Gibbs Free Energy

Gibbs Free Energy (G) Balances the relationship between enthalpy (ΔH) and entropy (ΔS) Balances the relationship between enthalpy (ΔH) and entropy (ΔS) State function State function Enthalpy of system minus the product of temperature times entropy of system Enthalpy of system minus the product of temperature times entropy of system G = H – TS G = H – TS Maximum amount of energy available to do work, “free” Maximum amount of energy available to do work, “free”

Change in Gibbs Free Energy (ΔG) ΔG = ΔH – TΔS ΔG = ΔH – TΔS Relates enthalpy and entropy to determine which has more importance in determining whether a reaction is spontaneous Relates enthalpy and entropy to determine which has more importance in determining whether a reaction is spontaneous Combines energy transfer as heat (ΔH) and energy released to contribute to disorder (ΔS) Combines energy transfer as heat (ΔH) and energy released to contribute to disorder (ΔS)

Change in Gibbs Free Energy (ΔG) ΔG = ΔH – TΔS ΔG = ΔH – TΔS ΔG < 0, spontaneous reaction ΔG < 0, spontaneous reaction Energy available to do work Energy available to do work ΔG > 0, nonspontaneous reaction ΔG > 0, nonspontaneous reaction Energy deficiency, no leftover energy and not enough energy for reaction Energy deficiency, no leftover energy and not enough energy for reaction

How can we apply the Gibbs equation to determine spontaneity of reaction? ΔHΔSΔGResult Spontaneous (all temperatures) Nonspontaneous (all temperatures) Spontaneous (low temperatures) Nonspontaneous (low temperatures) ΔG = ΔH – TΔS

Example 1: Predict whether the reaction will be spontaneous or nonspontaneous and under what temperature conditions (use chart and explain). Predict whether the reaction will be spontaneous or nonspontaneous and under what temperature conditions (use chart and explain). a)C 6 H 12 O 6 + 6O 2  6CO 2 + 6H 2 O ΔH = kJ a)Cl 2 (g)   2Cl (g) b)N 2 + 2F 2  N 2 F 4 ΔH = -7.1 kJ

Standard Free Energy Change (ΔG°) ΔG° = ΔH° – TΔS° ΔG° = ΔH° – TΔS° Free energy change for a reaction occurring under standard conditions (1 atm, 1M, 25°C) Free energy change for a reaction occurring under standard conditions (1 atm, 1M, 25°C)

Two Paths to Calculating ΔG 1)ΔG = ΔH – TΔS Determine ΔH. What methods can we use? Determine ΔH. What methods can we use? Determine ΔS. Determine ΔS. Then calculate ΔG Then calculate ΔG

Two Paths to Calculating ΔG 2) Use Standard Free Energy of Formation (ΔG f °) values to determine ΔG Standard Free Energy of Formation (ΔG f °) --- ΔG° for the formation of 1 mole of a chemical compound in its standard state. Standard Free Energy of Formation (ΔG f °) --- ΔG° for the formation of 1 mole of a chemical compound in its standard state. ΔG f ° for element formation in their most stable state = 0. ΔG f ° for element formation in their most stable state = 0. aA + bB  cC + dD aA + bB  cC + dD ΔG° =[c (ΔG f °) C + d(ΔG f °) D ] - [a (ΔG f °) A + b (ΔG f °) B ]

Example 2: A)Find ΔG for a chemical reaction given ΔH = -218 kJ and ΔS = -765 J/K at 32°C. B) At what temperature does this reaction become spontaneous? Assume only temperature changes.

Example 3: Calculate ΔG° rxn under standard conditions for the following reaction using ΔG f ° values. Fe 2 O 3 (s) + 2Al (s)  2Fe (s) + Al 2 O 3 (s)

Example 4: Calculate ΔG° at 298 °K for the following reaction using the Gibbs equation. 4HCl (g) + O 2 (g)  2Cl 2 (g) + 2H 2 O (g) ΔH°= kJ

A spontaneous reaction is NOT necessarily fast!!!! A spontaneous reaction is NOT necessarily fast!!!! Reaction rate involves kinetics ! ! Reaction rate involves kinetics ! !

Homework pp. 743 #19, 27, 33, 36-37, pp. 743 #19, 27, 33, 36-37, 40-41