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1 Reversible Reactions 2 reversible reaction - a chemical reaction in which the products formed react to produce the original reactants. 3 The reaction between NO2 and N2O4 is reversible. cooling N2O4 is formed 2NO2(g) → N2O4 (g) N2O4 decomposes when heated forming NO2 heating N2O4 (g) → 2NO2 (g) 4 reaction to the right (forward) 2NO2(g) → N2O4 (g) → reaction to the left (reverse) Ice water Hot water 5 Rates of Reaction 6 • The rate of a reaction is variable. It depends on: – concentrations of the reacting species – reaction temperature – presence or absence of catalysts – the nature of the reactants 7 Chemical Equilibrium 9 chemical equilibrium equilibrium A dynamic state The in state which in At equilibrium the concentrations of the two orthemore which rate opposing of the forward processes reaction are products and the reactants are not taking place equals the rate at of thethe same reverse time reaction and at the in changing. asame chemical rate. change. 10 A saturated salt solution is in equilibrium with solid salt. salt crystals are dissolving Na+ and Clare crystallizing NaCl(s) → Na+(aq) + Cl-(aq) → At equilibrium the rate of salt dissolution equals the rate of salt crystallization. 11 Le Chatelier’s Principle 12 Henri LeChatelier 13 In 1888, the French chemist Henri LeChatelier This generalization, known as set forth a far-reaching generalization on the LeChatelier’s Principle, states behavior of equilibrium systems. If a stress or strain is applied to a system in equilibrium, the system will respond in such a way as to relieve that stress and restore equilibrium under a new set of conditions. 14 Effect of Concentration on Equilibrium 15 • For most reactions the rate of reaction increases as reactant concentrations increase. • The manner in which the rate of reaction changes with concentration must be determined experimentally. 16 An equilibrium is disturbed when the concentration of one or more of its components is changed. As a result, the concentration of all species will change and a new equilibrium mixture will be established. 17 The system is at equilibrium results in C and results D in A and B increases the rate of being produced faster being used faster than the forward reaction than they are used.they are produced. A + B →C + D → Increasing the concentration of B 18 The system is again at equilibrium In the new equilibrium concentration of A has decreased concentrations of B, C and D have increased A+ B→C + D → After enough time has passed, the rates of the forward and reverse reactions become equal. 19 Effect of Concentration Changes on the Chlorine Water Equilibrium +increase decrease Cldecrease increase H H O O increase HOCl Cl 2 32 concentrationconcentration concentrationconcentration concentration Cl2(aq) + 2H2O(l) → HOCl(aq) + H3O+(aq) + Cl-(aq) → Equilibrium Equilibriumshifts shiftsto toright left 20 Effect of C22H O H33O22 Concentration Changes on pH Add 0.200 0.100 mol NaC2H3O2 NaC2H3O2(aq) → Na+(aq) C2H3O2(aq) + C H O (aq) Equilibrium shifts to left HC2H3O2(aq) + H2O(l) → H3O+(aq) + C C22H H33O O2-2(aq) (aq) → 1 L 0.100 M HC2H3O2 Equilibrium pH = 5.05 2.87 4.74 21 Effect of Pressure on Equilibrium 22 • Changes in pressure significantly affect the reaction rate only when one or more of the reactants or products is a gas and the reaction is run in a closed container. • The effect of increasing the pressure is to increase the concentrations of any gaseous reactants or products. 