CHEMISTRY. CHM202 Class #3 CHEMISTRY. Chapter 11. Chapter Outline for Class #3. Solutions: Properties and Behavior

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1 CHEMISTRY Fifth Edition Gilbert Kirss Foster Bretz Davies CHM202 Class #3 1 Chemistry, 5 th Edition Copyright 2017, W. W. Norton & Company CHEMISTRY Fifth Edition Gilbert Kirss Foster Bretz Davies Chapter 11 Solutions: Properties and Behavior 2 Chemistry, 5 th Edition Copyright 2017, W. W. Norton & Company Chapter Outline for Class # Interaction between Ions 11.2 Energy Changes during Formation and Dissolution of Ionic Compounds 11.3 Vapor Pressure of Solutions 3 3 1

2 Interaction between Ions Ion ion interaction: Interactive force between ions of opposite charge that results in an ionic bond Coulomb s law: (Q Q ) d E 1 2 E is the electrostatic potential between the two particles with charges Q 1 and Q 2, separated by a distance d. The value of E is negative when the charges have opposite signs indicating the attraction for each other. As the distance d increases the electrostatic potential decreases. 4 4 Size of Polyatomic Ions Electrostatic potential is determined by charge and ionic radius. 5 5 Size of Monoatomic Ions 6 6 2

3 Practice: Predicting Ion Ion Interactions Predict which salt in the following pairs exhibits the greater ion ion interactions. a) MgO and NaF b) MgO and MgS Collect and Organize: We are given two pairs of ionic compounds and want to predict which compound exhibits the greater ionic attractions. Ionic charges will be based on positions in the periodic table and ionic radii are available in Figure Practice: Predicting Ion Ion Interactions Predict which salt in the following pairs exhibits the greater ion ion interactions. a) MgO and NaF b) MgO and MgS Analyze: Ionic charge will dominate the attractive force the greater the charge, the greater the attractive force. If charges are similar, we can refer to ionic radii the larger the radii, the weaker the attractive force. 8 8 Practice: Predicting Ion Ion Interactions Predict which salt in the following pairs exhibits the greater ion ion interactions. MgO and NaF b) MgO and MgS Solve: MgO vs. NaF Mg +2 and O 2 Q 1 Q 2 = 4 Na +1 and F 1 Q 1 Q 2 = 2 In this case, ionic charge will dominate. MgO would exhibit the greater attractive force

4 Practice: Predicting Ion Ion Interactions Predict which salt in the following pairs exhibits the greater ion ion interactions. a) MgO and NaF b) MgO and MgS Solve: MgO vs. MgS Mg +2 and O 2 Q 1 Q 2 = 4 Mg +2 and S 2 Q 1 Q 2 = 4 In this case the cationic/anionic charges are the same, so interionic forces will be determined by ionic radii. Since S 2 is larger than O 2, the ionic attractions in MgO will be greater than in MgS Enthalpy of Solution Dissolution of ionic solids Enthalpy of solution ( H soln ) depends on: Energies holding solute ions in crystal lattice Attractive force holding solvent molecules together Interactions between solute ions and solvent molecules H soln = H ion ion + H dipole dipole + H ion dipole H hydration = H dipole dipole + H ion dipole When solvent is water: H soln = H ion ion + H hydration Dissolution of Ionic Compounds

5 Lattice Energy Lattice energy (U): The energy released when one mole of an ionic compound forms from its free ions in the gas phase: M + (g) + X (g) MX(s) k(q1q 2) U = d Where k is a proportionality constant depending on lattice structure H ion ion Lattice energy (U) energy released when crystal lattice is formed H ion ion = energy required to remove ions from crystal lattice: H ion ion = U H solution = H hydration U Comparing Lattice Energies k(q1q 2) U = d Lattice energy depends on: Ionic charge Ionic radius

