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- Solubility products -Thermochemistry -Thermodynamics Dr. G. Birungi 1 Solubility • Definition – The amount of a substance that dissolves in a given quantity of solvent at a given temperature to form a saturated solution (g/L) – Question: What happens when we add a solution of a salt to a solution of another salt? AgNO3(aq) + NaCl(aq) ?? KNO3(aq) + NaCl(aq) ?? • Consider quantitative predictions – A reaction will only happen between salts in solution if ions are removed from solution • Ions removed from solution by formation of a precipitate 2 Solubility- Rules – Salts of group 1 metals, NH4+, nitrates and acetates are soluble – Chlorides, bromides and iodides are soluble except those of Ag+, Pb2+, Hg22+. – Sulfates are soluble except those of Pb2+, Hg22+, Sr2+ and Ba2+. Ag2SO4 and CaSO4 are only slightly soluble. – Carbonates, phosphates and sulphites are insoluble (except those of group 1 and NH4+) – Sulfides (S2-) are insoluble except those of groups 1 and 2 and NH4+. – Hydroxides are insoluble except those of groups 1 2 and NH4+. Hydroxides of group 2(Ca2+, Sr2+ or Ba2+) are slightly soluble 3 Formation and Dissolution of Precipitates • Solubility: maximum amount of solute that will dissolve in a given amount of solvent (depends on solvent, temperature and pressure) No compound is infinitely soluble and no compound is perfectly insoluble. Solute Solubility (g solute/100 g solvent) Qualitative Solubility Description Less than 0.1 Insoluble 0.1 – 1 Slightly soluble 1 – 10 Soluble Greater than 10 Very soluble 1–4 Formation and Dissolution of Precipitates • Saturated solution – Contains maximum concentration of solute • Equilibrium between undissolved and dissolved solute. • Unsaturated solution – Contains less than the maximum solute • Supersaturated solution – Contains more than maximum amount of solute • Often reached by heating followed by cooling • Solutes (even those called “soluble”) have a limited solubility in a particular solvent. • Slightly soluble (often called “insoluble”) ionic compounds have a relatively low solubility – Reach equilibrium with little solute dissolved – Heterogeneous equilibrium 5 Solubility curve • Explore how environmental factors affect solubility (amount of solute present in solutions) – Example how does temperature affect solubility? – How does pressure affect solubility? 6 Solubility Curve 7 Any solution can be made saturated, unsaturated, or supersaturated by changing the temperature 8 Solubility curve - Gas 9 Effect of temperature and pressure summary • Solids and liquids – Solubility increases with temperature (decreases with a decrease in temp) – Liquids and solids exhibit practically no change of solubility with changes in pressure. • Gases – As the temperature increases, the solubility of a gas decrease as shown by the downward trend – There is an increase in solubility with an increase in pressure. (Henry's Law) 10 Why is this Important? Dissolving and Precipitation occurs around us: – Tooth enamel dissolves in acidic solution (tooth decay) – Precipitation of certain salts in kidneys causes kidney stones – Waters of Earth contains dissolved salts as water passes over and through the ground – Precipitation of CaCO3 from groundwater is responsible for cave formation. Let’s look at the factors that affect solubility! 11 Example • Use the solubility chart on the left to answer the following questions: – At 15 0C, which of the salts is: • Most soluble • Least soluble – What mass of solute will dissolve in 100mL of water at the following temperatures. • KNO3at 70°C • NaCl at 100°C • NH4Cl at 90°C 12 Solubility-Product Constant (Ksp) • Solubility-product constant (Ksp): equilibrium constant for equilibrium between slightly soluble ionic solid and a solution of its ions – Indicates how soluble the solid is in water • Solubility: quantity that dissolves to form a saturated solution (g/L) • Molar solubility: number of moles of solute that dissolves in forming a liter of saturated solution of solute (mol/L) • Solubility depends on concentrations of other ions and pH but Ksp is a constant. 