Week 12/Th: Lecture Units ‘31 & 32’

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Unit 30: Chemical Spontaneity -- entropy, 2nd Law of Thermo -- “free” energy -- spontaneity Unit 31: Phase Equilibria -- liquid / gas -- phase diagrams -- phase boundaries Unit 32: Solutions -- more concentration units Issues: Exam 3 next Monday !!! Homework Set 9 due on Saturday @ 08:00AM

Week 12/Th: Help Preparing for Exam

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★ Tuesday, November 13th, 6 - 8:30pm in N100 BCC (LRC Mock Exam, reservation required!) ★ Thursday, November 15th, 6 - 8:30pm in N100 BCC (LRC Q&A Review Session) ★ Sunday, November18th, 4 - 6pm in 138 Chemistry (Dr. Pollock)

★ Monday, November 19th, EXAM day for everyone

★ Tuesday, November 20th, NO LECTURE – Happy Thanksgiving!

Week 12/Th: Where to Take the Exam

© DJMorrissey, 2o12

Alternate Exam Monday, November 19, 6:45am-7:45am room 138 Chemistry

CEM 141 - Fall Semester 2012 – Exam 3 Rooms Monday, November 19th

7:15pm-8:15pm

Week 12/Th: Phase Diagram .. Generic

© DJMorrissey, 2o12

Last time we considered the equilibrium between the liquid and gas phases of a pure material. We can also look for equilibria between the other phases and we find a general pattern that we can display on a phase diagram for a pure material.

Note: watch out for relative scales on the Temperature and Pressure axes Land marks: A – triple point B – critical point C – Solid/Gas Equil. D – Solid/Liquid Equil. Lines – ΔP / ΔT

ΔS + ΔH +

Week 12/Th: Phase Equilibria

© DJMorrissey, 2o12

The boundaries between the phases of a pure material are points of equilibria between the phases. The energy content of the two phases is equal at equilibrium and the system does not change at the macroscopic level, ΔG = 0 ΔG = ΔH – TΔS = 0 ΔH = TΔS ΔH / T = ΔS for equilibria and for reversible change SULFUR

For example, NON-Hydrogen bonded liquids at their normal boiling points: ΔHVAP / TNB = ΔSVAP ~ 10.5 R

Week 12/Th: Phase Diagram .. Water

© DJMorrissey, 2o12

A: Larger due to H-bonds in liquid not present in gas phase. ΔHVAP / TNB = (40.7 kJ/mol) / 373 K = 0.109 kJ/mol-K ~ 13 * R

Q: Do you expect the ΔSVAP for water to be less than, about the same, or larger than the average value of 10.5 R and why? Q: What is ΔSVAP for water?

Figure: http://serc.carleton.edu/research_education/equilibria/phaserule.html

Week 12/Th: Solutions & Phase Change

© DJMorrissey, 2o12

There are a number of practical applications that rely on understanding how the phase diagram of pure water changes for aqueous solutions. Recall that the definition of a solution is a mixture of matter that is random all the way down to the smallest components. For example, all mixtures of gases are solutions, alloys are solutions, most mixtures of solids are not solutions. The solubility of a substance in a solvent is often given in terms of the number of grams that can be dissolved in a given volume. Most of the time we are concerned with aqueous solutions and all of the solutions can be characterized by the amounts of the different components that are present. We have already introduced two units of measure for this …

Week 12/Th: Concentration Units

© DJMorrissey, 2o12

Name Symbol Definition Use Unknown Quantity

Molarity M Moles solute per Liter of Solution

Lab reactions Stoichiometry

Moles of Solvent, Temperature dependent

molality m Moles solute per kg of Solvent

Physical properties (a number ratio)

Volume of solution

Mole fraction

X Moles solute per Total moles

Physical properties, gas mixtures (a number ratio)

Volume of solution

Weight percent

Wt % (w/w)

Mass of solute per Total mass

Commercial composition (easy)