23 Increase Pressure increases CO2 concentration CaCO3(s) → CaO(s) + CO2(g) → Equilibrium shifts to left 24 Decrease Pressure decreases CO2 concentration CaCO3(s) → CaO(s) + CO2(g) → Equilibrium shifts to right 25 In a system composed entirely of gases, a increase in the pressure of the container will cause the reaction and the equilibrium to shift to the side that contains the smallest number of molecules. 26 Increase Pressure N2(g) + 3H2(g) → 2NH3(g) → 1 mol 3 mol 6.02 x 1023 molecules 1.81 x 1024 molecules 2 mol 1.20 x 1024 molecules 2.41 x 1024 molecules Equilibrium shifts to the right towards fewer molecules. 27 Increase Pressure N2(g) + O2(g) → 2NO(g) → 1 mol 1 mol 6.02 x 1023 molecules 6.02 x 1023 molecules 2 mol 1.20 x 1024 molecules 1.20 x 1024 molecules Equilibrium does not shift. The number of molecules is the same on both sides of the equation. 28 Effect of Temperature on Equilibrium 29 The rate of the reaction that absorbs heat is increased to a greater extent, and the equilibrium shifts to favor that reaction. In a reversible reaction, the rates of both When the process temperature is endothermic, of a systemthe is the forward and the reverse reactions are raised, the(left forward rate oftoreaction right)increases. reaction is increased by an increase in temperature. increased. When the process is exothermic, the reverse (right to left) process is increased. 30 Heat may be treated as a reactant in endothermic reactions. oC moles CO At room temperature very little COCO forms. At 1000 moles 2 C(s) + CO2(g) + heat → 2CO(g) → Equilibrium shifts to right 31 Effect of Catalysts on Equilibrium 32 A catalyst is a substance that influences the rate of a reaction and can be recovered essentially unchanged at the end of the reaction. A catalyst does not shift the equilibrium of a reaction. It affects only the speed at which the equilibrium is reached. https://www.youtube.com/watch?v=DP05fxR4oUk 33 Energy Diagram for an Exothermic Reaction Activation A catalyst catalyst energy: speeds does not up the achange minimum reactionthe energy by energy lowering required ofthe a 34 for a reaction activation reaction. energy. to occur. AlCl3 PCl3(l) + S(s) → PSCl3(l) In the little Very presence thiophosphoryl of a catalystchloride the reaction is formed is complete in the in absence a few seconds. of a catalyst because the reaction is so slow. MnO2 2KClO3(s) → 2KCl + 3O2(g) Δ The laboratory preparation of oxygen uses manganese dioxide as a catalyst to increase the rate of the reaction. 35 Equilibrium Constants 36 At equilibrium the rates of the forward and reverse reactions are equal, and the concentrations of the reactants and products are constant. 37 The equilibrium constant (Keq) is a value representing the unchanging concentrations of the reactants and the products in a chemical reaction at equilibrium. 38 For the general reaction aA + bB → cC + dD → reactants products at a given temperature: The concentrations of A, B, C, and D represent the equilibrium concentrations. The brackets around [A], [B], [C], and [D] represent concentrations in Molarity. The products are written on the top of the fraction & the reactants on the bottom. The coefficients to balance the equation a, b, c, and d are written as exponents. 