13 Solubility-Product Constant (Ksp) Practice: Write an ionic equation for the dissolution, and the equation for the solubility product for: (a) Calcium carbonate (b) Magnesium hydroxide (c) Ag3PO4 • Magnitude of Ksp is a measure of how far to the right dissolution proceeds at equilibrium (saturation). – Used to compare solubility 14 Ksp of Selected Ionic Compounds (25 °C) Name, Formula Ksp Aluminum hydroxide, Al(OH)3 3 x 10-34 Cobalt(II) carbonate, CoCO3 1.0 x 10-10 Iron(II) hydroxide, Fe(OH)2 4.1 x 10-15 Lead(II) fluoride, PbF2 3.6 x 10-8 Lead(II) sulfate, PbSO4 Mercury(I) iodide, Hg2I2 Silver sulfide, Ag2S Zinc iodate, Zn(IO3)2 1.6 x 10-8 4.7 x 10-29 8 x 10-48 3.9 x 10-6 See Appendix D in your book for a much more extensive list. 1–15 Example Which of the following compounds will have the greatest molar solubility in water A) AgCl Ksp = 1.8 x 10-10 B) AgBr Ksp = 5.0 x 10-13 C) AgI Ksp = 8.3 x 10-17 Give an explanation/reason for your answer 16 Solubilities and Solubility Products • Ksp for a slightly soluble solid can be determined from its solubility – as long as there is no other reaction • Example 1: – The aqueous solubility for compound PbI2 is 0.54 grams/100 ml at 25 degrees Celsius. What is the Ksp of PbI2 at 25 oC ? 17 • Convert 0.54 grams of PbI2 to mols – 0.54 461.0 =0.001167137 mols of PbI2 • Covert ml to L – 100 1000 = 0.1 • Find molarity – 0.001167137 0.1 = 0.011714𝑀 𝑜𝑓 𝑃𝑏𝐼2 • 𝑃𝑏𝐼2 ⇋ 𝑃𝑏 2+ + 2𝐼− • 𝐾𝑠𝑝 = 𝑃𝑏 2+ [𝐼 − ]2 • 𝐾𝑠𝑝 = 0.011714 × (0.023427)2 = 6.4 x 10−6 18 Solubilities and Solubility Products • Example 2: – Determine the molar solubility of MgF2 from its solubility product (Ksp = 6.4 x 10-9). 19 Example 3 • Calculate the molar solubility of calcium fluoride, CaF2 (Ksp = 3.9 x 10-11) 20 Example 4 • The Ksp for LaF3 is 2.0 10-19. What is the solubility of LaF+ in water in moles per litre? 21 Example 5 • Calculate the molar solubility of AgCl at 250C, given the relevant Ksp value is 1.82 10 -10 22 Factors that Affect Solubility: Common Ion Effect • The presence of a common ion decreases the solubility of a slightly soluble ionic compound. • The shift in equilibrium that occurs because of the addition of an ion already involved in the equilibrium reaction. AgCl(s) ⇌ Ag+(aq) + Cl-(aq) adding NaCl( aq ) shifts equilibrium position 23 The effect of a common ion on solubility CrO42- added PbCrO4(s) Pb2+(aq) + CrO42-(aq) PbCrO4(s) Pb2+(aq) + CrO42-(aq) 24 Example Calculate the molar solubility of MgF2 in 0.10 mol dm-3 MgCl2 at 25 C. Ksp of MgF2 = 7.4 x 10-11 25 Example The solubility of Ca(OH)2 in water is 0.012 mol dm-3. What is its solubility in 0.10 mol dm-3 Ca(NO3)2? Ksp of Ca(OH)2 is 6.5 x 10-6 26 Le Châtelier’s Principle Determine the effects of solubility when each of the following is added to a mixture of the slightly soluble solid NiCO3 and water at equilibrium: (a)Ni(NO3)2 (b)KClO4 (c) K2CO3 (d) HNO3 27 Effect of pH on solubility • [H3O+] can have a profound effect on the solubility of an ionic compound. – Solubility of slightly soluble salts containing basic anions increases as [H+] increases • More basic anion…more solubility is influenced by pH • For example -The molar solubility of a sparingly soluble hydroxide increases as the pH is decreased 28 Effect of pH on solubility • Example 𝑀𝑔(𝑂𝐻)2 (𝑠) ⇌ 𝑀𝑔2+ (𝑎𝑞) + 2𝑂𝐻 − (𝑎𝑞) – If we add an acidic buffer, what will happen to the solubility of Mg(OH)2 ? Explain • By adding an acid we increase [H+] and consequently decrease [OH-] as Kw must be maintained • If we decrease the amount of OH- in the system, the position of equilibrium will shift to the right • If the position of equilibrium moves to the right then the concentration of Mg2+ increases • As [Mg(OH)2] = [Mg2+], the molar solubility of magnesium hydroxide increases as the pH is lowered. 