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Chemistry and the material world 123.102 Unit 4, Lecture 4 Matthias Lein Gibbs free energy Gibbs free energy to predict the direction of a chemical process. Exergonic and endergonic reactions. Temperature dependence of the Gibbs free energy. The chemical equilibrium. The law of mass action. Reaction quotients. Gibbs free energy and equilibrium constants. Equilibrium temperature. Consider the reaction: N2(g) + 3 H2(g) → 2 NH3(g) Let us calculate ΔH° and ΔS° for the following reaction to decide in which direction each of these factors will drive the reaction. ΔHf°(N2) = 0 kJ/mol, ΔHf°(H2) = 0 kJ/mol, ΔHf°(NH3) = -46.11 kJ/mol ΔSf°(N2) = 191.61 J/molK, ΔSf°(H2) = 130.68 J/molK, ΔSf°(NH3) = 192.54 J/molK ΔH° = -92.22 kJ ΔS° = -198.78 J/K Enthalpy favours the reaction. It is exothermic, heat is released. Entropy dis-favours the reaction. Order is increased, the number of particles decreases. The questions remains: Will the reaction occur spontaneously? The Gibbs free energy G is defined as: G = H -TS a measure that connects the two previously treated thermodynamic functions H, the enthalpy and S, the entropy. ΔG < 0 ΔG = 0 ΔG > 0 favored reaction (Spontaneous) Neither the forward nor the reverse reaction prevails (Equilibrium) disfavored reaction (Non-spontaneous) Josiah Willard Gibbs United States, 1839 – 1903 Just like reactions are called exothermic and endothermic depending on the heat produced (ΔH), reactions are also called: ΔG < 0 exergonic (Spontaneous) ΔG > 0 endergonic (Non-spontaneous) Wether a reaction is exergonic or endergonic depends on the reaction enthalpy, the reaction entropy and the temperature at which the reaction is carried out. Consider the reaction (at 25° C): N2(g) + 3 H2(g) → 2 NH3(g) Let us calculate ΔH° and ΔS° for the following reaction to decide in which direction each of these factors will drive the reaction. ΔHf°(N2) = 0 kJ/mol, ΔHf°(H2) = 0 kJ/mol, ΔHf°(NH3) = -46.11 kJ/mol ΔSf°(N2) = 191.61 J/molK, ΔSf°(H2) = 130.68 J/molK, ΔSf°(NH3) = 192.54 J/molK ΔH° = -92.22 kJ ΔS° = -198.78 J/K ΔG = ΔH° -TΔS° ΔG = -92,220 J - (298.15 K × -198.75 J/K) ΔG = -32.96kJ Answer: at 25° C the reaction occurs spontaneously (it is exergonic) How about the same reaction at 500° C? N2(g) + 3 H2(g) → 2 NH3(g) ΔH° = -92.22 kJ ΔS° = -198.78 J/K ΔG = ΔH° -TΔS° ΔG = -92,220 J - (773.15 K × -198.75 J/K) ΔG = 61.4kJ Answer: at 500° C the reaction does not occur spontaneously (it is endergonic) The entropic term becomes more and more important as the temperature rises. ΔG = ΔH° -TΔS° The sign of ΔG tells us the direction in which a reaction will proceed to reach an equilibrium and the magnitude of ΔG tells us how far from equilibrium the reaction still is (ΔG = 0 means equilibrium). Thus, as the reaction progresses, the magnitude of ΔG will become smaller and smaller until an equilibrium is reached. The chemical reaction is then finished. ΔG = ΔH° -TΔS° When we calculate ΔG values for reactions, we are comparing G values for the reactants and products. Therefore, the calculated ΔG value for the reaction only represents the true ΔG value for the very start of the reaction. Over the course of the reaction the magnitude of ΔG will go to zero. Equilibrium: N2O4(g) → 2 NO2(g) colourless brown Equilibrium: N2O4(g) → 2 NO2(g) Equilibrium: N2O4(g) → 2 NO2(g) Equilibrium: N2O4(g) → 2 NO2(g) Law of mass action (Kc is the equilibrium constant) This equation only describes the equilibrium. Equilibrium: N2O4(g) → 2 NO2(g) Systems that are not at equilibrium can be described by a similar Qc expression. Qc is called the reaction quotient Remember that Qc and Kc depend on temperature and pressure. Equilibrium: N2O4(g) → 2 NO2(g) The connection between the equilibrium constant and the Gibbs free energy: ΔG° = -RT ln K So, to calculate ΔG for a given reaction quotient: ΔG = ΔG° +RT ln Q (note that the expression becomes zero for Q = K) What does the magnitude of Kc mean? K = 1 means that the reaction quotient is balanced. There is approximately the same amount of educt and product in the reaction mixture. K > 1 means that that the numerator in the reaction quotient is larger. There is more product in the reaction mixture. K < 1 means that the denominator in the reaction quotient is larger. There is more educt in the reaction mixture. For example: 2 SO2(g) + O2(g) → 2 SO3(g) ΔG° = -1.40⋄102 kJ/mol What is the value of the equilibrium constant K? We know: ΔG° = -RT ln K (R = 8.314 J/molK, T = 298K) ln K = -ΔG° / RT = -(-1.40⋄102 kJ/mol) / (8.314 J/molK × 298K) = 56.5 = e56.5 So: K = 3⋄1024 The equilibrium temperature We know that at equilibrium: ΔG = 0 which means that ΔH – TΔS = 0 That means that if we know two quantities out of ΔH, ΔS and T, we can calculate the value of the third at the equilibrium. For example: Br2(l) → Br2(g) ΔH° = 30.9 kJ/mol, ΔS° = 93.2 J/molK At what temperature is the two phases of liquid and gaseous Br2in equilibrium? At equilibrium: ΔG = ΔH – TΔS = 0 so ΔH = TΔS and T = ΔH / ΔS T = 30900 J/mol / 93.2J/molK = 332 K 332 K = 59° C ● Today we covered: ● Gibbs free energy to predict the direction of a chemical process. Exergonic (ΔG < 0) and endergonic (ΔG > 0) reactions. Temperature dependence of the Gibbs free energy. The chemical equilibrium. The law of mass action. Reaction quotients. Gibbs free energy and equilibrium constants (ΔG° = -RT ln K). Equilibrium temperature (ΔH = TΔS). ●