39 For the general reaction aA + bB → cC + dD → at a given temperature c K eq d [C] [D] = a b [A] [B] 40 For the reaction 3H2 + N2 → 2NH3 → 2 K eq [NH3 ] = 3 [H2 ] [N2 ] 41 For the reaction 4NH3 + 3O2 → 2N2 + 6H2O → 2 K eq 6 [N2 ] [H2O] = 4 3 [NH3 ] [O2 ] 42 The magnitude of an equilibrium constant indicates the extent to which the forward and reverse reactions take place. H2 + I2 → 2HI → 2 K eq [HI] o = = 54.8 at 425 C [H2 ] [I2 ] At equilibrium more product At equilibrium more reactant than exists. than reactant product exists. COCl2 → CO + Cl2 → K eq [CO][Cl2 ] -4 o = = 7.6 x 10 at 400 C [COCl2 ] 43 When the molar concentrations of all species in an equilibrium reaction are known, the Keq can be calculated by substituting the concentrations into the equilibrium constant expression. 44 Calculate the Keq for the following reaction on concentrations of PCl5 = 0.030 mol/L, PCl3 = 0.97 mol/L and Cl2 = 0.97 mol/L at 300oC. PCl5(g ) → PCI3(g) + Cl2(g) → K eq [PCl3 ][Cl2 ] (0.97)(0.97) = 31 = = (0.030) [PCl5 ] 45 Ionization Constants 46 In addition to Kw, several other ionization constants are used. 47 Ka 48 When acetic acid ionizes in water, following equilibrium is established: + → HC H O H +C CH HO 3 → 2 2(aq) 22 33 the 2 Ka is the ionization constant for this equilibrium. + 2 [H ][C2H3O ] Ka = [HC2H3O2 ] Ka is called the acid ionization constant. Since the concentration of water is large and does not change appreciably, it is omitted from Ka. 49 At 25oC, a 0.100 M solution of HC2H3O2 is 1.34% ionized and has an [H+] of 1.34 x 10-3 mol/L. Calculate Ka for acetic acid. + → HC2H3O2(aq) H + C2H3O22 → Because each molecule of HC2H3O2 that + ionizes yields one H and one C 2H33O22 , the concentrations of the two ions are equal. + -3 [H ] = [C2H3O2 ] = 1.34 x 10 mol/L The moles of unionized acetic acid per liter are 0.100 mol/L – 0.00134 mol/L = 0.099 mol/L 50 Substitute these concentrations into the equilibrium expression and solve for Ka. + -3 [H ] = [C2H3O2 ] = 1.34 x 10 mol/L [HC2H3O2] = 0.099 mol/L + 2 [H ][C2H3O ] Ka = [HC2H3O2 ] -3 -3 (1.34x10 )(1.34x10 ) -5 = 1.8x10 (0.099) 51 What is the [H+] & pH in a 0.50 M HC2H3O2 solution? The ionization constant, Ka, for HC2H3O2 is 1.8 x 10-5. The equilibrium expression and Ka for HC2H3O2 are: + → HC2H3O2(aq) H + C2H3O22 → + 2 [H ][C2H3O ] Ka = [HC2H3O2 ] Because each molecule of HC2H3O2 that + ionizes yields one H and one C2 H33O22 , the concentrations of the two ions are equal. 52 What is the [H+] in a 0.50 M HC2H3O2 solution? The ionization constant, Ka, for HC2H3O2 is 1.8 x 10-5. The equilibrium expression and Ka for HC2H3O2 are: + → HC2H3O2(aq) H + C2H3O22 → + 2 [H ][C2H3O ] Ka = [HC2H3O2 ] + 2 [H ] = [C2H3O ] Let x = [H+] 2 = [C2H3O ] [HC2H3O2] at equilibrium is 0.50 – x. 53 Substitute these values into Ka for HC2H3O2. x= [H+] 2 = [C2H3O ] HC2H3O2 = 0.50 - x + 2 [H ][C2H3O ] Ka = [HC2H3O2 ] 2 (x)(x) x -5 Ka = = = 1.8 x 10 0.50 - x 0.50 - x Assume x is small compared to 0.50 - x. Then 0.50 – x 0.50 54 Substitute these values into Ka for HC2H3O2. x= [H+] 2 = [C2H3O ] HC2H3O2 = 0.50 - x + 2 [H ][C2H3O ] Ka = [HC2H3O2 ] 2 (x)(x) x -5 Ka = = = 1.8 x 10 0.50 0.50 2 x -5 = 1.8 x 10 0.50 Solve for x2. 2 x = 0.50 x 1.8 x 10 -5 55 Take the square root of both sides of the 2 -5 -5 -6 equation. x == 0.90 0.50 x 1.8 10 x=10 9. 0 x 10 -6 -3 x = 9.0 x 10 = 3.0 x 10 + [H ] = 3.0 x 10 -3 mol/L mol/L -3 pH = -log(3.0 x 10 ) = 2.52 Making no approximation and using the quadratic equation the answer is 2.99 x 10-3 mol/L, showing that it was justified to assume Y was small compared to 0.5. 