29 Effect of pH on solubility • Predict the effect on solubility of the following if a strong acid is added. – CaCO3(s) ⇌ Ca2+(aq) + CO32-(aq) – AgCl(s) ⇌ Ag+(aq) + Clˉ(aq) – Ca(OH)2(s) ⇌ Ca2+(aq) + 2OHˉ(aq) 30 If limestone (CaCO3) deposit is near surface…sinkhole 31 If limestone (CaCO3) deposit is well below the surface….caves A view inside Carlsbad Caverns, New Mexico 32 Cango Caves, South Africa 1–33 Sudwala Caves, South Africa 1–34 Predicting the Formation of a Precipitate Compare Qsp to Ksp to predict if a precipitate will form and, if not, what concentrations of ions will cause it to do so. Qsp = Ksp Qsp > Ksp Qsp < Ksp soln is saturated & no changes occur ppt forms until soln is saturated soln is unsaturated & no ppt forms 35 Practice • Example 1: Determine whether CaHPO4 will precipitate from a solution with [Ca2+] = 0.0001 mol dm-3 and [HPO42-] = 0.001 mol dm-3. • Example 2: Does silver chloride precipitate when equal volumes of a 2 x 10-4 mol dm-3 solution of AgNO3 and a 2 x 10-4 mol dm-3 solution of NaCl are mixed? AgNO3(aq) + NaCl(aq) AgCl(s) + NaNO3(aq) 36 Practice Will a precipitate form when 0.10 dm3 of 8.0 x 10-3 mol dm-3 Pb(NO3)2 is added to 0.40 dm3 of 5.0 x 10-3 mol dm-3 Na2SO4? Ksp for PbSO4 = 6.3 x 10-7 37 Concentration Necessary to Form a Ppt • We can also determine the concentration of an ion necessary for precipitation to begin. • Assume that precipitation begins when Qsp = Ksp • Example: If a solution contains 0.0020 mol CrO42- per dm3, what concentration of Ag+ ion must be added as AgNO3 before Ag2CrO4 begins to precipitate. (Neglect any increase in volume upon adding the solid silver nitrate.) 38 Thermochemistry • Thermochemistry is the study of energy changes that occur during chemical reactions • Focus is on heat and matter transfer between the system and the surroundings 39 Energy • The ability to do work or transfer heat. – Work is done when the energy applied causes an object that has mass to move. • Kinetic energy - energy of motion • Potential energy – energy due to position – energy associated with forces of attraction and repulsion between objects 40 Units of Energy • The SI unit of energy is the joule (J). kg m2 1 J = 1 s2 • An older, non-SI unit is still in widespread use called the calorie (cal). 1 cal = 4.184 J 41 Conservation of Energy • Energy is neither created nor destroyed. • In other words, the total energy of the universe is a constant; if the system loses energy, it must be gained by the surroundings, and vice versa. ∆ E = Efinal - Einitial 42 • The energy of a system depends only on its present state, not on the path by which the system arrived at that state. • So, E depends only on Einitial and Efinal. – In the system below, the water could have reached room temperature from either direction. 43 Exchange of Heat - System and Surroundings • When heat is released by the system to the surroundings, the process is exothermic • When heat is absorbed by the system from the surroundings, the process is endothermic. 44 Endothermicity & Exothermicity • Process is exothermic when H is negative. • Process is endothermic when H is positive. 45 Atoms & Molecules Possess Energy • Kinetic energy – proportional to the absolute temperature • Potential energy – chemical bonds hold atoms together, forces of attraction give rise to a compound’s potential energy (chemical energy) bonds break - increase PE of system bonds made - lower PE of system Changes in chemical energy occur during chemical reactions 46 Heat Capacity & Specific Heat Capacity • Heat capacity - the amount of energy required to raise the temperature of a substance by 1 K (1C). • Specific heat capacity - the amount of energy required to raise the temperature of 1 g of a substance by 1 K. 47 Heat Capacity & Specific Heat Capacity To calculate a quantity