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Lesson 16.2 Free EnergySuggested Reading Zumdahl Chapter 16 Sections 16.4, 16.6 - 16.9?Essential QuestionWhat is the relationship between spontaneity and free energy?Learning Objectives:.Relate free energy to spontaneity.Calculate the standard free energy change for a chemical reaction.Show the relationship between free energy and the equilibrium constant.?Predict the direction of a reaction using the concept of free energy.Describe the free energy change with respect to equilibrium for both spontaneous and non-spontaneous reaction.IntroductionIn the previous lesson you learned that the quantity ?H - T?S can serve as a criterion for the spontaneity of a reaction at constant temperature and pressure. If the value of this quantity is negative the reaction is spontaneous, if positive the reaction is non-spontaneous, and if equal the reaction is at equilibrium. It is convenient to define a new thermodynamic quantity in terms of H and S that will be directly useful as a criterion of spontaneity. For this purpose, American physicist J. Willard Gibbs (1839 -1903) introduced the concept of free energy, G, which is a thermodynamic quantity defined by the equation G =?H - TS, or??G =??H - T?S?This quantity gives a direct criterion for spontaneity. As a reaction proceeds at a given T and P, reactants form products and the enthalpy and entropy changes. These changes result in a change in free energy. The following rules are useful in predicting the spontaneity of a reaction.When??G is a large negative number (more negative than -10 kJ), the reaction is spontaneous as written, and reactants transform almost entirely to products at equilibrium (K > 1).When??G is a large positive number (larger than about 10 kJ), the reaction is non-spontaneous as written, and reactants do not give significant amounts of products at equilibrium (K < 1).When??G has a small negative or positive value (less than about 10 kJ), the reaction gives an equilibrium mixture with significant amounts of both reactant and products.Standard Free Energy ChangeRecall that for the purposes of tabulating thermodynamic data, certain standard states are choses, which are indicated by a superscript degree sign on the symbol of the quantity. The standard states are as follows:?Pure liquids and solids: 1 atm pressure, 298 KGases: 1 atm partial pressure, 298 KSolutions: 1 M concentration, 298 KThe standard free energy change,??G°, is the free energy change that occurs when reactants in their standard states are converted to products in their standard states.??G° =??H° - T?S°Thus if we know the standard enthalpy and entropy changes for a reaction we can calculate the standard free energy change.Example: Calculating ??G°?=??H°?- T?S°? What is the standard free energy change?for the Haber process at 25 C.N2(g) + 3H2(g) → 2NH3(g)Solution:Calculate??H° ?and??S°, then substitute these values into??G°?=??H°?- T?S°.You may find it helpful to use a table to organize your data.?Reaction?N2(g)?3H2(g)?2NH3?Enthalpy (kJ)?0?3 x 0 = 0?2 x (-45.9) = -91.8?Entropy (J/K)?191.5?3 x 130.6 = 391.8?2 x 193 = 386Now calculate each quantity.?H°?= -91.8 - (0) = -91.8 kJ?S°?= 386 - (191.5 + 391.8) = -197. J/KSubstitute into equation for free energy, using units of kJ/K for S.?G°?=?-91.8 kJ?- (298 K)(-.197 kJ/K) = -33.1 kJThus, this reaction is deemed spontaneous under these conditions.Standard Free Energies of FormationThe standard free energy of formation,??G°f, of a substance is defined similarly to standard enthalpy of formation. That is,??G°f?is the free energy change that occurs when 1 mole of substance is formed from its elements in their stablest states (reference forms) at 1 atm and a specified temperature (usually 298 K).For example, the standard free energy of formation of NH3(g) is the free energy change for the reaction 1/2N2(g) + 3/2H2(g) → NH3(g). The reactants, each at 1 atm, are converted to 1 mol of product at 1 atm. In the previous example, we found that the??G°?for two moles of ammonia is -33.1 kJ, so the??G°f ?for 1 mole should equal -33.1kJ / 2 = -16.6 kJ.?As in the case of standard enthalpies of formation, the standard free energies of formation of the elements in their stablest states are assigned a value of 0. Also, as with standard enthalpies,??G°?can be found from??G°f??using the equation below. Standard energies of formation can be found alongside values for standard enthalpies of formation and standard entropies in tables of thermodynamic data.Example: Calculating??G°?from Standard Free Energies of FormationCalculate??G°?for the combustion of 1 mol of ethanol at 25 deg C.C2H5OH(l) + 3O2(g) → 2CO2(g) + 3H2O(g)Solution:Obtain the values for??G°f ?from a table of thermodynamic data, and then calculate???G°?from??G°f?using the equation above. Always double check to make sure your equations are balanced!?C2H5OH(l) + 3O2(g) → 2CO2(g) + 3H2O(g)?? ??? ??? ??? ??? ??? ??? ??? ??? ??? ??? ??? ??? ??? ?G°f??kJ: ????-174.8? ? ? ? ? ?0 ? ? ? ?2 x -394.4 ? 3 x -228.6Substitute and solve.?G°?= [2(-394.4) + 3(?-228.6)?- (-174.8)]kJ = -1299.8 kJSince??G°?is a large negative number we can predict that this reaction is spontaneous.Free Energy Change During a Reaction For any spontaneous reaction, the total free energy of the substances in the reaction mixture decreases (exergonic reaction) as the reaction proceeds. For a non-spontaneous reaction, free energy increases during a reaction (endergonic reaction).This is analogous to what occurs during exothermic and endothermic reactions.Free Energy and Equilibrium ConstantsOne of the most important results of chemical thermodynamics is an equation relating the standard free energy change for a reaction to the equilibrium constant.?If we know the equilibrium constant, Keq, for a chemical change (or if we can determine the equilibrium constant), we can calculate?the standard state free energy change,?Go, for the reaction using the equation:In this equationR = 8.314 J mol-1?K-1?or 0.008314 kJ mol-1?K-1.T is the temperature on the Kelvin scale.Keq?is the equilibrium constant at the temperature T.This equation also take the form?Go?= -2.303 RT log KIf you want the free energy change for a reaction involving reactants and products in nonstandard states (G), you can obtain it from the standard free energy change using this equation:G =?Go?+ RT ln Qwhere Q is the reaction quotient.Example: Calculating K from the Standard Free Energy ChangeFind the value of the equilibrium constant K at 298 K for the reactionThe standard free energy change equals -13.6 kJ.Using?Go?= -2.303 RT log K, we can rearrange for K as followsYou can understand and use?Go?as a criterion for spontaneity in terms of its relationship with K. When the equilibrium constant is greater than one, the equilibrium mixture is mostly products,?Go?is negative, so the reaction is spontaneous.?When the equilibrium constant is less than one, the equilibrium mixture is mostly reactants,?Go?is positive, so the reaction is non-spontaneous. When K is near 0,?Go?is near 0 and there are appreciable amounts of both reactants and products.For a given chemical reaction, the minimum free energy of the system is found at the equilibrium position as shown in the diagram to the right.Change in Free Energy with TemperatureIf you must calculate the change in temperature at a temperature other than 298 K, you can use a simple method that gives approximate results. Precise calculation are complicated, so this is the preferred method under these circumstances. In this method you assume??H°?and??S°?are approximately constant with temperature and use?G°T?=??H°?- T?S°?(approximation)The following table summarizes the effect of temperature on spontaneity.Homework: Book questions pg. 783 questions 45, 47, 51, 61, 63 ................
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