CHEM 1120 1st Edition Lecture 2 Outline of Last Lecture II. Intermolecular ForcesA. 6 Types of Intermolecular Forces (IMFs)III. Introduction to Chapter 13, The Solution ProcessA. Solution Terms: Homogeneous, Solute vs Solvent, etcB. Intermolecular Interactions Involved in Solution FormationIV. Dissolution processV. SolubilityOutline of Current Lecture I. Solution Process and EntropyII. Gas Solubility, Pressure, and TemperatureIII. Units of ConcentrationIV. Beginning Colligative PropertiesCurrent LectureI. In our physical world on Earth, we assume objects will move toward lowerenergy levels. The same is true of most chemical reactions or processes. This means that most chemical processes are exothermic and release more heat than they absorb, maintaining the reaction. Although, this doesnot mean every single chemical process or reaction has to be exothermic;some can proceed “uphill” toward a higher energy level and be consideredendothermic. This additionally means that these reactions tend toward a higher entropy (S) and become more disordered in the process. Entropy = thermodynamic measure of the randomness or disorder of a systemThese notes represent a detailed interpretation of the professor’s lecture. GradeBuddy is best used as a supplement to your own notes, not as a substitute.Processes in which entropy decreases tend not to be spontaneous and processes in which entropy increases tend to be the opposite, spontaneous. Gases in general have a greater entropy than liquids or solids because of the randomness or high disorder of their particles.Lower Energy = Less Disorder = Lower Entropy = Not SpontaneousHigher Energy = More Disorder = Higher Entropy = SpontaneousII. As stated in the previous lecture notes from lecture 1, polar molecules dissolve other polar molecules, and nonpolar molecules dissolve other nonpolar molecules. In other words, the rule is: Like Dissolves Like. Henry’s Law = the solubility of a gas is directly proportional to the pressure of the gas over the solution(solubility)GAS = kH x PGAS kH is Henry’s constant (different for each gas and usually given to you on tests and such), solubility is in Moles/Liter, and the pressure of the gas is in atmPressure and temperature can have effects on the solubility of solutions. Generally, for most solids, the higher the temperature gets, the more soluble the solid becomes. Therefore, change in HLATT is bigger than change in HHYDR, so the reaction is endothermic. Take note that exceptions do exist. Gases do not have an HLATT because the molecules are already taken apart and separate. The change in HHYDR is almost always negative, so the reaction is exothermic, and no exceptions exist here.III. Although they are similar, concentration and strength are not the same and are not interchangeable words. Concentration is the quantity of solute in a given volume or mass of solution whereas strength is the electrolytic behavior of a solution (weak, strong, nonelectrolyte). We talk about a total of five concentration units:1. Molarity M = moles solute/Liters of solutionExample: A solution is prepared from exactly one gram of ethanol (C2H6O) and nine grams of water. Calculate the molarity. (solution density = 0.983 g/ml)Solution:1g C2H6O x (1mol/46.07g) = 0.02171 mol C2H6O10g x (1mL/0.983g) x (1L/1000mL) = 0.01017 L solutionM = (0.02171 mol/0.01017 L) = 2.13 M2. Molality m = moles solute/kg of solventExample: A solution is prepared from exactly one gram of ethanol (C2H6O) and nine grams of water. Calculate the molality.Solution: 1 g C2H6O x (1 mol/46.07g) = 0.02171 mol C2H6Om = (0.02171 mol/0.009kg) = 2.41 m3. Mass Percent mass % of component A = (mass of A/total mass of solution) x 100Example: A solution is prepared from exactly one mole of ethanol (C2H6O) and nine moles of water. What is the mass percent of each in the solution?Solution: mass C2H6O = 1 mole x 46.07g/mol = 46.07g mass water = 9 moles x 18.02 g/mol = 162.18gmass % ethanol = [46.07g/(46.07g + 162.18g)] x 100 = 22.12% ethanolmass % water = 100 - mass % ethanol = 100 - 22.12% = 77.88% water4. Volume Percent volume % of component A = (V of A/total V of solution) x 100(same as mass percent, but with volumes)5. Mole Fraction XA = moles of A/total moles of solutionExample: A solution is prepared from exactly one gram of ethanol (C2H6O) and one gram of water. What is the mole fraction of each in the solution?Solution: Molar masses = 46.07g/mol and 18.02g/mol1 g water x (1mol/18.02g) = 0.05549 mol water1 g C2H6O x (1 mol/46.07g) = 0.02171 mol C2H6OXWATER = 0.05549mol/ (0.05549mol + 0.02171mol) = 0.7188 XETHANOL = 1- XWATER = 1- 0.7188 = 0.2812**parts per million/parts per billion (these may come up in a few questions but will not be focused on)ppm of component A = (mass of A in solution/total mass of solution) x 10^6ppb of component A = (mass of A in solution/total mass of solution) x 10^9IV. Colligative Properties are solution properties that depend only on the number of solute particles, not on the nature or identity of the solute particles. An example of colligative properties: vapor pressure lowering, boiling point elevation, freezing point depression, osmosis. Vapor pressure (P) lowering is when the vapor pressure of a solution’s solvent is less than the vapor pressure of the pure solvent (constant temperature). Raolt’s Law:PSOLUTION = XSOLVENT + POSOLVENTExample: How many grams of glucose (C6H12O6) must be dissolved in 552 grams of water at 20 degrees C (v.p. = 17.5 torr) to lower its vapor pressure by 2.0 torr?Solution:The change in PSOLUTION = 2.0 torr, so 15.5 torr = XWATER x (17.5 torr)molar masses = 180.1559 g/mol and 18.02 g/mol(552 g water)/(18.02 g/mol) = 30.64 mol waterXWATER = (15.5torr/17.5 torr) = (30.64 mol water/30.64mol water + mol glucose)mol glucose =3.953(3.953 mol glucose x 180.1559 g/mol) = 712.3 g glucoseReal solutions may not consistently follow Raolt’s law just like gases stray from the Ideal Gas Law. Weak solute-solvent interactions will result in positive differences from Raolt’s law and strong solute-solvent interactions will result in negative differences from Raolt’s