of heat (q) heat = specific heat mass temperature change q = J = m T J g-1 K-1 g K s 48 Specific Heats… q = s m T If ∆T > 0, then q > 0 and heat is gained by the system • If ∆T < 0, then q < 0 and heat is lost by the system • Molar heat capacity is the product of specific heat times the molar mass of a substance (units are J mol–1 K–1) 49 Examples • Pg 140 • How much heat is needed to warm 250 g of water from 220C (about room temperature) to near its boiling point, 980C, given the specific heat capacity of water is 4.18 Jg-1K-1? – 7.9 104 J • What is the molar heat capacity of water? – 75.2 Jmol-1K-1 50 Calorimetry • Calorimetry is a technique used to measure heat exchange in chemical reactions • A calorimeter is the device used to make heat measurements. • Calorimetry is based on the law of conservation of energy. • qwater = - q surroundings 51 Coffee Cup Calorimeter • We indirectly measure the heat change for the system by measuring the heat change for the water in the calorimeter • The specific heat for water is well known (4.184 J/g/K), • We can measure H for the reaction with this equation: q = m s T 52 Calorimetry example • How much heat is given off when an 869 g iron bar cools from 94 °C to 5 °C? (specific heat of Fe = 0.444 J g-1 K-1) ? q = m x s x Dt Dt = tfinal – tinitial = 5°C – 94°C = -89°C q = 869 g x 0.444 J/g/K x -89 K = -34,000 J (q < 0, heat lost!) 53 Example In a neutralization reaction between a strong acid and a strong base, the following reaction forms water. H+(aq) + OH-(aq) H2O(l) If 25.00 mL of 2.50 M HCl and 25.00 mL of 2.50 M NaOH, both at 21.1 0C react and the final temperature is 37.8 0C, determine the heat of the neutralization reaction expressed per mole of H2O formed. 54 Solution: • Assume that 50.00 mL of water forms which absorbs all the heat. Heat of reaction = qneutr = -qcalorim qcalorim = m x s x ∆T = (50.00 mL x 1.00 g/mL) x 4.18 J/g/K x (37.8 – 21.1) K = 3.5 x 103 J qneutr = -qcalorim = -3.5 kJ 55 n(H+) = C x V = 2.50 mol dm-3 x 25.00 x 10-3 dm3 = 0.0625 mol H+ H+(aq) + OH-(aq) → H2O(l) n(H+) = n(OH-) = n(H2O) = 0.0625 mol Amount of heat produced per mole of H2O is qneutr = - 3.5kJ/0.0625 mol = - 56 kJ/mol H20 (Since q < 0, heat is lost, that is neutralisation reaction is exothermic!) [Do example 4.6 in textbook] 56 Enthalpy • Enthalpy (H) is used to quantify the heat flow into or out of a system in a process that occurs at constant pressure • Standard enthalpy of reaction (Hrxn): the enthalpy change for the transformation of reactants in their standard states to products in their standard states. • There are many types of heats of reaction, e.g. – heat of formation, – heat of combustion, – heat of hydrogenation. 57 Standard State • Standard state is the reference state for the material's thermodynamic state properties. • Required for comparison purposes. • Standard state of any substance - the physical state at which it is most stable at 1 bar and 298 K. • When a reaction occurs at constant pressure, the heat of reaction is equal to the enthalpy change. 58 Enthalpies of Reaction • The change in enthalpy, H, is the enthalpy of the products minus the enthalpy of the reactants: H = Hproducts − Hreactants 59 Hess’s Law • Hess’s law states that “If a reaction is carried out in a series of steps, H for the overall reaction will be equal to the sum of the enthalpy changes for the individual steps.” 60 Hess’s Law • The total enthalpy change depends only on the initial state of the reactants and the final state of the products. 61 Calculation of H We can use Hess’s law in this way: Horxn= nHof(products) - mHof(reactants) where n and m are the stoichiometric coefficients. 