lecture 07 2010

lecture 07 2010 - 36 The Gibbs Free Energy We have seen...

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Unformatted text preview: 36 The Gibbs Free Energy We have seen that the entropy, S, is a thermodynamic function of state that allows us to predict when a process is spontaneous or not. The entropy was invented as a variable that allows us to make this determination. ΔSuniverse = ΔSsystem + ΔSsurroundings ≥ 0 If ΔSuniverse > 0, then the process is spontaneous If ΔSuniverse = 0, the system is at equilibrium If ΔSuniverse < 0, then the reverse process is spontaneous Now, knowing what is going on in the entire universe is often not a particularly useful way to determine spontaneity. It would be better to have a function that can be evaluated more simply and depend only on the system. The appropriate variable is the Gibbs Free Energy, denoted G, and is appropriate for constant temperature and pressure situations. G = H – TS defines the Gibbs free energy. At constant temperature, ΔG = ΔH - TΔS defines the change in Gibbs Free energy. Note that this definition is consistent with ΔG = qP – qrev = qactual – qrev Recall that if qactual is qrev, then ΔG = 0. If qactual is qirrev, then ΔG < 0 The spontaneity criterion can be expressed only in terms of properties of the system: ΔG = 0 ΔG < 0 ΔG > 0 The system is at equilibrium The process is spontaneous The reverse process is spontaneous. Now, a process can be spontaneous even when ΔH is unfavorable: ΔG < 0 is facilitated by ΔH < 0 and ΔS > 0 But , even for some endothermic processes, entropy can cause the process to occur spontaneously 37 Compare the solubilities of NH4NO3 and NaOH. The first is strongly endothermic, but entropy is favorable because the crystalline lattice is broken, to make a more dissordered system. The beaker gets very cold in the dissolucation process. The second is strongly exothermic, and the solution gets very hot. Standard States The standard Gibbs Free Energy change ΔG° for a process is the change in Gibbs free energy for a hypothetical process in which the reactants in their standard states are converted into products in their standard states. As for enthaply, the standard state is the state of matter at specified temperature and pressure. So, we see that a process is spontaneous if the standard Gibbs Free Energy is negative. Reactants Free energy Products Δ G° So, we can define the standard Gibbs Free Energy change for any chemical reaction as follows: Δ G° Products Reactants -Σ Δ Gf° , summed over reactants +Σ Δ Gf° , summed over products Elements 38 So, in a process completely analogous to what we used for computing reaction enthalpy changes, we can use free energies of formation (for formation of a substance from the elements) to calculate ΔG° for any arbitrary reaction. So, ΔGf° = ΔHf° - TΔS° The Gibbs free energies of formation for elements in their standard states are zero The standard change in entropy, ΔS°, can be calculated from Third Law entropies. The Third Law of thermodynamics says that the entropies of ideal crystalline solids approach zero as T → 0 K. More later. Now, ΔG° has a great deal of physical significance. Let’s make a connection between a chemical reaction at equilibrium, and a chemical reaction in which reactants and products are in their standard states. ΔG aA + bB cC + dD Δ G° aA + bB cC + dD What is the relationship between these two different equilibria? We need a correction between a reaction under any conditions and at standard conditions. Remember that the reaction quotient Q is given by using this relation, we find that we can get the desired relationship from At equilibrium, ΔG = 0, so at equilibrium, Q = Keq, and therefore 39 The beauty and extraordinary characteristic of this relationship is that we can calculate the equilibrium state, an actual state, from the standard state Gibbs free energy change, a quantity that describes a hypothetical process, one that generally does not occur in Nature.!! THIS IS A BIG DEAL. Let’s calculate a couple of equilibrium constants. Let’s just calculate Keq for the solution equilibria that we demonstrated in class. Δ Hf° (kJ/mol) NaOH (s) NH4NO3 (s) Na+ (aq) NH4+ (aq) NO3- (aq) OH-(aq) -427 -366 -240 -132 -205 -230 S° (J / (K-mol) 64 151 59 113 146 -11 For NH4NO3: ΔH° = -132 –205 –(- 366) = 28 kJ/mol ΔS° = 146 + 113 –151 = 108 J /(K-mol) For NaOH: ΔH° = -240 –230 –(- 427) = - 43 kJ/mol ΔS° = 59 –11 –64 = -16 J /(K-mol) ...
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This note was uploaded on 02/13/2010 for the course CHM 29275 taught by Professor Farrar during the Spring '10 term at Rochester.

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