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### 77 Cards in this Set

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 equilibrium [reactants] and [products] doesn't change; rate of forward reaction is equal to the rate of the reverse reaction Keq given aA +bB -><- cC + dD Keq = [C]^c [D]^d/[A]^a [B}^b brackets = molar concentraion at equilibrium solids and liquids are not included (because concentraion doesnt change) If gaseous, then can use partial pressure mass action ration K eq is constant at given temp regardless of concentrations of product or reactant. Reaction will go in direction to equilibrium If Keq < 1 reaction favors the reactant If Keq = 1 reaction ba;ances the reacant and product Keq > 1 products favored Equilirium reached if ration of products and reactants remains constant & soulution is saturated with solute (system no longer changes with time) Q = reaction quotient aA =bB -><- cC + dD where Q = [C]^c [D]^d/[A]^a [B}^b Q = Keq no change Q < Keq Reaction proceed forward to produce more product Q > Keq Reaction proceeds in the reverse direction to attain more reactant Le Chatliers a system at equilibrium will try to neutralize any imposed change or stress (e.g., adding reactant) to reestablish equilibrium Eg., add more reactant, system will react by favoring forward reaction to consume reactant Changing volume of reaction container which is at equilibrium If gas, and decrease volume (increase pressure), then reaction favors the soulte with least number of moles and vice cersa + change H endothermic -chnage H exothermic changing heat in a system at equilibrium only effects endo thermic/ exothermic reaction If add heat, and the process is exothermic, reaction proceeds left If cool it down, reaction proceeds to right Keq and changes in temp changing temp will effect Keq; increases with endothermic, decreases with exothermic addition of inert or unreactive gas to system in equilibrium has no effect adding a catalyst increases reaction rate but does not affect the quilibrium; adding catalyst to a system that is already in quilibrium will have n oeffect change in H is negative exothermic and heat is on the product side Change in H is positive endothermic heat on reactant side solubility product constant Ksp the extent to which a salt will disolve in water can be determined from its solunility product constant Ksp molar solubility number of moles of a salt that will saturate a liter of water (asked to find out just how much of a solid can disosle in H20 given Ksp = 2.7 x 10 ^-9 for Li3PO4, how many moles of this salt would be required to form a saturated 1 liter aq solution equil = li3PO4 -><- 3Li + PO4 let x = [PO4] and [Li] = 3x Ksp = (3x)^2x = 27x^4= 2.7 x 10^-9 -> x = 3.2^-3 ol needed Qsp < Ksp more salt can be dissolved Qsp = Ksp solution is saturated Qsp> Ksp excess salt will precipitate Common ion effect reduce the solubility of ion is solution already has common ion; adding more ions that are in product will drive reaction to the left Arrenhius acid acids ionize in water to produce hydrogen H+ ions HCL -> H+ + Cl- Arrenhius base ionize in H20 to produce OH- ions NaOH -> Na+ +OH- Bronset Lowry Acid acids are proton donor Bronsted Lowry Base Bases are proton acceptors Lewis Acid electron pair acceptor Lewis Base electron pair donor strong acid on e that dissociates completly ( or nearly so ) in H20 E.g., HCl (aq) + H2) (l) -> H30 (aq) + Cl- (aq) (HF is a example of weak acid) Ka (equilirbium expression for acid dissocation) "acid ionization or dissocation constant" strength of acid is directly related to how much the products are favored over the reactants Ka = [H30] [A-]/[HA] Ka > 1 products favored acid is strong Ka < 1 reactants favored and acis is weak Comparing the Ka strengths The larger Ka, the stronger the acid Common strong acids HI, HBr, HCl, HClO3, HClO4, H2SO4, HNO3 Weak acid any acid that is not a strong acid from the list Kb Base ionization constant Kb = [HB][OH]/[B] Larger Kb, stronger base Strong base Group 1 hydroxides (LiOH, NaOH, KOH, RbOH, CsOH) CaOH2 CH3NH2 Sr(OH)2 Ba(OH)2 Metal amids (NaNH2 Weak base NH3 (ammonia) and amines Conjugate base of a strong acid has no basic properties (Cl- will not tend to accept proton, and doesnt act like a base The conjugate base of a weak acis weak base (F- has some tendency to accept proton) polyprotic has more than 1 proton to donatel; generally, everytime a polyprotic acid donates a proton, the resulting species will be a weaker acid than its predecessor amphoteric can act like a base or acid autoionization (self-ionization) of water, reacts with self in a Bronster Lowry acid-base reaction, 1 molecules acting as the acid and the other as athe base Kw =1.0*10^-14 = [H3O][OH] varies with temp (like other equillibrium constants), this is at 25 degrees C Kw is constatnt at a given temp regardless of the H3O concentration If the introduction of an acid increases the concentration of H3O ions, then equilibrium is disturbed the reverse reaction is favored, decreasing the concentration of OH- ions If introduce base to increase its concentration, equilibium is disturbed reverse reaction is favored, decreasing the concentration of H3O ions, however the produce to [H3O] and [OH ]will remain equal to Kw Kw = [OH][H3O] Increase one concentration decreases the other pH = -log[H+] means that [H+] = 10^-pH E.g., pH = 7 , [H] = 10^-7 trend for acidity down and to the left pOH = -log[OH] [OH] = 10^-pOH pH + pOH =14 [H] = 6.2x10^-5 pH is between 4 and 5 lowest concentration of H3O has highes pH solution has pH of 3.5, what is [H] [H] = 10^-3.5 = 10^.5 x 10^-4 = sqrt(10) x 10^-4 = 3.2 *10^-4 KaxKb= Kw pH calculation for strong acid pH comes directly from molarity of solution; so .01 M solution of HCl will have [H} = .01 = pH 2 weak acid concentration of a H ions will be much less than the undissociated ions when K < 10 ^-4 x added to or subtracted from a number is neglible with weak acids , use ICE HCN -><- H+ + CN- Initial: .2 0 0 at equil: (.2-x)M x x know Ka = 4.9 x 10^-10 x^2/(.2-x) equivl to: x^2/.2 = 4.9 x 10^-10 "p" take negative log Ka1 < Ka2 So pKa1 > pKa2 Ka2 is more acidic pKa + pKb 14 weak acid may have a weak or strong conj base 10^x = 3.16 x = log(3.16) x between 0 and 1 (10^0 or 10^1) so x equal .5 log (AB) = log (A) + log (B) log (10^2) = 2 log (10^3) = 3 log (10^2 x 10^3) = 5 in a series of oxyacids more acids means stronger acid rate constant (kinetics) K and equilibrium (thermodynamics) are not related formally are not related formally If H2) is the solvent do NOT include it in the equillibrium expression condensation is exothermic there's no net consumption or production of H+ in an acetal reaction H is a catalysit?? If the reacion is does not involve gases ignore the effect of pressure (if it does include gases, do include pressure such that reaction will favor products or reactants that have least gas moles