Density and/or Volume of Solution

(molar masses)

Week 12/Th: Example in Notes

© DJMorrissey, 2o12

The density of a 19% solution of HCl is 1.0929 g/mL. (1) Calculate the molarity of the solution. (2) What is the mole fraction of HCl in this solution? 1L of solution = 1 L * 1000 mL/L * 1.0929 g/mL = 1092.9 g Mass of HCl in 1L = 1092.9g * 19/100 = 207.7 g Moles of HCl in 1L = 207.7g / 36.46 g/mol = 5.695 mol Thus 5.695 moles in 1L à 5.695 Molar

Week 12/Th: Example in Notes

© DJMorrissey, 2o12

The density of a 19% solution of HCl is 1.0929 g/mL. (1) Calculate the molarity of the solution. (2) What is the mole fraction of HCl in this solution? 1L of solution = 1 L * 1000 mL/L * 1.0929 g/mL = 1092.9 g Mass of HCl in 1L = 1092.9g * 19/100 = 207.7 g Moles of HCl in 1L = 207.7g / 36.46 g/mol = 5.695 mol Mole Fraction of HCl = moles HCl / Total moles Moles water in 1L = (1092.9 – 207.7)g / 18.01 g/mol = 49.11 mol water Χ(HCl) = 5.695 / (5.695 + 49.11) = 0.1039

Week 12/Th: Another Concentration Example

© DJMorrissey, 2o12

Concentrated nitric acid is listed as 70.7% (by mass) with a density of 1.408 g/cm3; calculate other concentration units. 1L of solution = 1 L * 1000 mL/L * 1.408 g/mL = 1408 g Mass of HNO3 in 1L = 1408g * 70.7/100 = 995.5 g a) Moles HNO3 in 1L = 995.5g / 63.02g/mol = 15.80 mol à 15.8 M molality = moles solute / kg of solvent Mass of water in 1 L = 1.408 – 0.9955 kg = 0.4125 kg b) molality = 15.8 moles HNO3 / 0.4125 kg water = 38.3 m

Exam Review: HNO3 -- Lewis Structure?

-- # of lone pairs on N? -- bond order?

-- electron arrangement? -- 3D structure?

Week 12/Th: Another Concentration Example

© DJMorrissey, 2o12

Concentrated nitric acid is listed as 70.7% (by mass) with a density of 1.408 g/cm3; calculate other concentration units. 1L of solution = 1 L * 1000 mL/L * 1.408 g/mL = 1408 g Mass of HNO3 in 1L = 1408g * 70.7/100 = 995.5 g a)  Moles HNO3 in 1L = 995.5g / 63.02g/mol = 15.80 mol à 15.8 M Mass of water in 1 L = 1.408 – 0.9955 kg = 0.4125 kg b) molality = 15.80 moles HNO3 / 0.4125 kg water = 38.3 m Mole fraction HNO3 = moles HNO3 / total moles Moles water = 412.5 g / 18.01 g/mol = 22.90 mol c) X(HNO3) = 15.8 moles / (15.80 + 22.90) moles = 0.408

15.8M 38.3m X(HNO3) = 0.408

Week 12/Th: A Michigan Example

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Dow Chemical got started in Michigan extracting bromine from the brine water in the ground near Midland with NaBr at ~ 4.0x10-3 M. (a) How many grams of bromine per 1.0 m3

of solution? (b) What is the volume of liquid bromine?

The reaction is: 2NaBr (aq) + Cl2 (g) à 2NaCl (aq) + Br2 (l) 1m3 of solution à 1 m3 * 1000 L/m3 * 4.0x10-3 mol/L = 4.0 mol NaBr a) Mass of Br2 = 4.0mol NaBr * (1mol Br2 /2 mol NaBr) * (79.9*2) g/mol Br2 = 3.2x102 g b) Volume of Br2 Volume = mass / density = 3.2x102 g / 2.928 g/mL = 1.1x102 mL