pH or pOH if [OH-] Henderson-Hasselbach Equation o Used to solve for new pH of a buffer solution [ conj.base] o pH= pK a +log [ weak acid] o can use this or just do a new ICE table