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Section 19.3 Free Energy, the Reaction Quotient and the Equilibrium Constant

Section 19.3 Free Energy, the Reaction Quotient and the Equilibrium Constant. Bill Vining SUNY College at Oneonta. Recall Q, K and Reaction Favorability. The system is not at equilibrium. Reactants will be consumed, and product

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Section 19.3 Free Energy, the Reaction Quotient and the Equilibrium Constant

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  1. Section 19.3Free Energy, the Reaction Quotient and the Equilibrium Constant Bill Vining SUNY College at Oneonta

  2. Recall Q, K and Reaction Favorability The system is not at equilibrium. Reactants will be consumed, and product concentration will increase until Q = K. The reaction will proceed in the forward direction as written (reactants  products). Q < K The system is at equilibrium, and no further change in reactant or product concentration will occur. Q = K The system is not at equilibrium. Products will be consumed, and reactant concentration will increase until Q = K. The reaction proceeds to the left as written (reactants  products). Q > K

  3. G, Go and Q Q represents the current state of the system. G determines if the reaction system in its current state is spontaneous in either direction. Go determines if the system is product-favored or reactant-favored. G < 0 Reaction is spontaneous in the forward direction (reactants  products). G = 0 Reaction is at equilibrium. G > 0 Reaction is spontaneous in the reverse direction (reactants  products). Go< 0 Reaction is product-favored. Go= 0 Reaction is neither reactant- or product-favored.. Go > 0 Reaction is reactant-favored.

  4. Connecting G, Go and Q G = Go + RT lnQ Use standard thermodynamic data (linked) to calculate G at 298.15 K for the following reaction, assuming that all gases have a pressure of 16.91 mm Hg. 2NO(g) + O2(g) 2NO2(g) Gfo (kJ/mol) NO(g) 86.6 O2(g) 0 NO2(g) 51.3

  5. Connecting G, Go and Q G = Go + RT lnQ Use standard thermodynamic data (linked) to calculate G at 298.15 K for the following reaction, assuming that all gases have a pressure of 16.91 mm Hg. 2NO(g) + O2(g) 2NO2(g) Gfo (kJ/mol) NO(g) 86.6 O2(g) 0 NO2(g) 51.3

  6. G, Go and the Equilibrium Constant, K At equilibrion, G = 0 so 0 = Go + RT ln Q Go = RT ln Q and Use standard thermodynamic data to calculate K at 298.15 K for the following reaction. 2NO(g) + O2(g) 2NO2(g) Gfo (kJ/mol) NO(g) 86.6 O2(g) 0 NO2(g) 51.3 Go = -70.6 kJ

  7. G, Go and the Equilibrium Constant, K A student determines the value of the equilibrium constant to be 1.11E6 for the following reaction.3Fe2O3(s) + H2(g)2Fe3O4(s) + H2O(g)Based on this value of Keq: G° for this reaction is expected to be (greater, less) than zero. Calculate G° and the free energy change for the reaction of 1.67 moles of Fe2O3(s) at standard conditions at 298K.

  8. G, Go, Q, and K

  9. G, Go, Q, and K

  10. G, Go, Q, and K Can a chemical system have a positive value of Go and yet still be favored to reactin the forward direction?

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