Lec 28 March 16 Mon

Lec 28 March 16 Mon - 3.19 D Non-standard free energy(Sect 19-6(when you no longer have 1atm/1M conditions Δ G o is the driving force of a

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Unformatted text preview: 3.19 D. Non-standard free energy (Sect. 19-6) (when you no longer have 1atm/1M conditions) -- Δ G o is the driving force of a reaction done at standard pressure or concentration conditions (not necessarily at 25 ºC)-- but relative amounts of reactants or products could also drive a reaction (remember LeChatelier’s Principle)-- so non-standard Δ G depends on Δ G o and concentrations For non-standard conditions, now use : Δ G = Δ Gº + RT ln Q 1. If Q < K rxn → Δ G < 0 2. If Q > K rxn ← Δ G > 0 3. If Q = K Δ G = 0 Therefore, when P ≠ 1 atm and conc ≠ 1 M at equilibrium: Δ G = 0 = Δ Gº + RT ln Q and Q = K at eq. Δ Gº = – RT ln K • The equilibrium constant for a reaction is related to the change in standard free energy. (Remember K has concentrations divided by 1 M so that K has no units) • If you know the equilibrium constant, you now have a third way of determining Δ Gº , and this way is good for any temperature. • When a reaction is at equilibrium, usually Δ Gº ≠ 0 because usually eq ≠ 1 M. For reactions, generally, K ≠ 1. • When is Δ Gº = 0 ? Only when there is equilibrium when products and reactants are both under standard conditions (e.g. finding phase transition at 1 atm) Δ G = – RT ln K + RT ln Q = – RT ln (K/Q) 3.20 Ex. The dissociation of the weak acid HF (KEx....
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This note was uploaded on 03/25/2009 for the course CHEM 102 taught by Professor Drk during the Spring '09 term at University of Alberta.

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Lec 28 March 16 Mon - 3.19 D Non-standard free energy(Sect 19-6(when you no longer have 1atm/1M conditions Δ G o is the driving force of a

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