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Spontaneous Reactions and Gibbs Free EnergyFree Energy the energy given off by a reaction that is available to do workspontaneous reaction a reaction that proceeds to form theproducts without outside intervention All spontaneous reactions release free energy and are said to be exergonic. ( - G ) Whether or not a reaction is spontaneous depends on its entropy and enthalpy.Entropy measure of the disorder of a system + S is favorable ( S ) All systems strive toward increased entropy prefer greater disorder Enthalpy heat content of a system H is favorable ( H ) All systems strive to release energy prefer to be exothermic In order to determine the spontaneity of a reaction we use the equation for Gibbs free energy which is related to both the change in entropy (S) and the change in enthalpy (H). G = H T S G is the Gibbs Free Energy H is the Enthalpy T is the temp. in Kelvin S is the Entropy If G is negative, the reaction is spontaneous. (Rxn is exergonic.)If G is positive, the reaction is not spontaneous.(Rxn is endergonic.)If G is zero, the reaction is at equilibrium.A reaction that is not spontaneous, that is, a reaction with a + G at a given temperature may become spontaneous if the temperatureis changed. H S G Temperature (favorable) (favorable) spontaneous at ALL temperatures + or spontaneous at LOW temperatures + or spontaneous at HIGH temperatures (unfavorable) (unfavorable) spontaneous at NO temperatures 1. State whether the values for H and S would be positive or negative and favorable or unfavorable. Use these estimates to predict whether G would be sponataneous at all temps, high temps, low temps, or no temps. a. 2 H2 (g) + O2 (g) 2 H2O (g) + 242 kJ Exothermic reaction so H is negative; favorableReactants more disordered so S is negative; unfavorable Spontaneous at low temperaturesb. 6 CO2 (g) + 6 H2O (l) + heat C6H12O6 (s) + 6 O2 (g) Endothermic reaction so H is positive; unfavorableReactants more disordered so S is negative; unfavorable Spontaneous at no temperatures G is the Gibbs Free Energy H is the Enthalpy G = H T S T is the temp in Kelvin S is the Entropy Ho = Ho(products) Ho (reactants) So = So (products) So (reactants) 2. Using the values for standard entropy and standard enthalpy, determine whether the formation of Carbon Dioxide gas from the reaction of solid. Carbon (graphite) with Oxygen gas is spontaneous (standard temp = 25oC) C (s, graphite) + O2 (g) CO2 (g) Substance Hfo (kJ/mol) So (J/K mol)C (s, graphite) 0.0 5.69 = .0057 kJ/molO2 (g) 0.0 205 = .205 kJ/mol CO2 (g) - 393.5 214 = .214 kJ/mol So = So (products) So (reactants) S CO2 SC SO2 = .214 kJ/mol (.0057 kJ/mol + .205 kJ/mol) = .003 kJ/mol Ho = Ho(products) Ho (reactants)H CO2 HC HO2 = 393.5kJ/mol (0.0 kJ/mol + 0.0kJ/mol) =393.5 kJ/mol T = 25oC +

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