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Complex ion equilibria

Complex ion equilibria. Problem: Find the solubility of AgCl in 1.00 M NH 3. The four step procedure. (1) Write down the reaction or reactions. K sp. AgCl ( s ) ⇋ Ag + ( aq ) + Cl – ( aq ). K sp = 1.8 × 10 –10. K f. Ag + ( aq ) + 2NH 3 ( aq ) ⇋ Ag(NH 3 ) 2 + ( aq ).

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Complex ion equilibria

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  1. Complex ion equilibria Problem: Find the solubility of AgCl in 1.00 M NH3 The four step procedure (1) Write down the reaction or reactions. Ksp AgCl (s) ⇋ Ag+ (aq) + Cl– (aq) Ksp = 1.8 × 10–10 Kf Ag+ (aq) + 2NH3 (aq)⇋ Ag(NH3)2+ (aq) Kf = 1.6 × 107 (2) Write an equilibrium constant expression (or expressions) Ksp = [Ag+][Cl–] Kf = [Ag(NH3)2+]/([Ag+][NH3]2) (3) The Notorious Table CHEM 114 Fundamental Chemistry Ksp = (x – y)x Kf = y/[(1.00 – 2y)2(x – y)] KspKf = xy/[(1.00 – 2y)2] ~ y2/[(1.00 – 2y)2] y = 0.0484639, x = 0.0484639 x – y = 3.71 × 10–10

  2. Complex ion equilibria CHEM 114 Fundamental Chemistry

  3. Heat flowing from a hot object to a cold object. • Gas expanding into a vacuum • A sidewalk absorbing sunlight and warming up. • Tequila, triple sec and lemon juice mixing to give a margarita. Spontaneous processes • Hydrogen reacting with oxygen to give water. • A spontaneous process is unidirectional without an external agent. • What determines the direction of spontaneous processes? CHEM 114 Fundamental Chemistry

  4. Expansion of an ideal gas • Before • After Note that the chances of all the molecules being on one side goes down as we increase the number of molecules. If we want it to go down linearly with the size of the system, we need to take the log of W. CHEM 114 Fundamental Chemistry Nowadays we write S = k ln W. k = R/NA

  5. Entropy of an ideal gas, and the second law S2 – S1 = n[R ln(V2/V1)+ Cp,mln(T2/T1)] The entropy of a system increases as the volume increases. The entropy of a system increases as the temperature increases The Second Law of Thermodynamics: Entropy increases in all spontaneous processes Clausius formulation: No process is possible whose sole result is the transfer of heat from a body of lower temperature to a body of higher temperature. Kelvin formulation: No process is possible in which the sole result is the absorption of heat from a reservoir and its complete conversion into work. Las Vegas formulation of the two main laws of thermodynamics: First Law: you can never win (ΔU = q + w = 0) Second Law: you can never break even (for a ‘heat engine’, w < qabsorbed) CHEM 114 Fundamental Chemistry

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