62 Heat of Combustion • The heat of combustion is the heat released in the reaction of one mole of a substance in its standard state with oxygen. • The equation for the combustion of a compound must show the reaction of one mole of the compound with sufficient oxygen to convert all of the carbon and hydrogen present to gaseous CO2 and liquid H2O. • C6H6(l) + 7½O2(g) 6CO2(g) + 3H2O(l) H° = -3274 kJ • CH4(g) + 2O2(g) CO2(g) + 2H2O(l) H° = -890.2 kJ 63 Example The heat of combustion of ethane gas, C2H6(g), is –1560.1 kJ mol-1. If H for CO2(g) and H2O(l) are –393.5 kJ mol-1 and –285.8 kJ mol-1 respectively, calculate the heat of formation of ethane. combustion equation: C2H6(g) + 3½O2(g) 2CO2(g) + 3H2O(l) Horxn = nHof (products) - mHof (reactants) = [(2 mol)Hof (CO2) + (3 mol)Hof (H2O)] [(1 mol)Hof (C2H6) + (3½ mol)Hof (O2)] -1560.1 kJ = (2 x -393.5) + (3 x -285.8) -Hof (C2H6) - 0 kJ Hof (C2H6) = -84.3 kJ mol-1 64 Enthalpy of Formation • The standard heat of formation of a compound (H°f): the heat change when one mole of the compound in its standard state is formed from its elements in their standard states at a specified temperature. • The standard enthalpy of formation of an element in its standard state is taken to be zero. 65 Enthalpies of Formation 66 Standard Enthalpies of Formation Standard enthalpies of formation, H0f, are measured under standard conditions (25°C and 1.00 atm pressure). 67 Calculate the standard enthalpy of formation of CS2 (l) given that: C(graphite) + O2 (g) S(rhombic) + O2 (g) CS2(l) + 3O2 (g) CO2 (g) H0rxn = -393.5 kJ SO2 (g) H0rxn = -296.1 kJ CO2 (g) + 2SO2 (g) H0rxn = -1072 kJ 1. Write the enthalpy of formation reaction for CS2 C(graphite) + 2S(rhombic) CS2 (l) 68 2. Add the given rxns so that the result is the desired rxn. (molar ratios!) C(graphite) + O2 (g) CO2 (g) H0rxn = -393.5 kJ 2S(rhombic) + 2O2 (g) 2SO2 (g) H0rxn = -296.1x2 kJ 0 + CO2(g) + 2SO2 (g) CS2 (l) + 3O2 (g) Hrxn = +1072 kJ C(graphite) + 2S(rhombic) CS2 (l) H0rxn = -393.5 + (2x-296.1) + 1072 = 86.3 kJ 69 Benzene (C6H6) burns in air to produce carbon dioxide and liquid water. How much heat is released per mole of benzene combusted? The standard enthalpy of formation of benzene is 49.04 kJ/mol. 2 C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l) 0 Hrxn = nH0f (products) -mH0 f (reactants) 0 Hrxn = [12H0f (CO2) +6H0 f(H2O)] - [2H0f (C6H6)] 0 Hrxn = [ 12x–393.5 + 6x–285.8 ] – [ 2x49.04 ] = -6535 kJ -6535 kJ = - 3267 kJ/mol C6H6 2 mol 70 Application of Hess’s Law Find H of NO2(g) given that ½N2(g) + ½O2(g) NO(g) 2NO2(g) 2NO(g) + O2(g) Desired equation: ½ N2(g) + O2(g) NO2(g) Hf = 90.25 kJ …..(1) Hf = 114.14 kJ …(2) H = ? 2NO(g) + O2(g) 2NO2(g) NO(g) + ½ O2(g) NO2(g) H = -114.14 kJ H = -57.07 kJ..(-2)/2 + ½O2(g) NO(g) NO(g) + ½O2(g) NO2(g) ½N2(g) + O2(g) NO2(g) H = 90.25 kJ H = -57.07 kJ H = 33.18 kJ ½N2(g) 71 Bond Energies • bond breaking - endothermic • bond formation – exothermic • Bond energies indicate the amount of energy required to break a particular bond and are therefore positive. H = [sum of bond energies of all bonds in reactants (bonds broken)] - [sum of bond energies of all bonds in products (bonds formed)] 72 Example - hydrogenation reaction C2H4(g) + H2(g) C2H6(g) Bond energies of reactants = EC=C + 4EC-H + EH-H = 606.1 + 4(409.64) + 430.5 kJ = 2675.16 kJ Bond energies of products = EC-C + 6EC-H = 334.4 + 6(409.64) kJ = 2792.24 kJ H = 2675.16 - 2792.24 kJ = -117.08 kJ mol-1 73 Thermodynamics Spontaneous Change: Entropy and Free Energy 74 • In previous chapters we have studied: – How fast does the change occur – How is rate affected by concentration and temperature – How much product will be present • Why does a change occur in the first place? 75 Chemical Thermodynamics…What are the Driving Forces? • Methane burns but not the reverse CH4 + 2O2 CO2 + 2H2O • Steel chain rusts, but a rusty one will not become shiny • A cube of sugar dissolves in a cup of coffee after a few seconds of stirring, but not the reverse 76 Spontaneity • How can we tell if a reaction will proceed or not? – Some chemical and physical changes take place by themselves, given enough time. • A spontaneous chemical reaction is one that, given sufficient time, will achieve chemical equilibrium, with an equilibrium constant greater than 1, by reacting from left to right. 