14.1-14.4: Chemical Equilibrium Equilibrium amazonaws com
14.1-14.4: Chemical Equilibrium Equilibrium: balanced state of a system in which there are no observable changes w/ time Dynamic eq: balanced state of a system in which 2 opposing processes occur at = rates (physical + chemical processes [photochromic lenses]) Chemical eq: 1. chemical eq is dynamic; rate of forward reaction = rate of reverse reaction; no net change in concentration of reactants/products 2. eq is independent of direction of approach if T, V, + mass are the same 3. catalysts do not affect eq concentrations, but decrease time needed to reach eq Homogeneous eq: all reactants + products are in same phase Rate of forward = rate of reverse reaction aA+bBcC+dD k(forward)[A]^a[B]^b = k(reverse)[C]^c[D]^d k(forward)/k(reverse) = [C]^c[D]^d / [A]^a[B]^b therefore: K(c) = [C]^c[D]^d / [A]^a[B]^b k: rate constant K: eq constant [x]: moles/L (for g and solutes in dilute solutions only) ^x: stoichiometric coefficients Pure liquids, pure solids, and solvents do NOT appear in K(c) expression K(c) = [products]/[reactants] If Kc >> 1, reaction is product-favored; more product at eq If Kc 1000: reaction essentially goes to completion 1 < K < 1000: [products] > [reactants] at eq 0.001 < K < 1: [reactants] > [products] at eq K < 0.001: essentially no reaction occurs Kc of reverse reaction is reciprocal of Kc of forward reaction Kc of reaction that was multiplied by a N is the initial Kc raised to a power equal to that N Overal Kc of multistep reaction is product of steps: Kc(step 1) x Kc(step 2) The Kc for a net reaction w/ 2+ steps is the product of the eq constants for the individual steps Dilute solutions: NH3 (aq) + H2O (l) NH4+ (aq) + OH- (aq) If [H2O] >> [NH4+], [H20] >> [OH-], so [H2O] is like a constant Kc = [NH4+][OH-]/[NH3][H2O]const K’c = Kc x [H2O]const = [NH4+][OH-]/[NH3] Heterogeneous eq: not all reactants and products are in the same phase Pure liquids, pure solids, and solvents do NOT appear in K(c) expression What counts are g + aq
Ideal Gas: PV = nRT For gas a: PaV = naRT; Pa = naRT/V = [a]RT Pa = partial pressure of gas a na/V = molarity = [a] [a] = Pa/RT Gas-phase reaction: aA (g) + bB (g) cC (g) + dD (g) Kc = [C]^c [D]^d / [A]^a [B]^b Px = [x]RT Kp = PC^c PD^d / PA^a PB^b = ([C]RT)^c ([D]RT)^d / ([A]RT)^a ([B]RT)^b = [C]^c [D]^d / [A]^a [B]^b times (RT)^(c+d) / (RT)^(a+b) = Kc(RT)^(c+d-a-b) Kp = Kc(RT)^(change in n) R = 0.082 L atm/mol K T in Kelvin Change in n = moles of g product minus moles of g reactants Kc: calculated w/ concentrations in mol/L Kp: calculated with partial pressure in atm Ex: 3Fe (s) + 4H2O (g) Fe3O4 (s) + 4H2 (g) Kc = [H2]^4/ [H2O]^4 Kp = P(H2)^4/P(H2O)^4 Calculating Kp ex: N2 (g) + 3H2 (g) 2NH3 (g), Kc = 5.8 x 10^5 at 25 degree C Kp = Kc(RT)^(change in n) T = 25 + 273 = 298 K Change n = 2 – (3+1) = -2 Kp = 5.8 x 10^5 ((0.0821 L atm/molK)(298))^-2 = 9.7 x 10^2 mol^2/L^2 atm^2
14.5-14.6: Chemical Eq Q = [products]/[reactants] Changes as a reaction moves to equilibrium Kc = [products]eq/[reactants]eq Constant! (if T is constant) At eq: Q = Kc If Q < Kc, Q increases to reach eq Amt of reactant decreases to make more product Reaction moves forward = reaction spontaneous as written If Q > Kc, Q decreases to reach eq Amt of product decreases to make more reactant Reaction moves back = reaction spontaneous in reverse Le Chatelier’s principle: don’t push me ‘cause I’ll push back If a change is applied to a system at eq, the eq will shift to reduce the effect of the change Concentration, pressure (or volume) for gas-phase eq, and T will affect eq 1. Eq shifts toward products when reactant is added or product removed Eq shifts toward reactants when product is added or reactants removed 2. For g phase ONLY: Eq shift to side w/ fewer moles when P increases or when