Thermo Notes # 4Entropy & the Free EnergyFriday, February 3CHEM 102HT. HughbanksProblem The “Mond Process” is usedcommercially to produce pure nickel.Ni(s) + 4 CO(g) → Ni(CO)4(g) First, try to predict the signs of ∆S° and ∆H°. Then use data on next slide to calculatevalues.Data for Problem Ni(s) + 4 CO(g) → Ni(CO)4(g) ∆S° , ∆H° = ??Ni(s) CO(g) Ni(CO)4(g)∆Hf° 0 −110.52 −602.9S298° 29.87 197.56 410.6∆Hf° in kJ mol-1, S298° in J K-1 mol-1.Watch units!Answers for Problem Ni(s) + 4 CO(g) → Ni(CO)4(g) ∆S° = -409.5 J/K ∆H° = −160.8 kJFree Energy One more state function .... We know ∆Suniverse > 0 for a spontaneouschange, but ∆Suniv = ∆S(sys) + ∆Ssurr andkeeping track of the surroundings isinconvenient, to say the least. We are still looking for a state function of thesystem that will predict spontaneity. Define a new function that satisfies this need.Call it “free energy.” (sometimes “Gibbs freeenergy”)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 functionthat predicts spontaneity. Consider a process that occurs at constanttemperature. ∆G = ∆H – T∆S This is the central equation in chemicalthermodynamics!ΔG & Spontaneity ∆G = ∆H - T∆S(remember, when not shown, ΔX = ΔXsys) Compare this with ΔSuniv . ∆Suniv = ∆Ssys + ∆Ssurr = ∆S + ∆Ssurr∆Ssurr = qsurr / T = − qsys / T= − ΔH / T (const P,T) So: ∆Suniv = ∆S − ∆H/T  Or: T∆Suniv = T∆S − ∆H∆G & Spontaneity∆G = ∆H – T∆ST∆Suniv = T∆S – ∆HSo ∆G = – T∆Suniv From this, we see that ∆G and ∆Sunivwill always have opposite signs. (T > 0) Spontaneous process → ∆Suniv > 0, so Spontaneous process → ∆G < 0∆G & Spontaneity ∆G is thus the function we have beenseeking: a state function of the system sign tells us whether a process (reaction orphase change) is spontaneous ∆G is generally the most usefulthermodynamic function for a chemist.ΔG - Change in Free Energy Predictor of spontaneity. A spontaneous reactionhas ∆G < 0. Also tells us the maximum amount of energywhich can be produced and used to do work. So∆G is useful in determining amounts of fuelneeded, etc. We saw that when ∆Suniv = 0, the system andsurrounding are at equilibrium. Likewise, when∆G = 0, the system is at equilibrium – a concept wewill emphasize for the next several weeks!Spontaneity: Role of ∆H & ∆S The form of ΔG shows us the role of ΔH and ΔS in determining spontaneity.∆G = ∆H − T∆ S ∆H < 0 → exothermic → favors spontaneity ∆S > 0 → entropy increases → favors spontaneitySpontaneity: Role of T ΔG tells us whether or not a reactionwill occur spontaneously.∆G = ∆H − T∆S Usually, we assume that ∆H and ∆S donot depend on T. This means that onlythe T∆S term varies with T. Effect of temperature depends on signs.Spontaneity: Role of ∆H, ∆S, T!H !S" +AlwaysSpontaneous+ "NeverSpontaneous+ +Spontaneous atsufficiently high T" "Spontaneous atsufficiently low T∆G: Using Tabulated Data Thermodynamic tables (Appendix E) usuallyinclude ∆Gf° values. These are defined and used just like ∆Hf°’s. refer to formation reactions same standard state convention ∆G°rxn = Σ n ∆Gf°products - Σ n ∆Gf°reactants Can also use ∆H°, ∆S° to find ∆G°Tabulated Data For N2(g): ∆Hf° = 0, ∆Gf° = 0, S° = 191.61 J K-1 mol-1 Note that the data give S° and ΝΟΤ ΔSf° Also, note that ∆Gf° IS NOT EQUAL TO∆Hf° − T S°Problem - Mond ProcessNi(s) + 4 CO(g) → Ni(CO)4(g) (From earlier problem, ∆H° = –160.8 kJand ∆S° = –409.5 J/K ) Given ∆Gf° = –137.168 kJ/mol for CO,find ∆Gf° for Ni(CO)4(g). Use ∆H° and ∆S° to find the range oftemperatures at which the reaction isspontaneous.Mond ProcessNi(s) + 4 CO(g) → Ni(CO)4(g) ∆H° and ∆S° are both negative, so at lowT the reaction is spontaneous. At high T, ∆G° becomes positive, soreaction proceeds spontaneously to theleft. At high T, Ni(CO)4 decomposes.Mond ProcessNi(s) + 4 CO(g) → Ni(CO)4(g) For Ni purification:First react impure Ni with pure CO toform 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. (Tboil = 42°C for Ni(CO)4.)Mond ProcessNi(s) + 4 CO(g) → Ni(CO)4(g) 2nd step in purification: Remove gas phase from reactionchamber. This is a mixture of CO andNi(CO)4. Run reaction “backwards” to producepure nickel as a precipitate.Mond ProcessNi(s) + 4 CO(g) → Ni(CO)4(g) To produce Ni, we want the reaction togo from right to left. This means highT needed. Typically run at 230°C, where ∆G = ∆H – T∆S = + 45.18 kJ Finally, remove pure Ni from vessel.Reaction Free Energies - More Example (prob. 751): Use data in Appendix2A to calculate ∆G˚ for each of thefollowing reactions under standardconditions (at 25 K):(a) 2 SO2(g) + O2(g) → 2 SO3(g)(b) 2 CaCO3(s) → CaO(s) + CO2(g)(c) 2 C8H18(l) + 25 O2(g) → 16 CO2(g) +18 H2O(l)Data358+6.4–249.9C8H18(l)69.91–237.13–285.83H2O(l)213.74–394.36–393.51CO2(g)39.75–604.03–635.09CaO(s)92.9–1128.8–1206.9CaCO3(s)256.76–371.06–395.72SO3(g)205.14O2(g)248.22–300.19–296.83SO2(g)Sm˚ (J/mol–1 K-1)∆Gf˚ (kJ/mol-1)∆Hf˚(kJ/mol-1)∆H, ∆S, & ∆G vs. T The most important temp. dependencein ∆G can be seen directly:∆G = ∆H - T∆S ∆H and ∆S actually do depend on T:(∆X = Xfinal(T) – Xinitial(T); X = H or S)For infinitesimal changes for a single substance:dH = CPdT ; dS =dqT=CPdTT∆H, ∆S, & ∆G vs. T ∆H and ∆S at temperatures other than298 K can be obtained by integration,provided we know the heat capacities:H (T ) = CPd!T298T" ; S(T ) =CPd!T!T298T" We then account for the changes in H’sand S’s for all the substances in the“reactants” and “products” and pluginto ∆G = ∆H -