lecture19_March_29

# lecture19_March_29 - 4.4.D Non-standard free energy...

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4.4.D Non-standard free energy (Section 20.4) -when system is not at standard conditions (i.e., 1 bar or 1 M), we use Δ G = Δ Gº + RT ln Q Q = Reaction Quotient rxn at equilibrium At equilibrium: Δ Gº = RT ln K 59 - Δ Gº is free energy change for reaction at standard P or concentration conditions Q < K Δ G < 0 rxn proceeds to right; forward reaction Q > K Δ G > 0 rxn proceeds to left; reverse reaction Q = K Δ G = 0 (not necessarily at 298K) Δ G = 0, Q = K Δ G = RT ln K + RT ln Q = RT ln Q/K

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The dissociation of the weak acid HF (K = 7.2 × 10 -4 ) is HF (aq) H + (aq) + F - (aq) Find Δ G at 298 K when the initial concentrations are [HF] = 0.67 M, [F - ]= 0.52 M, [H + ]= 0.10 M Will the reaction proceed towards more HF dissociation? Δ G = Δ Gº + RT ln Q could use Δ f as T = 298 Δ G = + 11.6 kJ/mol = + 11.6 kJ Units? Calculations for 1 mole of HF (aq) as, so can leave out the “/mol”. No But Δ Gº in table for HF(g) not HF(aq)! 60 Δ Gº = RT ln K = (8.314 J/Kmol)(298 K)ln(7.2 × 10 -4 ) = + 17.9 kJ/mol = (8.314 J/Kmol)(298K) ln[(0.52)(0.10)/(0.67)] = 6.33 kJ/mol
61 4.4.E T-dependence of equilibrium constant (Section 17.6) Using: Δ G º = Δ H º ± T Δ S º and Δ Gº = ± RT ln K Divide by RT: ± Δ H

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## This note was uploaded on 04/27/2011 for the course ENGG 130 taught by Professor Lubell during the Spring '08 term at University of Alberta.

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lecture19_March_29 - 4.4.D Non-standard free energy...

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