V decreases Eq shift to side w/ more moles when P decreases or when V increases K does not change if P or V change; eq position may change If V doubles, all concentrations divide by 2 Kc is not changed by solvent addition/removal, but eq position may change 3. eq shifts in the direction that counteracts the change in T If T is lowered on endothermic eq system, eq will shift to left If T is lowered on exothermic eq system, eq will shift to right 14.8: Industrial applications of Le Chatelier’s Principle Haber-Bosch Process: N2 + 3H2 2NH3 + heat (exothermic) Increasing the pressure favors the forward reaction in which 4 mol of g molecule is converted to 2 mol Decreasing the concentration of NH3 favors the forward reaction in order to replace the NH3 that has been removed 15.3: Solubility and Equilibrium Solutions form by efficient intermolecular forces like dipole-dipole, ion-dipole, London, or H bonds
The process of dissolving is favored if the solute-solute interactions are weaker than the solutesolvent interactions The solubility of a substance is the max amt that dissolves in a given amt of solvent at a specific T Ionic compounds dissolve by electrostatic attraction of opposite charges (ion-dipole interactions) like NaCl Ion electrostatic interaction (crystal structure of NaCl) < ion-dipole electrostatic interactions (hydration shells of Na+ and Cl-) The process of forming solution: 1. separating solvent molecules (endothermic) 2. separating solute particles (endothermic) 3. forming new solute-solvent associations (exothermic) Unsaturated solutions - [solute dissolved] < solubility = [solute dissolved]max at a T -Q [solute dissolved]eq - Q > Kc - ex: sugar in water, honey, CO2 in water
16.1-16.2: acids and bases Bronsted Lowry (BL): Acid: proton donor Base: proton acceptor (at least one lone pair of e or pi bond) Ionization of acid in water: HA + H2O A- + H3O+ Ionization of base in water: B + H2O BH+ + OHWater acts as solvent, but also as a weak donor/acceptor of protons Amphiprotic: substance that can act as an acid or base, depending on the reaction partner 16.3-16.4: the autoionization of water; the pH scale H2O (l) + H2O (l) OH- (aq) + H3O+ (aq) Kw = [H3O+][OH-] = ionization constant for water At 25 degree C: [H3O+] = 10^-7 M = [OH-] Kw = [H3O+][OH-] = 10^-7 x 10^-7 = 10^-14 mol^2/L^2 Kw increases as T increases. H3O+ and OH- are present in all aq solutions At 25 degree C: Neutral aq solutions: [H3O+] = 10^-7M = [OH-] Acidic aq solutions: [H3O+] > 10^-7M > [OH-] Basic aq solutions: [H3O+] < 10^-7M < [OH-] For all three, [H3O+][OH-] = 10^-14 pH: a mathematical parameter to express [H3O+] as positive integers pH = -log[H3O+] If pH = 4, then [H+] = 10^4 At 25 degree C: Neutral aq solutions: pH = -log[10^-7] = -(-7.00)log10 = 7.00 x 1 = 7.00 Acidic aq solutions: pH < 7.00 Basic aq solutions: pH > 7.00 pOH = -log[OH-] At 25 degree C: pKw = pH + pOH = 14.00 Find pH and pOH of acid with density and % w/w 1. use % w/w and density to find volume
2. find moles of acid using given % as g and molecular weight 3. find M (moles/L) 4. take log of M to get one answer 5. subtract step 4 from 14 to get other part of answer Can measure pH with pH indicators (bromothymol blue from yellow acid to blue base; methyl orange from red acid to orange neutral; methyl red from red acid to yellow base; phenolphthalein from white acid to red base), pH meters 16.2; 16.5: Ionization Constants of Acids and Bases Acid strength: tendency of acid to donate proton The stronger a Bronsted Lowry acid, the easier it dissociates into protons and the corresponding conjugate base HA + H2O A- + H3O+ Keq = [H3O+][A-]/[HA][H2O] but H2O is like constant, so: Acidity constant = Ka = [H2O]Keq = [H3O+][A-]/[HA] Ka = [H+][A-]/[HA] The larger the Ka, the stronger the acid Strong acid: practically complete dissociation Eq shifted to right; product-favored; strong electrolytes Ka >> 1 Ex: mineral acids like HCl, H2SO4, HNO3 Weak acid: practically no dissociation Eq shifted to left; reactant-favored; weak electrolyte Ka Ka2 Carboxylic acids are BL acids (proton donors) Bases: B + H2O BH+ + OHKb = [BH+][OH-]/[B] (basicity constant, base ionization constant, dissociation constant) The larger the Kb, the stronger the base Amines are BL bases (proton acceptor) Strong acids make weak conjugate base Weak acids form strong conjugate base