77 Spontaneous Reaction Cu(s) + Cl2(g) CuCl2(s) spontaneous 2H2(g) + O2(g) 2H2O(g) • spontaneous reaction but occurs only if you ignite the mixture 78 Nonspontaneous Reaction O3(g) O(g) + O2(g) nonspontaneous *This does not mean that it does not occur at all. It means that, when equilibrium is achieved, not many O3 molecules have broken down into products.* That is [O][O 2 ] K 0.0063 1 at 25 C [O3 ] 79 Chemical Thermodynamics • Thermodynamics lets us predict whether a process will occur given enough time. • Spontaneous process: occurs by itself without ongoing input of energy from outside system – Freezing of water at 1 atm and –5 °C – Burning or falling may need a little “push” to get started, but will continue without external aid • Nonspontaneous process: does not occur unless we make it happen; system must be supplied with continuous input of energy – A book rises only if something else supplies energy 80 Spontaneity • Spontaneous does not mean instantaneous and has nothing to do with the rate. – Given enough time it will occur by itself – Chemical reaction proceeding toward equilibrium is an example of a spontaneous change. If a change is spontaneous in one direction, it is not spontaneous in the other. 81 Mechanical Systems • Some spontaneous processes occur with loss of energy. • change in the direction which lowers their energy • position of equilibrium is when the energy “available” is minimized (energy has different forms: thermal, potential, kinetic) 82 System and Surroundings • System – portion of the universe we wish to study. • Surroundings – everything else. Universe = System + Surroundings Example: In the chemistry lab, system is usually a flask, beaker, etc and the surrounds are the rest of the laboratory. 83 Chemical Systems • The enthalpy (H) is a measure of the total energy of a “system” (that part of the universe we are considering – the rest is called the “surroundings”). • Can we use the enthalpy to predict the direction of chemical change? • Some scientists though that the sign of H determined spontaneity. 84 Exothermic Physical: CaCl2(s) Ca2+(aq) + 2Cl-(aq) Chemical: 8Al(s) + 3Fe3O4(s) 4Al2O3(s) + 9Fe(l) reactants H products The enthalpy (energy) of the chemical system is lowered. 85 Endothermic Physical: H2O(s) H2O(l) H2O(g) NH4Cl(s) NH4+(aq) + Cl-(aq) Chemical: Ba(OH)28H2O(s) + 2NH4NO3(s) Ba(NO3)2(aq) + 2NH3(g) + 10H2O(l) products H reactants Some spontaneous chemical reactions are endothermic (enthalpy increases). 86 Enthalpy • H cannot be used to predict if a reaction or process will go. • The reason is that it represents the total energy of the system. We need to examine the available energy (the energy available to do useful work). The unavailable energy per degree kelvin is known as the entropy (S) of a system. 87 Identifying Spontaneous Processes Predict whether the following are spontaneous? 1. When a piece of metal heated to 150 C is added to water at 40 C, water gets hotter. 2. Water at room temperature decomposes into H2(g) and O2(g). 3. Reaction of sodium metal and chlorine gas to form sodium chloride. 4. Reaction of nitrogen atoms to form N2 molecules at 25 C and 1 atm. How can we tell the direction of a spontaneous change when it is not as obvious? 88 Direction of Chemical Change G = H – TS – G ≡ available energy, Gibbs energy – H ≡ total energy, enthalpy – S ≡ unavailable energy per Kelvin, entropy – T ≡ temperature • The direction of chemical change is the direction which lowers the Gibbs energy. 