In a saturated solution, equilibrium exists between dissolved and undissolved solute. Slightly soluble/insoluble ionic compounds reach equil w/ very little solute dissolved. We can use this equilibrium to determine ion concentrations for these compounds. N.b. for soluble ionic compounds [ion] will be the same as the M. Ex: If [Pb(NO 3 ) 2 ] is M then [Pb 2+ ] = M; [NO 3 - ] = M To determine [ion] for a slightly soluble compound we can use the ion product expression, Q sp, as we would for any equilibrium. PbF 2 (s)  Pb 2+ (aq) + 2F - (aq) Q sp = [Pb 2+ ][F - ] 2

For a saturated solution, equilibrium exists between dissolved and undissolved solute, so Q sp attains a constant value called the solubility-product constant, K sp. Ex: Write the solubility-product constant expression for an aq solution of calcium phosphate. The K sp value indicates how far the dissociation proceeds at equilibrium. That is, the greater the value of K sp, the more dissociation is likely to occur. Like any K eq it can be calculated from the [equilibrium] of each ion. Ex: The solubilty of lead(II) fluoride in water at 25 0 C is g/100 mL of solution. Calculate the K sp of lead(II) fluoride.

The K sp for a substance can also be used to determine its solubility. Ex: calcium hydroxide (lime) is a major component in both plaster and cement. Calculate its molar solubility if the K sp = 6.5 x Ex: The K sp of copper(II) iodate at 25 0 C is 1.40 x Calculate the molar solubility. As long as we compare compounds with the same number of ions in the formula, K sp values indicate the relative solubilties: the higher the K sp the greater the solubilty (see Table 18.2).

A particular application of Le Chatelier’s principle is known as the common ion effect. Adding a common ion decreases the solubility of a slighly soluble compound. Imagine a saturated solution of lead(II) chromate with a K sp = 2.3 x If we add Na 2 CrO 4 the [CrO 4 2- ] is increased, shifting the equilbrium back to the left decreasing the [Pb 2+ ]. PbF 2 (s)  Pb 2+ (aq) + CrO 4 2- (aq) Ex: What is the solubility of calcium hydroxide in 0.10 M calcium nitrate? By comparing the values of Q sp and K sp it can be determined whether combining solutions of specific concentrations will, or will not for a precipitate.

If Q sp = K sp solution is saturated, no change will occur. Ex: Does a precipitate form when 50.0 mL of M BaCl 2 is added to 50.0 mL M Na 2 SO 4 ? Ex: A solution consists of 0.20 M magnesium chloride and 0.10 M copper(II) chloride. What [OH - ] would be necessary to separate the metal ions? K sp of Mg(OH) 2 = 6.3 x ; K sp Cu(OH) 2 = 2.2 x If Q sp > K sp a precipitate will form. If Q sp < K sp solution is unsaturated, no precipitate will form. K sp BaSO 4 = 1.1 x

 Thermodynamics allows us to predict if a process is spontaneous – now, never use that again; call it Thermodynamically Favored.  TF means that the process will occur under the defined conditions.  Ball rolls downhill, not up  Iron rusts, rust is not spontaneously converted to Fe & O 2  At temperatures above 273K H 2 O(s) will melt, not freeze  TF (spontaneity) does not refer to how fast a reaction proceeds.  What determines if a process is TF: 1. Change in enthalpy 2. Change in entropy 3. Temperature

 Entropy (S) is the disorder of a system  dependent upon temperature  +ΔS = more disorder, less order = greater # of microstatesmicrostates  -ΔS = less disorder, more order  Universal tendency greater entropy (probability)  Which has a higher entropy?  Cube of sugar in coffee / cube of sugar  I 2(g) / I 2(s)  PCl 5 (g) / PCl 3 (g) + Cl 2 (g)  C(s) + CO 2 (g) / 2 CO(g)

 We can relate enthalpy & entropy quantitatively in a relationship known as…..  ΔG = ΔH – T ΔS ΔG = Change in Gibb’s Free Energy[kJ/mol] ΔH = change in Enthalpy (endo is + & exo is -) [kJ/mol] ΔS = change in Entropy [J/mol. K] T = Temperature in Kelvin  -ΔG = Thermodynaically favorable.  +ΔG = not favorable but if reversible the reverse rxn is TF  ΔG = 0 Equilibrium