Acid strength from big to small: H2SO4, HBr, HCl, HF, NH4+, OH-, H2, CH4 Strong acids: HCl HBr HI HNO3 HClO4 H2SO4 Weak acid: HF Base strength from big to small: CH3-, H-, NH3, F-, Cl-, Br-, HSO4Strong bases: LiOH NaOH KOH Ca(OH)2 Ba(OH)2 Sr(OH)2 Weak base: NH3 CH3COOH (aq) + H2O (l) H3O+ (aq) + CH3COO- (aq) Percent ionization = [H3O+]eq/[CH3COOH]initial x 100 Ka can tell us: - acid strength: extent of prton transfer; percent ionization, not concentration - strong acid: strong tendency to release H+; if Ka > 10, assume complete dissociation - strong base: weak tendency to release H+; reaction doesn’t go to completion
16.2; 16.6: Acid-Base behavior From left to right across period, decreasing size so increasing electronegativity and increasing acidity Down a column/group, increasing size and increasing acidity Positive/negative charges are stabilized when spread over a larger volume Larger ions are more stable Oxoacid strength (H-O-X): The more electronegative the X atom, the weaker the O-H bond (the less e- around H nucleus), so the stronger the oxoacid HClO > HBrO > HIO The more O atoms, the more electronegative the X group, so the weaker the O-H bond, so the stronger the oxoacid HClO4 > HClO3 > HClO2 > HClO 16.7: Problem solving using Ka, Kb, and percent ionization x = [-b+/- root(b^2-4ac)]/2a 0.1 M CH3COOH (1% ionized) means that at eq, [H3O+] = [CH2COO-] = 1% x 0.1 M = 0.001 M ionized acid [CH3COOH] = 0.1 – 0.001 M = 0.099 M acid that didn’t ionize Ex: pH of a 0.1M solution of lactic acid is 2.34 at 25 deg C. Determine Ka. -log [H3O+] = 2.43; [H3O] = 0.0037 M HA + H2O H3O+ + AI: 0.1; blank 0; 0 Chnge: 0.0037; blank 0.0037; 0.0037 Eq: 0.1 – 0.0037; blank 0.0037; 0.0037 Ka = [H3O+][A-]/[HA] = 1.4 x 10^-4 Approximation can be applied to weak acids/bases if: [acid]initial > 100 Ka or [base]initial > 100 Kb Contribution of water autoionization cannot be ignored if pH of initial solution is b/w 6-8 Ex: what is pH of 0.1 M H2SO4 solution? A: 1-log2 because pH = -log (2 x 10^-1) Ex: what is [H3O+] in aqueous 10^-6 M H2SO4 solution? A: 2 x 10^-6 + 10^-7 HA (aq) + H2O (l) H3O+ (aq) + A-(aq) Ka = [H3O+][A-]/[HA] A- (aq) + H2O (l) HA (aq) + OH- (aq) Kb = [HA][OH-]/[A-]
Ka x Kb = Kw The larger the Ka, the smaller the Kb pKa = – log Ka pKb = – log Kb The smaller the pKa, the stronger the acid At 25 degree C: Ka x Kb = Kw = 1.0 x 10^-14 pKa + pKb = 14
Ideal Gas: PV = nRT For gas a: PaV = naRT; Pa = naRT/V = [a]RT Pa = partial pressure of gas a na/V = molarity = [a] [a] = Pa/RT Gas-phase reaction: aA (g) + bB (g) cC (g) + dD (g) Kc = [C]^c [D]^d / [A]^a [B]^b Px = [x]RT Kp = PC^c PD^d / PA^a PB^b = ([C]RT)^c ([D]RT)^d / ([A]RT)^a ([B]RT)^b = [C]^c [D]^d / [A]^a [B]^b times (RT)^(c+d) / (RT)^(a+b) = Kc(RT)^(c+d-a-b) Kp = Kc(RT)^(change in n) R = 0.082 L atm/mol K T in Kelvin Change in n = moles of g product minus moles of g reactants Kc: calculated w/ concentrations in mol/L Kp: calculated with partial pressure in atm Ex: 3Fe (s) + 4H2O (g) Fe3O4 (s) + 4H2 (g) Kc = [H2]^4/ [H2O]^4 Kp = P(H2)^4/P(H2O)^4 Calculating Kp ex: N2 (g) + 3H2 (g) 2NH3 (g), Kc = 5.8 x 10^5 at 25 degree C Kp = Kc(RT)^(change in n) T = 25 + 273 = 298 K Change n = 2 – (3+1) = -2 Kp = 5.8 x 10^5 ((0.0821 L atm/molK)(298))^-2 = 9.7 x 10^2 mol^2/L^2 atm^2