89 Criterion for Spontaneity ΔG = ΔH – TΔS A process is spontaneous in the direction in which the Gibbs energy decreases: i.e. G is negative. At constant temperature, G < 0 for a spontaneous process G > 0 for a nonspontaneous process G = 0 for a process at equilibrium 90 For ΔG to be negative: ΔH negative ΔS positive negative negative but ΔH more negative than (-TΔS) (lower T) positive positive but (-TΔS) more negative than ΔH (higher T) 91 Spontaneity • melting of ice –endothermic –highly organised structure becomes less wellordered • octane combustion 2C8H18(g) + 25O2(g) → 16CO2(g) + 18H2O(g) –exothermic –27 molecules in the gas phase are converted to 34 molecules 92 Spontaneity Clearly two factors should be considered when trying to predict spontaneity: • a decrease in energy or enthalpy • an increase in disorder or entropy Sometimes these effects reinforce one another, but at other times they oppose one another. The final outcome is determined by their relative magnitudes. 93 Spontaneous Process • Matter changes from a more ordered to a less ordered state. A change in order is a change in the number of ways of arranging the particles, and is a key factor in determining the direction of a spontaneous process. 94 Disorder and Entropy • There is a natural tendency for a system to become disordered – For parts of a system to have more ways of being arranged. • Creating order requires work • Disorder increases when a process results in more ways for the atoms, ions, or molecules in the system to be arranged. 95 The number of ways to arrange a deck of playing cards 96 Entropy • The driving force for a spontaneous process is an increase in the entropy of the universe. • Entropy, S, can be viewed as a measure of randomness, or disorder. – Number of microstates of system is number of ways it can disperse its thermal energy among various modes of motion of all its particles – units of J K-1 mol-1 97 Third Law of Thermodynamics • A perfect crystal has zero entropy at a temperature of absolute zero Ssys = 0 at 0 K • “Perfect” means that all the particles are aligned flawlessly in the crystal structure with no defects of any kind. • At absolute zero, all particles in the crystal have their minimum energy, and there is only one way they can be arranged. • Because S is explicitly known (= 0) at 0 K, S values at other temperatures can be calculated. 98 Absolute Entropies • Elements in their standard states have an entropy greater than zero, i.e. S° > 0. • The entropy of any pure substance can be measured at a given temperature. • The absolute entropy of one mole of a substance in its standard state is called the standard molar entropy, S°. 99 Standard Molar Entropies • Standard molar entropy (S°): J K-1 mol-1(Appendix D) – Predicting relative values of S° – Affected by temperature, physical state, dissolution, and atomic or molecular complexity. 1. Temperature changes: S° increases as temp rises 2. Physical states and phase changes: -when a more ordered phase changes to a less ordered one, S +ve -S° increases as the substance changes from a solid to a liquid to a gas: Ssolid < Sliquid << Sgas 100 Ssolid < Sliquid << Sgas 101 Standard Molar Entropies 3. Dissolution of a solid or liquid: the entropy of a dissolved solid or liquid solute is greater than the entropy of the pure solute -Type of solute and solvent affect overall entropy change. 4. Dissolution of a gas: a gas is so disordered to begin with that it becomes more ordered when it dissolves in a liquid -Entropy of a solution of a gas in a liquid is always less than entropy of pure gas. 5. Atomic size or molecular complexity: S° (KCl) < S° (CaCl2) < S° (GaCl3) 102 Entropy and vibrational motion NO NO2 N 2 O4 103 Examples 1. Choose the member with the higher entropy: (a) 1 mol of SO2(g) or 1 mol of SO3(g) (b) 1 mol of CO2(s) or 1 mol of CO2(g) (c) 1 mol of KBr(s) or 1 mol of KBr(aq) (d) Seawater in midwinter (2 C) or midsummer (23 C) 2. Predict whether S is positive or negative for the following: (a) H2O(l) → H2O(g) (b) Ag+(aq) + Cl-(aq) → AgCl(s) (c) 4Fe(s) + 3O2(g) → 2Fe2O3(s) (d) N2(g) + O2(g) → 2NO(g) 104 Entropy changes for reversible phase transitions heat transferred ΔS temp at which change occurs q T 105 Melting of ice at normal boiling point H2O(s, 1 atm) → H2O(l, 1 atm) q ΔS T ΔH fusion Tmp 6.02 kJ mol -1 273.15 K 22.0 J K -1 mol -1 What is ΔS° when one mole of water freezes? 