Free Energy and Spontaneity CHEM 107 T. Hughbanks Free Energy One more state function... We know S universe > 0 for a spontaneous change, but... We are still looking for a state function of the system that will predict spontaneity. Define a new function that satisfies this need. Call it free energy. (sometimes Gibbs free energy )
Free Energy: Definition Define the free energy by: G = H TS G is a state function, since H, T, & S are. If T & P are variables we control, G is the function that predicts spontaneity. Consider a process that occurs at constant temperature. G = H T S This is the central equation in chemical thermodynamics! G & Spontaneity G = H - T S (for any X, when not shown, X = X sys ) Compare this with S univ. So: S univ = S sys + S surr = S + S surr S surr = q surr / T = q sys / T = H/T (const P) S univ = S H/T Or: T S univ = T S H
G & Spontaneity G = H T S T S univ = T S H So G = T S univ From this, we see that G and S univ will always have opposite signs. (T > 0) Spontaneous process S univ > 0, so Spontaneous process G < 0. G & Spontaneity G is thus the function we have been seeking: a state function of the system sign tells us whether a process (reaction or phase change) is spontaneous G is generally the most useful thermodynamic function for a chemist.
G - Change in Free Energy Predictor of spontaneity. A spontaneous reaction has G < 0. Also tells us the maximum amount of energy which can be produced and used to do work. So G is useful in determining amounts of fuel needed, etc. Spontaneity: Role of H & S The form of G shows us the role of H and S in determining spontaneity. G = H T S H < 0 exothermic favors spontaneity S > 0 entropy increases favors spontaneity
Spontaneity: Role of T G tells us whether or not a reaction will occur spontaneously. G = H T S Usually, we assume that H and S do not depend on T. This means that only the T S term varies with T. Effect of temperature depends on signs. Spontaneity: Role of H, S, T ΔH ΔS + Always Spontaneous + Never Spontaneous + + Spontaneous at sufficiently high T Spontaneous at sufficiently low T
2 NO 2 N 2 O 4, S < 0, H < 0 low T: N 2 O 4 favored; high T: NO 2 favored N 2 O 4 : colorless O 120 pm N 134 O NO 2 : brown G: Using Tabulated Data Thermodynamic tables (Appendix E) usually include G f values. These are defined and used like H f s. refer to formation reactions same standard state convention G rxn = n G f products n G f reactants Can also use H and S to find G
Tabulated Data For N 2 (g): H f = 0, G f = 0, S = 191.61 J K -1 mol -1 Note that the data give S and NOT S f Also, note that G f TO H f TS IS NOT EQUAL Problem - Mond Process Ni(s) + 4 CO(g) Ni(CO) 4 (g) (From earlier problem, H = 160.8 kj and S = 409.5 J/K ) Given G f = 137.17 kj/mol for CO, find G f for Ni(CO) 4 (g). Use G and S to find the range of temperatures at which the reaction is spontaneous.
Ni(s) + 4 CO(g) Ni(CO) 4 (g) (1) H = 160.8 kj ; S = 409.5 J/K G = H T S = 160.8 kj (298 K)( 0.4095 kj/k) = 38.77 kj/mol C(graphite) + ½ O 2 (g) CO(g) (2) G f = 137.17 kj/mol (given) Find G f : Ni(s) + 4 C + 2 O 2 (g) Ni(CO) 4 (g) (3) Ni(s) + 4 CO(g) Ni(CO) 4 (g) H = 160.8 kj ; S = 409.5 J/K Use G and S to find the range of temperatures at which the reaction is spontaneous.
Mond Process Ni(s) + 4 CO(g) Ni(CO) 4 (g) H and S are both negative, so at low T the reaction is spontaneous. At high T, G becomes positive, so reaction proceeds spontaneously to the left. At high T, Ni(CO) 4 decomposes. Mond Process Ni(s) + 4 CO(g) Ni(CO) 4 (g) For Ni purification: First react impure Ni with pure CO to form Ni(CO) 4. (This works only for Ni, other metals are not as reactive with CO.) Need low T for reaction to go to right. Run at 50 C. (T boil = 42 C for Ni(CO) 4.)
