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 FACT: ∆G o rxn is the change in free energy when pure reactants convert COMPLETELY to pure products.  FACT: Product-favored systems have K eq > 1. 

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Presentation on theme: " FACT: ∆G o rxn is the change in free energy when pure reactants convert COMPLETELY to pure products.  FACT: Product-favored systems have K eq > 1. "— Presentation transcript:

1  FACT: ∆G o rxn is the change in free energy when pure reactants convert COMPLETELY to pure products.  FACT: Product-favored systems have K eq > 1.  Therefore, both ∆G˚ rxn and K eq are related to reaction favorability. Thermodynamics and K eq

2 K eq is related to reaction favorability and so to ∆G o rxn. The larger the value of K the more negative the value of ∆G o rxn ∆G o rxn = - RT lnK ∆G o rxn = - RT lnK where R = 8.31 J/Kmol Thermodynamics and K eq

3 Calculate K for the reaction N 2 O 4 --->2 NO 2 ∆G o rxn = +4.8 kJ ∆G o rxn = +4800 J = - (8.31 J/K)(298 K) ln K ∆G o rxn = - RT lnK Thermodynamics and K eq K = 0.14 When ∆G o rxn > 0, then K 0, then K < 1

4 ∆G, ∆G˚, and K eq ∆G is change in free energy at non-standard conditions. ∆G is change in free energy at non-standard conditions. ∆G is related to ∆G˚ ∆G is related to ∆G˚ ∆G = ∆G˚ + RT ln Q where Q = reaction quotient ∆G = ∆G˚ + RT ln Q where Q = reaction quotient When Q K, reaction is spontaneous. When Q K, reaction is spontaneous. When Q = K reaction is at equilibrium When Q = K reaction is at equilibrium When ∆G = 0 reaction is at equilibrium When ∆G = 0 reaction is at equilibrium Therefore, ∆G˚ = - RT ln K Therefore, ∆G˚ = - RT ln K

5 But systems can reach equilibrium when reactants have NOT converted completely to products. In this case ∆G rxn is < ∆G o rxn, so state with both reactants and products present is MORE STABLE than complete conversion. ∆G, ∆G˚, and K eq Figure 19.10

6 Product favored reactionProduct favored reaction –∆G o and K > 1–∆G o and K > 1 In this case ∆G rxn is < ∆G o rxn, so state with both reactants and products present is MORE STABLE than complete conversion.In this case ∆G rxn is < ∆G o rxn, so state with both reactants and products present is MORE STABLE than complete conversion. ∆G, ∆G˚, and K eq

7 Product-favored reaction. 2 NO 2 ---> N 2 O 4 ∆G o rxn = – 4.8 kJ Here ∆G rxn is less than ∆G o rxn, so the state with both reactants and products present is more stable than complete conversion. ∆G, ∆G˚, and K eq

8 Reactant-favored reaction. N 2 O 4 --->2 NO 2 ∆G o rxn = +4.8 kJ Here ∆G o rxn is greater than ∆G rxn, so the state with both reactants and products present is more stable than complete conversion. ∆G, ∆G˚, and K eq

9  K eq is related to reaction favorability.  When ∆G o rxn < 0, reaction moves energetically “downhill”  ∆G o rxn is the change in free energy when reactants convert COMPLETELY to products. Thermodynamics and K eq


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