6 Born Haber Cycle and Lattice Energy Born Haber Cycle: Series of steps with corresponding H that describe the formation of an ionic solid from its elements e.g., Na(s) + ½ Cl 2 (g) NaCl(s) H f = kj Steps: 1. Sublimation of one mole Na(s) Na(g) = H sub 2. Bond breaking of one-half mole of Cl 2 (g) = ½ H BE 3. Ionization of one mole Na(g) atoms = IE 1 4. Ionization of one mole Cl(g) atoms = EA 1 5. Formation of one mole NaCl(s) from ions(g) = U Born Haber Cycle for NaCl H f = H sub + ½ H BE + IE 1 + EA 1 + U Calculating U Na + (g) + e (g) Na(g) Na(g) Na(s) Cl (g) Cl(g) + e (g) Cl(g) ½Cl 2 (g) Na(s) + ½Cl 2 (g) NaCl(s) Na + (g) + Cl (g) NaCl(s) H IE1 H sub H EA ½ H BE H f U U = H f ½ H BE H EA H sub H IE

7 Born Haber Cycle: H hydration H solution,nacl(aq) = H hydration,nacl(g) U NaCl(s) H hydration,nacl(g) = H hydration,na + (g) + H hydration,cl (g) Enthalpies of Hydration Practice: Calculating Enthalpy of Hydration Sodium fluoride is placed in drinking water to promote healthy teeth. Calculate the enthalpy of hydration of sodium fluoride. Collect and Organize: From Table 11.4, the enthalpy of hydration for Na + and F are 418 kj/mol and 502 kj/mol, respectively

8 Practice: Calculating Enthalpy of Hydration Sodium fluoride is placed in drinking water to promote healthy teeth. Calculate the enthalpy of hydration of sodium fluoride. Analyze: H hydration,naf(aq) = H hydration,na + (g) + H hydration,f (g) Practice: Calculating Enthalpy of Hydration Sodium fluoride is placed in drinking water to promote healthy teeth. Calculate the enthalpy of hydration of sodium fluoride. Solve: H hydration,naf(aq) = H hydration,na + (g) + H hydration,f (g) H hydration,naf(aq) = ( 418 kj/mol) + ( 502 kj/mol) H hydration,naf(aq) = 920 kj/mol Practice: Calculating Enthalpy of Hydration Sodium fluoride is placed in drinking water to promote healthy teeth. Calculate the enthalpy of hydration of sodium fluoride. Think About It: The value makes sense following the relationship between the enthalpy of hydration of the individual ions to the compound

9 Vapor Pressure of Solutions Vapor pressure: Pressure exerted by a gas in equilibrium with liquid Rates of evaporation and condensation are equal Factors Affecting Vapor Pressure Temperature As T increases, KE increases, vapor pressure increases Intermolecular forces Stronger forces, higher KE needed to enter gas phase vapor pressure decreases Presence of nonvolatile solute Affects rate of evaporation, decreases vapor pressure of solution compared to pure solvent Raoult s Law Raoult s law Vapor pressure of solution is equal to the vapor pressure of the pure solvent multiplied by the mole fraction of the solvent in the solution. P solution = X solvent P solvent Vapor pressure lowering A colligative property of solutions Ideal solution One that obeys Raoult s law

10 Practice: Raoult s Law What would be the vapor pressure of water over a solution that contains 62.1 g of ethylene glycol (62.1 g/mol) in g of water at 30 C? (The vapor pressure of pure water at 30 C is 31.8 mmhg.) Collect and Organize: 62.1 g ethylene glycol g of water Vapor pressure of pure water is 31.8 mmhg Practice: Raoult s Law 29 What would be the vapor pressure of water over a solution that contains 62.1 g of ethylene glycol (62.1 g/mol) in g of water at 30 C? (The vapor pressure of pure water at 30 C is 31.8 mmhg.) Analyze: Raoult s law: P solution = X H2O P H2O 1 mol 62.1 g = 1.00 mol ethylene glycol 62.1 g 1 mol g = mol H2O g 29 Practice: Raoult s Law What would be the vapor pressure of water over a solution that contains 62.1 g of ethylene glycol (62.1 g/mol) in g of water at 30 C? (The vapor pressure of pure water at 30 C is 31.8 mmhg.) Solve: P solution = X H2O P H 2O mol H O mol H O mol glycol P solution 2 = x 31.8 mm Hg = 29.7 mm Hg

11 Practice: Raoult s Law What would be the vapor pressure of water over a solution that contains 62.1 g of ethylene glycol (62.1 g/mol) in g of water at 30 C? (The vapor pressure of pure water at 30 C is 31.8 mmhg.) Think About It: The mole fraction is less than 1.00 so the pressure of a solution should be less than that of pure water. Since the solution was mostly water, the pressure should not go down by much