56 Calculate the percent ionization in a 0.50 M HC2H3O2 solution. The ionization of a weak acid is given by HA → H+ + A- → The percent ionization is given by + - concentration of [H ] or [A ] x 100= percent ionized initial concentration of [HA] 57 Calculate the percent ionization in a 0.50 M HC2H3O2 solution. The ionization of acetic acid is given by + → HC2H3O2(aq) H + C2H33O22 → The percent ionization of acetic acid is given by concentration of [H+ ] or [C2H3O-2 ] x 100= percent ionized initial concentration of [HC2H3O2 ] -3 + [H ] was 3.0 x 10 mol / L previously calculated x 100= 0.60% percent ionized -3 as 3.0 x 10 mol/L 0.50 mol/L 58 Ionization Constants (Ka) of Weak Acids at 25oC Acid Formula Ka Acid Formula Ka Acetic HC2H3O2 1.8 x 10-5 Hydrocyanic HCN 4.0 x10-10 Benzoic HC7H5O2 6.3 x 10-5 Hypochlorous HOCl 3.5 x10-8 Carbolic HC6H5O 1.3 x 10-10 Nitrous HNO2 4.5 x10-4 Cyanic HCNO 2.0 x 10-4 Hydrofluoric HF 6.5 x10-4 Formic HCHO2 1.8 x 10-4 59 Ionic Equilibria Problems 60 Use Ka as a measure of acid strength. Let 1 = Hi and 5 = Lo Acid Ka order HC2H3O2 1.8 10–5 H2S 6 10–8 HF 7 10– 4 HC2Cl3O2 2 10–1 HCN 7.2 10–10 61 Use Ka as a measure of acid strength. Let 1 = Hi and 5 = Lo Acid Ka HC2H3O2 1.8 10–5 order 3 H2S 6 10–8 4 HF 7 10– 4 2 HC2Cl3O2 2 10–1 1 HCN 7.2 10–10 5 62 Find the pH in 0.1 M HCN if Ka = 7.2 10–10 Step 1: ionize the acid + → HCN H CN → Step 2: write the Ka expression + - [H ][CN ] Ka = [HCN] 63 Step 3: let x = [H+] = [CN–] [HCN] = 0.1 M – x ≈ 0.1 M Step 4: substitute into Ka expression [x][x] 10 Ka = 7.2 10 [0.1] Step 5: solve for x: x = 7.2 10 2 11 x 7.2 10 11 8.5 10 6 64 Step 6: solve for pH 6 pH - log 8.5 10 5.07 65 Find [H+] in 0.1 M H2CO3 if Ka = 4.0 10–17 Step 1: ionize the acid → H2CO3 → 2H CO + -2 3 Step 2: write the Ka expression + 2 -2 3 [H ] [CO ] Ka = [H2CO3 ] 66 Step 3: let x = [H+] , [CO3–2] = ½ x [H2CO3] = 0.1 M – ½ x ≈ 0.1 M Step 4: substitute into Ka expression 2 x 2 3 [x] [ ] x 17 Ka = 4.0 10 [0.1] 0.2 Step 5: solve for x: x = 8.0 10 3 18 x 8.0 10 3 18 2.0 10 6 67 Find the pH of a 0.001 M solution of bombastic hydroxide (BmOH) if Kb = 1.6 10–10 Step 1: ionize the base + → BmOH Bm OH → Step 2: write the Kb expression + - [Bm ][OH ] Kb = [BmOH] 68 Step 3: let x = [Bm+] = [OH–] [BmOH] = 0.001 M – x ≈ 0.001 M Step 4: substitute into Kb expression [x][x] 10 Kb = 1.6 10 [0.001] Step 5: solve for x: x = 1.6 10 2 13 x 1.6 10 13 4.0 10 7 69 Step 6: solve for pOH 7 pOH - log 4.0 10 6.40 Step 7: solve for pH pH 14 6.40 7.60 70 • On the yellow “Equilibria in Aqueous Solutions” Self-Test, you may omit problems 11 – 16. • On the pink “Redox and Electrochemistry: Self-Test, you may omit all problems except 5 & 6. 71 Find the Ka for a 0.01 M HCN solution if the pH = 6.3 Step 1: ionize the acid + → HCN H CN → Step 2: write the Ka expression + - [H ][CN ] Ka = [HCN] 72 Step 3: use pH to find [H+] [H+] = 10–pH = 10–6.3 = 5.0 10–7 [H+] = [CN–], [HCN] = 0.01- 5.0 10–7 ≈ 0.01 Step 4: substitute into Ka expression 7 7 [5.0 10 ][5.0 10 ] 11 Ka = 2.5 10 [0.01] 73 Solubility Product Constant 74 The solubility product constant, Ksp, is the equilibrium constant of a slightly (sparingly) soluble salt. 75 Silver chloride is in equilibrium with its ions in aqueous solution. AgCl(s) → Ag+(aq) + Cl-(aq) → The equilibrium constant+is -- K sp = [Ag ][Cl ]] [Ag ][Cl The product of Keq=and K eq [AgCl(s) is a constant.[AgCl(s )] + Rearrange + - Keq x [AgCl(s)] = [Ag ][Cl ] = Ksp The The amount concentration of solid AgClof does solid not AgClaffect is a the equilibrium. constant. 76 Ksp for Sparingly Soluble Solids To find Ksp: 1. ionize the salt 2. write the Ksp expression 3. determine the M’s of the ions 4. substitute into the Ksp expression 77 The solubility of AgCl in water is 1.3 x 10-5 mol/L. AgCl(s) → Ag+(aq) + Cl-(aq) Because each formula unit of AgCl that dissolves yields one Ag+ and one Cl- , the concentrations of the two ions are equal. → [Ag+] = [Cl-] = 1.3 x 10-5 mol/L Ksp = [Ag+][Cl-] The Ksp has no denominator. = (1.3 x 10-5)(1.3 x 10-5) = 1.7 x 10-10 78 Molar Solubility for Sparingly Soluble Solids 1. 