106 Calculating the Change in Entropy of a Rxn • Chemists are especially interested in learning to predict and calculate change in entropy as a reaction occurs. • Calculations are similar for enthalpy changes: – calculated from enthalpies of formation of reactants & products. Hrxn° = npHf(products) - nrHf(reactants) – calculated from standard molar entropies of reactants & products. Srxn° = npS(products) - nrS(reactants) 107 Example Calculate ΔS° for the reaction SiCl4(g) + 2Mg(s) → 2MgCl2(s) + Si(s) Substance: SiCl4(g) Mg(s) MgCl2(s) Si(s) S°/J K-1 mol-1: 330.73 32.68 89.62 18.83 A. B. C. D. E. -254.96 J K-1 -198.02 J K-1 198.02 J K-1 254.96 J K-1 471.86 J K-1 108 Example Calculate ΔS° for the reaction 2Cl2(g) + SO2(g) → SOCl2(g) + Cl2O(g) Substance: Cl2(g) S°/J K-1 mol-1: 223.0 A. B. C. D. E. SO2(g) 248.1 SOCl2(g) 309.77 Cl2O(g) 266.1 -118.2 J K-1 -104.8 J K-1 104.8 J K-1 118.2 J K-1 1270.0 J K-1 109 Gibbs Energy • If a process is nonspontaneous in one direction (G > 0), it is spontaneous in the opposite direction (G < 0) • Standard Gibbs Energy Change (Go) Go = Ho – TSo – Used frequently to find any one of these variables when given other two 110 Example Calculate Grxn° at 25 °C for the decomposition of potassium chlorate, one of the common oxidizing agents in explosives, fireworks, and match heads. 4KClO3(s) 3KClO4(s) + KCl(s) KClO3 KClO4 KCl Hof So -397.7 kJ mol-1 143.1 J K-1 mol-1 -432.8 kJ mol-1 151.0 J K-1 mol-1 -436.7 kJ mol-1 82.59 J K-1 mol-1 111 Example Determine the standard Gibbs energy change at 298 K for the reaction: 2NO(g) + O2(g) → 2NO2(g) 112 Calculating Change in Entropy/Enthalpy of Rxn Hrxn° = npHf(products) - nrHf(reactants) Srxn° = npS(products) - nrS(reactants) • Can also calculate Gorxn using standard Gibbs energy of formation (Gof) G = standard Gibbs energy change that occurs if reactants in their standard state are converted to products in their standard state. Grxn° = npGf(products) - nrGf(reactants) Use Gfvalues to calculate Grxn° for previous examples. 113 Gibbs energy of formation • Gof: Gibbs energy change that accompanies the formation of exactly 1 mole of the pure substance from free elements in their most stable states under standard state conditions. C(s) + O2(g) → CO2(g) Gof = -394.4 kJ mol-1 • All elements in their standard states have Gof = 0. 114 Gibbs Energy and Equilibrium G = -RT ln K •Allows us to calculate standard Gibbs energy change of reaction (G) from its equilibrium constant, or vice versa. • As Gbecomes more positive, K becomes smaller – reaction reaches equilibrium with less product and more reactant • As Gbecomes more negative, K becomes larger – reaction reaches equilibrium with more product and less reactant 115 • K > 1 (lnK > 0) → ΔGrxn° < 0 • K < 1 (lnK < 0) → ΔGrxn° > 0 • K = 1 (lnK = 0) → ΔGrxn° = 0 116 A⇌ B The Relationship Between G° and K at 25 °C G0(kJ) K 100 3x10-18 50 2x10-9 10 2x10-2 1 7x10-1 0 1 -1 1.5 -10 5x101 -50 6x108 -100 3x1017 -200 1x1035 Essentially no forward reaction; reverse reaction goes to completion Forward and reverse reactions proceed to same extent REVERSE REACTION 9x10-36 FORWARD REACTION 200 Significance Forward reaction goes to completion; essentially no reverse reaction 117 Examples 1. Determine the value of the equilibrium constant at 298.15 K for the reaction C2H6(g) ⇌ H2(g) + C2H4(g) G298= 101.0 kJ 2. Determine the value of K at 100 °C for I2(g) + Cl2(g) 2ICl(g) For this rxn, H298= -26.9 kJ & S298= 11.3 J K-1 (Assume that H298and S298do not vary with T.) 118 Example Calculate G (kJ mol-1) for following process: AgCl(s) ⇌ Ag+(aq) + Cl¯(aq) Ksp = 1.6 x 10-10 at 25 C 119 Gibbs Energy and Cell Potential ΔGrxn° = -nFEcell° 120