Why does the ΔG have to be negative…? 2 nd Law of Thermodynamics: In any spontaneous (TF) process, there is always an increase in the entropy of the universe. ΔS univ = ΔS sys + ΔS surr Significance of ΔH depends on Temp ΔS surr depends on direction of heat flow; exo ΔS surr is +; endo ΔS surr is -; magnitude depends on T (effect greater at lower T); ΔS surr = - ΔH/T ΔG = ΔH – TΔS; -ΔG/T = - ΔH/T + ΔS; -ΔG/T = ΔS surr + ΔS = ΔS univ so ΔS univ = -ΔG/T Therefore if ΔG is negative ΔS univ is positive and the entropy of the universe increases. Q.E.D.

Between entropy and enthalpy, there can be four possible outcomes: Result + - always +never spontaneous - + always -always spontaneous + ++ for low T, and - for high Tdepends on T - -- for low T, and + for high Tdepends on T

 Predict the sign for ΔS for each of the following:  The thermal decomposition of calcium carbonate  The oxidation of sulfur dioxide in air to produce sulfur trioxide At what temperature is the following process spontaneous? Br 2 (l) → Br 2 (g) ΔH 0 = 31.0 kJ/mol ΔS 0 = 93.0 J/mol. K ΔS favors vaporization, ΔH favors the condensation. These opposite tendencies will exactly balance out when they are in dynamic equilibrium. This occurs at the boiling point for bromine. Remember at equilibrium ΔG = 0 so 0 = ΔH – T ΔS ΔH = T ΔS and T = ΔH / ΔS Now solve for T. T = 333 K

 Methanol is synthesized by the reaction CO(g) + 2H 2 (g)  CH 3 OH(g) a. Calculate the standard free-energy change for this reaction at 25 0 C. b. Is the reaction TF at this temp? c. If so, at what temperature does the reverse rxn become TF? n.b. Selected Thermodynamic Values Appendix A in text. CO(g)H 2 (g)CH 3 OH(g) ΔH 0 f (kJ/mol) S 0 (J/mol. K)

STANDARD FREE ENERGIES OF FORMATION ∆G f ө of a substance is the free energy change for the formation of 1 mol of the substance, in its standard state, from the most stable form of the constituent elements in their standard states. Ex: N 2 (g) + 3H 2 (g)  2NH 3 (g) ∆G ө = kJ So ∆G f ө = kJ/2 mol = kJ/mol Note: ∆G f ө for the most stable form of an element = 0. These values allow us to calculate ∆G rxn ө the same way we solved for ∆H rxn ө, using ∆H f ө values: ∆G ө = ∆G f ө (products) - ∆G f ө (reactants)

Example: Determine ΔG ө for: Fe 2 O 3 (s) + 3CO(g)  2 Fe(s) + 3CO 2 (g) ΔG ө values are: Fe 2 O 3 (s) = kJ/mol; CO(g) = kJ/mol; CO 2 = kJ/mol Answer is: -29 kJ So at 25 o C this reaction is Thermodynamically favorable

Reactions often carried out at non-standard conditions. Problem is overcome using this relationship: ΔG = ΔG 0 + RT ln Q (may be Q c, Q p, or Q sp ) Example: Calculate the free energy change for ammonia synthesis at 22 0 C given the following partial pressures: atm N 2, atm H 2, 2.0 atm; ΔG = kJ/mol

Common use of ΔG is to calculate the value for an equilibrium constant, K. ΔG = ΔG 0 + RT ln Q; at equil Q = K and ΔG = 0 so at equilibrium ΔG 0 = -RT ln K Example: For the reaction 2 CO(g) + O 2 (g)  2 CO 2 (g) ΔG 0 = kJ/mol; What is K p at 25 0 C? Example: Ag + (aq) + NH 3 (aq)  Ag(NH 3 ) 2 + (aq) ΔG 0 = kJ/mol; Calculate the value for K c. Example: Calculate the equilibrium vapor pressure of H 2 O(g) at 25 0 C. ΔG 0 f = kJ/mol for H 2 O(l); ΔG 0 f = kJ/mol for H 2 O(g)