2. 3. 4. 5. 6. ionize the salt write the Ksp expression let x equal the amount of salt that dissolves find concentrations of ions in terms of x substitute in the Ksp expression solve for x 79 The Ksp value for lead sulfate is 1.3 x 10-8. Calculate the solubility of PbSO4 in moles per liter. The equilibrium equation of PbSO4 is 2+ 2→ PbSO4 Pb + SO4 → The Ksp of PbSO4 is 2+ 2K sp = [Pb ][SO4 ] Because each formula unit of PbSO4 that 22+ dissolves yields one Pb and one SO 4 , the concentrations of the two ions are equal. 2+ 24 Let x = [Pb ] = [SO ] 80 The Ksp value for lead sulfate is 1.3 x 10-8. Calculate the solubility of PbSO4 in moles per liter. Substitute x into the Ksp equation. 2+ 24 K sp = [Pb ][SO ] K sp = (x)(x) = 1.3 x 10 2 x = 1.3 x 10 -8 x = 1.3 x 10 -8 -8 -4 = 1.1 x 10 mol/L The solubility of PbSO4 is 1.1 x 10-4 mol/L. 81 Solubility Product Constants (Ksp) at 25oC Compound Ksp Compound Ksp AgCl 1.7 x 10-10 CaF2 3.9 x 10-11 AgBr 5 x 10-13 CuS 9 x 10-45 AgI 8.5 x 10-17 Fe(OH)3 6 x 10-38 AgC2H3O2 2 x 10-3 PbS 7x 10-29 Ag2CrO4 1.9 x 10-12 PbSO4 1.3 x 10-8 BaCrO4 8.5 x 10-11 Mn(OH)2 2.0 x 10-13 BaSO4 1.5 x 10-9 82 Practice 83 The Ksp value for silver carbonate is 6.2 x 10-12. Calculate the solubility of Ag2CO3 in moles per liter. + 2→ Ag2CO3 2Ag + CO3 → + 2 23 K sp = [Ag ] [CO ] If x = amount of salt that dissolves, then [Ag+] = 2x, [CO3–2] = x 84 2 K sp = [2x] [x] = 4x3 4x3 = 6.2 x 10-12 x3 = 1.6 x 10-12 3 -12 x = 1.6 x 10 x = 1.2 x 10-4 mol/L The solubility of silver carbonate is 1.2 x 10-4 mol/L 85 The Ksp value for calcium phosphide is 1.7 x 10-40. Calculate the solubility of Ca3P2 in moles per liter. Solution in lecture! 86 Acid-Base Properties of Salts 87 hydrolysis is the term used for the general reaction in which a water molecule is split. 88 Salts that contain the anion of a weak acid undergo hydrolysis. The net ionic equation for the hydrolysis of sodium acetate is → C2H3O (aq) + HOH(l) HC2H3O2 (aq) + OH (aq) → 2 The water The solution molecule is splits. basic. 89 Salts that contain the cation of a weak base undergo hydrolysis. The net ionic equation for the hydrolysis of ammonium chloride is + → NH (aq) + HOH(l ) NH3 (aq) + H3O (aq) → + 4 The Thesolution water molecule is acidic. splits. 90 Salts derived from a strong acid and a strong base do not undergo hydrolysis. Na+ + Cl- + HOH(l ) → no reaction 91 Ionic Composition of Salts and the Nature of the Aqueous Solutions They Form Type of salt Nature of Aqueous Solution Examples Weak base-strong acid Acidic NH4Cl, NH4NO3 Strong base-weak acid Basic NaC2H3O2 Weak base-weak acid Close to Neutral NH4C2H3O2, NH4NO2 Strong base-strong acid Neutral NaCl, KBr 92 Buffer Solutions: The Control of pH 93 A buffer solution resists changes in pH when diluted or when small amounts of acid or base are added. 94 A weak acid mixed with its conjugate base form a buffer solution. Sodium acetate when mixed with acetic acid forms a buffer solution. 95 A weak acid mixed with its conjugate base form a buffer solution. If a small amount of HCl is added, the acetate ions of the buffer will react with the H+ of the HCl to form unionized acetic acid. H (aq) + C2H3O (aq) HC2H3O2 (aq) + 2 96 A weak acid mixed with its conjugate base form a buffer solution. If a small amount of NaOH is added, the acetic acid molecules of the buffer will react with the OH– of the NaOH to form water. OH + HC2H3O2 (aq) H2O(l ) + C2H3O (aq) - 2 97 HC2H3O2 NaOH NaC2H3O2 HCl NaC2H3O2 + H2O NaCl + HC2H3O2 99