14.5-14.6: Chemical Eq Q = [products]/[reactants] Changes as a reaction moves to equilibrium Kc = [products]eq/[reactants]eq Constant! (if T is constant) At eq: Q = Kc If Q < Kc, Q increases to reach eq Amt of reactant decreases to make more product Reaction moves forward = reaction spontaneous as written If Q > Kc, Q decreases to reach eq Amt of product decreases to make more reactant Reaction moves back = reaction spontaneous in reverse Le Chatelier’s principle: don’t push me ‘cause I’ll push back If a change is applied to a system at eq, the eq will shift to reduce the effect of the change Concentration, pressure (or volume) for gas-phase eq, and T will affect eq 1. Eq shifts toward products when reactant is added or product removed Eq shifts toward reactants when product is added or reactants removed 2. For g phase ONLY: Eq shift to side w/ fewer moles when P increases or when V decreases Eq shift to side w/ more moles when P decreases or when V increases K does not change if P or V change; eq position may change If V doubles, all concentrations divide by 2 Kc is not changed by solvent addition/removal, but eq position may change 3. eq shifts in the direction that counteracts the change in T If T is lowered on endothermic eq system, eq will shift to left If T is lowered on exothermic eq system, eq will shift to right 14.8: Industrial applications of Le Chatelier’s Principle Haber-Bosch Process: N2 + 3H2 2NH3 + heat (exothermic) Increasing the pressure favors the forward reaction in which 4 mol of g molecule is converted to 2 mol Decreasing the concentration of NH3 favors the forward reaction in order to replace the NH3 that has been removed 15.3: Solubility and Equilibrium Solutions form by efficient intermolecular forces like dipole-dipole, ion-dipole, London, or H bonds
The process of dissolving is favored if the solute-solute interactions are weaker than the solutesolvent interactions The solubility of a substance is the max amt that dissolves in a given amt of solvent at a specific T Ionic compounds dissolve by electrostatic attraction of opposite charges (ion-dipole interactions) like NaCl Ion electrostatic interaction (crystal structure of NaCl) < ion-dipole electrostatic interactions (hydration shells of Na+ and Cl-) The process of forming solution: 1. separating solvent molecules (endothermic) 2. separating solute particles (endothermic) 3. forming new solute-solvent associations (exothermic) Unsaturated solutions - [solute dissolved] < solubility = [solute dissolved]max at a T -Q [solute dissolved]eq - Q > Kc - ex: sugar in water, honey, CO2 in water
16.1-16.2: acids and bases Bronsted Lowry (BL): Acid: proton donor Base: proton acceptor (at least one lone pair of e or pi bond) Ionization of acid in water: HA + H2O A- + H3O+ Ionization of base in water: B + H2O BH+ + OHWater acts as solvent, but also as a weak donor/acceptor of protons Amphiprotic: substance that can act as an acid or base, depending on the reaction partner 16.3-16.4: the autoionization of water; the pH scale H2O (l) + H2O (l) OH- (aq) + H3O+ (aq) Kw = [H3O+][OH-] = ionization constant for water At 25 degree C: [H3O+] = 10^-7 M = [OH-] Kw = [H3O+][OH-] = 10^-7 x 10^-7 = 10^-14 mol^2/L^2 Kw increases as T increases. H3O+ and OH- are present in all aq solutions At 25 degree C: Neutral aq solutions: [H3O+] = 10^-7M = [OH-] Acidic aq solutions: [H3O+] > 10^-7M > [OH-] Basic aq solutions: [H3O+] < 10^-7M < [OH-] For all three, [H3O+][OH-] = 10^-14 pH: a mathematical parameter to express [H3O+] as positive integers pH = -log[H3O+] If pH = 4, then [H+] = 10^4 At 25 degree C: Neutral aq solutions: pH = -log[10^-7] = -(-7.00)log10 = 7.00 x 1 = 7.00 Acidic aq solutions: pH < 7.00 Basic aq solutions: pH > 7.00 pOH = -log[OH-] At 25 degree C: pKw = pH + pOH = 14.00 Find pH and pOH of acid with density and % w/w 1. use % w/w and density to find volume
