The Gibbs Free Energy

Transcription

1 40 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. S universe = S system + S surroundings 0 If S universe > 0, then the system is at equilibrium If S universe = 0, the system is at equilibrium If S universe < 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, G = H - T S defines the change in Gibbs Free energy. Note that this definition is consistent with G = q P q rev = q actual q rev Recall that if q actual is q rev, then G = 0. If q actual is q irrev, then G < 0 The spontaneity criterion can be expressed only in terms of properties of the system: G = 0 G < 0 G > 0 The system is at equilibrium The process is spontaneous The reverse process is spontaneous. Now, a process can be spontaneous even when H is unfavorable: G < 0 is facilitated by H < 0 and S > 0 But, even for some endothermic processes, entropy can cause the process to occur spontaneously

2 41 DEMO: Compare the solubilities of NH 4 NO 3 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 G 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 G Products So, we can define the standard Gibbs Free Energy change for any chemical reaction as follows: Reactants G Products -Σ G f, summed over reactants Elements +Σ G f, summed over products

3 42 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 G for any arbitrary reaction. So, G f = H f - T S The Gibbs free energies of formation for elements in their standard states are zero The standard change in entropy, S, 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, G 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. G aa + bb cc + dd 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 c [C] [D] Q = a [A] [B] d b using this relation, we find that we can get the desired relationship from G = G + RT ln Q = G RT log10 Q

4 43 At equilibrium, G = 0, so at equilibrium, Q = K eq, and therefore G = RT ln K eq 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 K eq for the solution equilibria that we demonstrated in class. H f (kj/mol) S (J / (K-mol) NaOH (s) NH 4 NO 3 (s) Na + (aq) NH + 4 (aq) NO - 3 (aq) OH - (aq) For NH 4 NO 3 : H = (- 366) = 28 kj/mol S = = 108 J /(K-mol) For NaOH: H = (- 427) = - 43 kj/mol S = = -16 J /(K-mol) We can define the enthalpic and entropic factors of the equilibrium constant as Enthalpic factor = e H RT and the entropic factor is given as e S R Let s evaluate both of these terms for the different dissolution processes.

5 44 For NH 4 NO 3 : Remember that this process is entothermic the beaker gets cold - H /RT = /[(8.31)(298)] = e - H /RT = e = The enthalpic factor in the equilibrium constant is small. Endothermic reactions, all other factors being equal, are expected to have small K eq. However, the entropic term is important: S /R = 108/8.31 = e S /R = So, K eq = ( )( ) = 5.7 So, the favorable entropic term make the dissolution of NH 4 NO 3 occur. The favorable entropy arises from the fact that the ordered crystalline structure of the solid is replaced by the more disordered solution. That is enough to overcome the unfavorable heat effect. For NaOH: Remember that this process is exothermic the beaker gets hot. - H /RT = /[(8.31)(298)] = e - H /RT = e = The enthalpic factor in the equilibrium constant is very large. Exothermic reactions, all other factors being equal, are expected to have large K eq. Now, let s consider the entropic term. S /R = -10/8.31 = -1.2 e S /R = So, K eq = ( )(0.15) = So, the favorable enthalpic term make the dissolution of NaOH occur.

6 45 It is a little surprising that the entropy change for dissolving NaOH is slightly negative. You would expect the solution to be highly disordered. However, small ions like Na + and OH - actually interact with the solvent and create ordered solvation shells around the ions. We say that small ions like Na + and OH - are structure-making ions. These structure-making effects are much smaller for larger, more complex ions like NO 3 -