CHEM 1120 1st Edition Lecture 2 Outline of Last Lecture II Intermolecular Forces A 6 Types of Intermolecular Forces IMFs III Introduction to Chapter 13 The Solution Process A Solution Terms Homogeneous Solute vs Solvent etc B Intermolecular Interactions Involved in Solution Formation IV Dissolution process V Solubility Outline of Current Lecture I Solution Process and Entropy II Gas Solubility Pressure and Temperature III Units of Concentration IV Beginning Colligative Properties Current Lecture I In our physical world on Earth we assume objects will move toward lower energy 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 does not mean every single chemical process or reaction has to be exothermic some can proceed uphill toward a higher energy level and be considered endothermic 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 system These 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 Spontaneous Higher Energy More Disorder Higher Entropy Spontaneous II 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 atm Pressure 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 solution Example 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 C2H6O 10g x 1mL 0 983g x 1L 1000mL 0 01017 L solution M 0 02171 mol 0 01017 L 2 13 M 2 Molality m moles solute kg of solvent Example 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 C2H6O m 0 02171 mol 0 009kg 2 41 m 3 Mass Percent mass of component A mass of A total mass of solution x 100 Example 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 18g mass ethanol 46 07g 46 07g 162 18g x 100 22 12 ethanol mass water 100 mass ethanol 100 22 12 77 88 water 4 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 solution Example 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 mol 1 g water x 1mol 18 02g 0 05549 mol water 1 g C2H6O x 1 mol 46 07g 0 02171 mol C2H6O XWATER 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 6 ppb of component A mass of A in solution total mass of solution x 10 9 IV 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 POSOLVENT Example 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 water XWATER 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 glucose Real 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 law