2. find moles of acid using given % as g and molecular weight 3. find M (moles/L) 4. take log of M to get one answer 5. subtract step 4 from 14 to get other part of answer Can measure pH with pH indicators (bromothymol blue from yellow acid to blue base; methyl orange from red acid to orange neutral; methyl red from red acid to yellow base; phenolphthalein from white acid to red base), pH meters 16.2; 16.5: Ionization Constants of Acids and Bases Acid strength: tendency of acid to donate proton The stronger a Bronsted Lowry acid, the easier it dissociates into protons and the corresponding conjugate base HA + H2O A- + H3O+ Keq = [H3O+][A-]/[HA][H2O] but H2O is like constant, so: Acidity constant = Ka = [H2O]Keq = [H3O+][A-]/[HA] Ka = [H+][A-]/[HA] The larger the Ka, the stronger the acid Strong acid: practically complete dissociation Eq shifted to right; product-favored; strong electrolytes Ka >> 1 Ex: mineral acids like HCl, H2SO4, HNO3 Weak acid: practically no dissociation Eq shifted to left; reactant-favored; weak electrolyte Ka Ka2 Carboxylic acids are BL acids (proton donors) Bases: B + H2O BH+ + OHKb = [BH+][OH-]/[B] (basicity constant, base ionization constant, dissociation constant) The larger the Kb, the stronger the base Amines are BL bases (proton acceptor) Strong acids make weak conjugate base Weak acids form strong conjugate base
Acid strength from big to small: H2SO4, HBr, HCl, HF, NH4+, OH-, H2, CH4 Strong acids: HCl HBr HI HNO3 HClO4 H2SO4 Weak acid: HF Base strength from big to small: CH3-, H-, NH3, F-, Cl-, Br-, HSO4Strong bases: LiOH NaOH KOH Ca(OH)2 Ba(OH)2 Sr(OH)2 Weak base: NH3 CH3COOH (aq) + H2O (l) H3O+ (aq) + CH3COO- (aq) Percent ionization = [H3O+]eq/[CH3COOH]initial x 100 Ka can tell us: - acid strength: extent of prton transfer; percent ionization, not concentration - strong acid: strong tendency to release H+; if Ka > 10, assume complete dissociation - strong base: weak tendency to release H+; reaction doesn’t go to completion
16.2; 16.6: Acid-Base behavior From left to right across period, decreasing size so increasing electronegativity and increasing acidity Down a column/group, increasing size and increasing acidity Positive/negative charges are stabilized when spread over a larger volume Larger ions are more stable Oxoacid strength (H-O-X): The more electronegative the X atom, the weaker the O-H bond (the less e- around H nucleus), so the stronger the oxoacid HClO > HBrO > HIO The more O atoms, the more electronegative the X group, so the weaker the O-H bond, so the stronger the oxoacid HClO4 > HClO3 > HClO2 > HClO 16.7: Problem solving using Ka, Kb, and percent ionization x = [-b+/- root(b^2-4ac)]/2a 0.1 M CH3COOH (1% ionized) means that at eq, [H3O+] = [CH2COO-] = 1% x 0.1 M = 0.001 M ionized acid [CH3COOH] = 0.1 – 0.001 M = 0.099 M acid that didn’t ionize Ex: pH of a 0.1M solution of lactic acid is 2.34 at 25 deg C. Determine Ka. -log [H3O+] = 2.43; [H3O] = 0.0037 M HA + H2O H3O+ + AI: 0.1; blank 0; 0 Chnge: 0.0037; blank 0.0037; 0.0037 Eq: 0.1 – 0.0037; blank 0.0037; 0.0037 Ka = [H3O+][A-]/[HA] = 1.4 x 10^-4 Approximation can be applied to weak acids/bases if: [acid]initial > 100 Ka or [base]initial > 100 Kb Contribution of water autoionization cannot be ignored if pH of initial solution is b/w 6-8 Ex: what is pH of 0.1 M H2SO4 solution? A: 1-log2 because pH = -log (2 x 10^-1) Ex: what is [H3O+] in aqueous 10^-6 M H2SO4 solution? A: 2 x 10^-6 + 10^-7 HA (aq) + H2O (l) H3O+ (aq) + A-(aq) Ka = [H3O+][A-]/[HA] A- (aq) + H2O (l) HA (aq) + OH- (aq) Kb = [HA][OH-]/[A-]
Ka x Kb = Kw The larger the Ka, the smaller the Kb pKa = – log Ka pKb = – log Kb The smaller the pKa, the stronger the acid At 25 degree C: Ka x Kb = Kw = 1.0 x 10^-14 pKa + pKb = 14