Mond Process Ni(s) + 4 CO(g) Ni(CO) 4 (g) 2 nd step in purification: Remove gas phase from reaction chamber. This is a mixture of CO and Ni(CO) 4. Run reaction backwards to produce pure nickel as a precipitate. Mond Process Ni(s) + 4 CO(g) Ni(CO) 4 (g) To produce Ni, we want the reaction to go from right to left. This means high T needed. Typically run at 230 C, where G = H T S = + 45.18 kj Finally, remove pure Ni from vessel.
G & phase changes G & phase changes: Example: H 2 O(l) H 2 O(g) For P = 1 atm; T < 373 K (100 C) G > 0 (we know water doesn t boil below 100 C) For P = 1 atm; T > 373 K G < 0 For P = 1 atm; T = 373 K G = 0 At the boiling point, the liquid and vapor are in equilibrium. The boiling point is defined as the temp. at which the liquid and vapor phases are at equilibrium at 1 atm.
G : H 2 O(l) H 2 O(g) more From Appendix E: H 2 O(g) H 2 O(l ) H f (kj/mol) 241.8 285.8 S (J/mol K) 188.8 69.9 For the vaporization process at 298 K: H 298 = H f (gas) - H f (liq.) S 298 = S (gas) - S (liq.) = 241.8 ( 285.8) = +44.8 kj = 188.8 69.9 = 118.9 J/K = 0.119 kj/k G 298 = H 298 (298 K) S 298 = +8.9 kj G : H 2 O(l) H 2 O(g) more We saw that at 298 K, G > 0 ( H 298 / S 298 ) = (44.8/0.119) K = 376 K this is close to 373 K, the boiling point! Why? rearrange: H 298 = (376 K) S 298 H 298 - (376 K) S 298 = 0 recall: H 373 -(373 K) S 373 = G 373 = 0 (why?) if H 373 H 298 and S 373 S 298 then G 373 = 0 H 298 - (373 K) S 298
Generalization H and S often don t change much for a process or reaction from H 298 and S 298 Often, can estimate for G at other temps.: G(T) H 298 T S 298 P-T Phase diagram for H 2 O
Phase diagram for CO 2 note: boiling and freezing points are not defined for CO 2. after Figure 13-16, not to scale Free Energies and Phase Diagrams
Phase Diagrams All phase changes are determined by G, phases are in equilibrium when G = 0. P-T phase diagrams tell us what phases are stable at various values of T and P. A line separating two regions of a phase diagram represents the temperatures and pressures where G = 0 for the change from one phase to another. Example: recent exam Aluminum metal has a melting point of 660.3 C. Use the table below to calculate an estimate for the boiling point of aluminum in C Substa nce H f (kj mol 1 ) G f (kj mol 1 ) S (J mol 1 K 1 ) C p (J mol 1 K 1 ) Al(s) 28.33 24.35 Al(l) 10.56 7.20 39.55 24.21 Al(g) 326.4 285.7 164.54 21.38 Explain why this is just an estimate. A one- or twosentence explanation is all that is necessary.
Example - from old exam Solid tin exists in two common forms (allotropes), white and gray tin. The standard molar enthalpy and molar free energy for the transformation between these two allotropes are given below Sn white Sn gray H = -2.090 kj/mol; G = 0.130 kj/mol (a) Under standard conditions (298 K, 1 atm), which allotrope is thermodynamically stable? (b) The standard molar entropy, S, of gray tin is 44.14 J mol -1 K -1. What is the standard molar entropy of white tin? (c) The melting point of tin is 505 K. For temperatures below 505 K, give the temperature ranges (if any) over which white and gray tin are the most stable allotropes.