Chem 104 1st Edition Lecture 11 Outline of Last Lecture I. Phase diagrams and their purposeII. Solids and their properties and structuresA. Unit cells: the basicsIII. Solutions and solubilityOutline of Current Lecture I. Continuation on solutions, their solubility, and their propertiesA. Heat of solution and their energetics of a solution formation and solution processesII. Heat of hydration and its equationIII. Solution equilibrium and solubility limitIV. SolidsA. The temperature dependency of solids solubility in waterB. A solid’s purification by crystallizationCurrent LectureI. Chemicals can be dissolved in similar solvents only if the solvents are in a similar structure. Remember: like dissolves like.A. Some compounds dissolve in water and release heat. Other compounds absorb heatfrom their surroundings. This is the heat of solution. To make a solution you have to overcome all attractions between solute-solute and solvent-solvent particles in orderto form new attractions between solute and solvent molecules. The overall change in heat for making a solution depends on the relative sizes of the change in heat for the three processes. This is expressed in the equation:∆ Hsolution=∆ Hsolute+∆ Hsolvent+∆ HmixThe solute and the solvent are +, the mix is -.B. Energy must be added to overcome all solute-solute attractions. Then, do the same for solvent-solvent attractions. Finally, a new solute-solvent attraction forms throughreleasing energy. If the total energy cost for breaking attractions between particles insolvent and solute is less than the energy released in making the new attractions then the process is exothermic. If it is greater than the energy released it is endothermic.II. Heat of hydration is the energy added to overcome attractions between water molecules and energy released in forming attractions between water molecules and ions. A reminder: the attractiveforces in water are called hydrogen bonds. Between ions they’re called ion-dipole attractions. For heat of solutions the equations are below:∆ Hsolute=−∆ Hlattice energy∆ Hsolution=∆ Hhydration−∆ Hlattice energy∆ Hsolution=−∆ Hlatticeenergy+∆ Hsolvent+∆ Hmix∆ Hsolution=∆ Hhydration−∆ Hlattice energyA. Example:What is the lattice energy of KI if ∆Hsolution= +21.5 kJ/mol and the ∆Hhydration= -583 kJ/molGiven: ∆Hsol’n= +21.5 kJ/mol, ∆Hhydration= -583 kJ/molFind: ∆Hlattice, kJ/mol∆ Hlattice=∆ Hhydration−∆ Hsolution¿(−583kJmol)−(+21.5kJmol)¿−604.5=−605 kJ /molIII. Solution equilibrium is the dissolution of a solute in a solvent. Solubility limit is a solution that has solute and solvent in dynamic equilibrium and is also saturated. Solutions can be made saturated at non room conditions then allowed to come to room conditions slowly. If you add more solute it will not dissolve. A solution with less solute than saturation is unsaturated. Supersaturated solutions have more solute than saturation. These are unstable and lose all then solute about saturation when disturbed.IV. Solubility is given in grams of a solute of solute that will dissolve in 100g of water. For most solids, the solubility of the solid increases as the temperature increases. Solubility curves can be used to predict a solution is saturated, unsaturated, or supersaturated based on a particular amount of solute it contains.V. Solids can be purified by crystallization, this is known as
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