CHM1046 FINAL EXAM PART II Chapter 16 16 5 Indicator substance that changes color at or near the equivalence point depends on pH o endpoint point where indicator changes color used to determine the equivalence point o if In HIn 1 solution will be intermediate color pH pKa o if In HIn 1 solution will be color of In pH pKa 1 o if In HIn 0 1 solution will be color of HIn pH pKa 1 16 6 Solubility equilibrium is based on the assumption that solids dissolve in water to give the basic particles from which they are formed o EX Mg OH 2 s Mg OH 2 s Mg2 aq 2OH aq representing dissolution of an ionic compound o measure of the solubility of a compound o EX Ksp Mg2 OH 2 Ksp solubility product constant equilibrium expression for a chemical equation o solids are omitted from equilibrium expression b c concentration of solid is Molar Solubility mol L number of moles of solute dissolved in 1 liter of saturated constant solution o Ksp is NOT the molar solubility o EX use ICE table Mg OH 2 s Mg2 aq 2OH aq solid s 2s s 2s o Calculate S from Ksp Ksp 2 06x10 13 given Ksp Mg2 OH 2 Ksp s 2s 2 2s3 4s3 2 06x10 13 4s3 5 15x10 14 s3 s 3 72x10 5 M Solubility g L number of grams of solute dissolved in 1 liter of a saturated solution 16 7 Q reaction quotient o EX Q Mg2 OH 2 Q Ksp towards products Ksp is value of product at equilibrium ONLY Q is value of product under any conditions o solution is unsaturated more of the solid ion compound can dissolve in solution Q Ksp reaction at equilibrium will not proceed in any direction o solution is saturated solution is holding equilibrium amount of the dissolved ions and additional solid will not dissolve in the solution Q Ksp towards reactants o solution is supersaturated under most circumstances the excess solid will precipitate out of a supersaturated solution EX Will precipitate form when we mix 0 010M AgNO3 0 015M KI o compare Q Ksp o determine possible cross products 1 5x10 4 o Q Ksp AgI should precipitate o AgI Ksp 8 51x10 17 for new mixed solution Q Ag I 0 010 0 015 If given different ions in a solution the salt with the lowest Ksp value will precipitate first 16 9 The presence of a common ion will decrease the solubility of the salt Chapter 17 17 3 Physical processes that lead to increase in Entropy S 0 o Solid Liquid o Liquid Vapor o Solvent Solute Solution o System at T1 System at T2 T2 T1 Calculating Entropy Change o EX C3H8 g 5O2 g 3CO2 g 4H2O g H 2044 kJ 1 calculate the entropy change in the surroundings associated with this rxn occurring at 25 C 2 determine the sign of the entropy change for the system 3 determine the sign of the entropy change in the universe will this rxn be spontaneous 1 T 273 25 298 K Ssurr Hrxn T 2044 kJ 298K 6 86 kJ K 6 86x103 kJ K 2 C3H8 g 5O2 g 3CO2 g 4H2O g 6 moles 7 moles Ssys is positive 3 Suniv Ssys Ssurr Thereforce S is positive and reaction is spontaneous 17 4 First Law of Thermodynamics Energy can be neither created nor destroyed it can only be converted from one to another o Spontaneous Physical and Chemical Processes waterfall funs downhill a lump of sugar dissolves in a cup of coffee at 1 atm water freezes below 0 C and ice melts above 0 C heat flows from hotter to colder object a gas expands in an evacuated bulb iron exposed to oxygen and water forms rust o State Functions properties that are determined by the state of the system regardless of how that condition was achieved EX energy enthalpy pressure volume temperature entropy Second Law Of Thermodynamics the entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process o for spontaneous process Suniv Ssys Ssurr 0 o for equilibrium process Suniv Ssys Ssurr 0 Third Law of Thermodynamics the entropy of a perfect crystalline substance is zero at the absolute zero of temperature 17 5 Gibbs Free Energy o for a constant temperature and constant pressure processes o G o G Hsys T Ssys direction o G 0 rxn is at equilibrium o G 0 rxn is spontaneous in the forward direction o G 0 rxn is nonspontaneous as written the rxn is spontaneous in the reverse Gibbs Free Energy and Chemical Equilibrium o G G0 RTlnQ o R is gas constant 8 314 J K mol o T is absolute temperature K o Q is reaction quotient At Equilibrium G 0 Q K 0 G0 RTlnK G0 RTlnK EX CCl4 g C s graphite 2Cl2 g H 95 7 kJ S 142 2 J K 1 calculate G at 25 C and determine whether the reaction is spontaneous 2 if the reaction is not spontaneous at 25 C determine at what temperature if any the reaction becomes spontaneous o 1 T 273 25 298 K G H T S 95 7x103 J 298 K 142 2 J K 95 7x103 J 42 4x103 J 53 3x103 J Rxn is not spontaneous o 2 G H T S 0 95 7x103 J T 142 2 J K T 95 7x103 J 142 2 J K 673 K 17 6 G0 RTlnK o When K 1 lnK is negative G0 rxn is therefore positive under standard conditions when Q 1 the reaction is spontaneous in the reverse direction o When K 1 lnK is positive G0 rxn is therefore negative under standard conditions when Q 1 the reaction is spontaneous in the forward direction o When K 1 lnK is zero G0 rxn is therefore zero the reaction happens to be at equilibrium under standard conditions EX Use tabulated free energies of formation to calculate the equilibrium constant for the following reaction at 298 K N2O4 g 2NO2 g o standard free energies found in book o N2O4 g G0 f 99 8 kJ mol NO2 g G0 f 51 3 kJ mol o G0 rxn np G0 f products nr G0 f reactants 2 G0 f NO2 G0 f N2O4 2 51 3 kJ 99 8 kJ 2 8 kJ Chapter 18 Electrochemical processes are oxidation reduction reactions in which energy released by a spontaneous reactions is converted to electricity or electrical energy is used to cause a nonspontaneous reaction to occur Oxidation number charge the atom would have in a molecule if electrons were completely transferred o 1 free elements uncombined state have an oxidation number of zero Na Be K Pb H2 O2 P4 0 o 2 in monoatomic ions the oxidation number is equal to the charge of the ion Li Li 1 Fe3 Fe 3 O2 O 2 o 3 the oxidation number of oxygen is usually 2 in H2O2 and O2 2 it is 1 o 4 the oxidation number …
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