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CHM 320 Lecture 7 Chapt. 6Chapter 6 HomeworkDue Friday, February 10Chapt. 6: 6-5,6-7, 6-8, 6-14, 6-16, 6-19, 6-24, 6-34, 6-40, 6-46, 6-52, 6-53CHM 320 Lecture 7 Chapt. 6Chapter 6: Chemical EquilibriumChemical equilibrium provides the basis for determining what species are present as well as how much of each species is present.CHM 320 Lecture 7 Chapt. 6The thermodynamic equilibrium constanta A + b B = c C + d DReactants ProductsK = [C]c[D]d/ [A]a[B]bNote: = same as ↔K = kforward/kreverseWhen K > 1, then the reaction is favored (will most likely occur). K is a thermodynamic term, so if the kinetics are not good then the reaction may take a very long time to occur.ForwardReverseCHM 320 Lecture 7 Chapt. 6Standard State IssuesK = [C]c[D]d/ [A]a[B]b• Since K values are dimensionless, so all concentrations are actually a ratio to the the standard state concentration (1 M for solutions and 1 bar for gases).• To keep units consistent, all concentrations are expressed in M and all pressures in bars.• For all pure solids or liquids at std. state, the concentration is 1 M. So, for pure solids and liquids, the concentration is 1 and can be dropped out of the calculation.CHM 320 Lecture 7 Chapt. 6What is, exactly, the term inside brackets?K = [C]c[D]d/ [A]a[B]b• [C] is the ratio of activities of component C at equilibrium to standard state activity of C.• At the present, we will assume the concentration is equivalent to the activity.CHM 320 Lecture 7 Chapt. 6Manipulating equilibrium constants• Add the reactions, multiply the equilibrium constants• Reverse the reaction, take 1/K• Multiply reaction by n, take KnCHM 320 Lecture 7 Chapt. 6Example: Determining the equilibrium constant of two RxAg+ + Cl-= AgCl(aq)K1= 2.0 x 103AgCl(s)= Ag++ Cl-K2= 1.8 x 10-10AgCl(s)= AgCl(aq)K3= ??(reaction of interest)K3= K1x K2= (2.0 x103) (1.8 x 10-10) = 3.6 x 10-7What does the K values tell you about AgCl??CHM 320 Lecture 7 Chapt. 6Thermodynamic Terms Associated w/EquilibriumEnthalpy => Related to the heat of the reactions.Symbol is ΔH = Hproducts–HreactantsWhen ΔH is negative - the reaction is exothermic (heat given off).When ΔH is positive - the reaction is endothermic (heat absorbed).Entropy => Related to the disorder of the reaction.Symbol is ΔS = Sproducts–SreactantsWhen ΔS is positive - the products are more disordered than reactants.When ΔS is negative - the products are less disordered than reactants.CHM 320 Lecture 7 Chapt. 6Gibbs Free Energy: ΔG = ΔH - T ΔS Gibbs Free Energy combines Enthalpy and Entropy.If ΔG is negative, then reaction is favored.So…if ΔH = - and ΔS = + , then Rxis favored.if ΔH = + and ΔS = - , then Rxis NOT favored.What if ΔH = - AND ΔS = - , ΔG must be calculated.Relationship of free energy and equilibrium constant:K = e-(ΔG°/RT) where R is gas constantT is temperature (K)CHM 320 Lecture 7 Chapt. 6Le Chatelier’s Principle• If, to a system at equilibrium, a stress be applied, the system will react so as to relieve the stress.• Examples: add reactants, take away productshttp://onsager.bd.psu.edu/~jircitano/equilibrium.htmlCHM 320 Lecture 7 Chapt. 6a A + b B = c C + d DIf more A is added:Le Chatelier’s Principle in ActionThen the consequence is the Rxproceeds “more” to the right (in the forward direction) to offset the additional “A”.The reaction quotient, Q, can be used to compare the effects of Le Chatelier’s Principle to the equilibrium state (K).Q = [C]c[D]d/ [A]a[B]b Note: activities do not have to be at equilibriumCHM 320 Lecture 7 Chapt. 6Relationship between K and Q• K: ratios of activities at equilibrium to the standard state activity • Q: ratios of activities under non-equilibrium conditions to the standard state activityCHM 320 Lecture 7 Chapt. 6Solubility Product => KspKspis the equilibrium constant that describes the reaction of a solid dissolving (dissolution). CaSO4(s)= Ca2+aq+ SO42-aqK = [Ca2+] [SO4-2] / [CaSO4] but the [CaSO4(s)] is 1 at std state, so K = [Ca2+] [SO4-2] = KspKsp‘s: CaSO4= 2.4 x 10-5SrSO4= 3.2 x 10-7BaSO4= 1.1 x 10-10CHM 320 Lecture 7 Chapt. 6Using Kspto determine concentrations in solutionRx=> La(IO3)3= La3++ 3IO3-How many grams of La(IO3)3will dissolve in 250.0 mL of water?Kspfor La(IO3)3= 1.0 x 10-11CHM 320 Lecture 7 Chapt. 6Common Ion Effect: A salt will be less soluable if one of its constituents ions is already present in the solution.CaSO4(s)= Ca2+aq+ SO42-aqRecall: For CaSO4, K = [Ca2+] [SO4-2] = Kspif you add CaSO4water, you will have [Ca2+] = [SO4-2].However, if you have one of the ions already present in the water, (for example Ca2+) then less of the CaSO4will dissolve so[Ca2+] > [SO4-2].CHM 320 Lecture 7 Chapt. 6How does the presence of 0.050 M LiIO3change how many grams of La(IO3)3will dissolve in 250 mL?Rx=> La(IO3)3= La3++


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NAU CHM 320 - CHM 320 Homework

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