Ch06-TCF

Ch06-TCF - Quantitative Chemical Analysis Chapter 6...

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Quantitative Chemical Analysis Quantitative Chemical Analysis Chapter 6: Chemical Equilibrium Chemical Equilibrium
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For the reaction aA + bB cC + dD We write the equilibrium constant K = [C] c [D] d [A] a [B] b All concentrations are referenced to their standard states solutes: 1 M gases: 1 bar (0.987 atmosphere) pure solids: concentration of pure solid pure liquids: concentration of pure liquid
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Thus; solute: [C] moles/liter [1] mole/liter = [C] ( unitless ) = [P] ( unitless ) [1] bar gas: [P] bars = [1] ( unitless ) pure solid: [concentration puresolid] [concentration pure solid] pure liquid: [concentration pure liquid] [concentration pure liquid] = [1] ( unitless ) And K is dimensionless P.S . Solvents of dilute solutions are generally taken to be pure liquids, so . .. [solvent] = 1.
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Equilibrium and Thermodynamics For the system: A + B C + D NOT at equilibrium: At equilibrium: [A] q [B] q [C] q [D] q Q = K = [C] e [D] e [A] e [B] e Gibbs free energy of system is defined … G = G o + RT ln Q Now, at equilibrium, G = 0 and Q = K so … G o = RT ln K ( R = 8.314 J/K . mole)
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•N ow , G rxn = H rxn T S rxn ; ( Reactants Products ) H = enthalpy, S = entropy Enthalpy ~ heat exothermic H rxn < 0 endothermic H rxn > 0 Entropy ~ order/disorder S rxn < 0 products more ordered S rxn > 0 products less ordered Gibbs free energy of reaction exergonic G rxn < 0 reaction lies to the right as written endergonic G rxn > 0 reaction lies to the left as written
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Lets take an example of an acid-base equilibrium Dilute solution, 25 o C From data tables, K = 1.75 x 10 5 , what is G o of dissociation? G o = (8.314)(298) ln (1.75x10 5 ) 1,000 = (2.48)( 10.953) = 27.1 5 kJ/mole mantissa Positive G o , so equilibrium lies to the left
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If exactly 1.00 formal concentration of HOAc in pure water, what are the actual concentrations of all species?
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Ch06-TCF - Quantitative Chemical Analysis Chapter 6...

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