CHEM 106 1nd Edition Lecture 10Outline of Last Lecture I. Absolute EntropyII. 3rd law of ThermodynamicsIII. Reactions + Spontaneity Outline of Current Lecture II. Free EnergyIII. Calculating ΔG°fIV. Temperature and SpontaneityV. Coupled ReactionsVI. Limitations of Thermodynamics Current LectureChapter 14: Thermodynamics (continued)From last lecture:ΔG = ΔH - T ΔSΔG < 0, spontaneous ΔG > 0, not spontaneousQuick review from Chapter 5:E → internal energy (page 200)Σ kinetic + potential energy of all components of system ΔE = q + w w → useful workQ → heat (no work)ΔG → energy available to do useful workCapital letters (e.g. ΔG, ΔH, ΔE, etc) indicate a state function Small letters not state functions – path dependent!Ex: Piston EngineIgnite fuel in piston, gases expand to move piston (w), also produces heat (wasted)W = -P ΔV section 5.2Conceptually, free energy tries to quantify useful work with energy from chemical reactions.ΔG provides insight into the maximum amount of useful energy possible from a given reaction.ΔG = ΔH - T ΔSΔH = ΔG + T ΔSΔG → energy available to do workΔH → enthalpy change from making and breaking bondsT ΔS → energy dispersed; disorderCalculating ΔG°f ΔG°f → standard free energy of formation From the elements for 1 mol of the substance ΔG°rxn = Σ nproducts ΔG°f, products – Σ nreactants ΔG°f, reactantsJust like with ΔH, ΔS, need balanced chemical equationAs with ΔH°f, remember that ΔG°f for elements in standard state is zero (e.g. O2(g), Cdiamond, Cu(s))<Reminder: S° is absolute!>Ex: Calculate ΔG°f for the following:H2O(g) + C(s) → H2(g) + CO(g)From appendix 4: (ΔG°f units are in kJ/mol)H2O (g): -228.6 C(s): 0 H2(g): 0 CO(g): -137.2ΔG°rxn = [1 mol H2 * 0 kJ/mol + 1 mol CO * -137.2 kJ/mol] – [1 mol H2O * -228.6 kJ/mol + 1 mol C * 0]= 91.4 kJIs this spontaneous?Temperature & Spontaneity For Chem 106, we assume that ΔS & ΔH are independent of TBut ΔG does vary with TΔG = ΔH - T ΔSSo, how can we know ΔG at other temperatures*need to calculate ΔH° & ΔS° @ 298K<Note title of Table, Appendix 4>Then use those ΔH & ΔS values to calculate ΔG @ any TEx: Phase EquilibriaH2O(l) ↔ H2O(g)ΔH° = 44.4 kJ ΔS° = 118.9 J/kSpontaneous? @ T = 298 K ΔG°f = 8.57 kJ no353 K +2 kJ no373K 0 kJ equilibrium393 K -2.7kJ yes*Trend: ΔS > 0 as T ↑, ΔG ↓Reaction becomes spontaneous at higher TΔG = 0 when there is no net change from the reaction.e.g. H2O (l) in equilibrium with H2O(g)We can calculate the T where ΔG = 0:ΔH = T ΔST = Δ HΔ S be careful with unitsFor H2O(l) ↔ H2O(g)T = 44,400 J118.9 J / k = 373 KFor H2O(g) + C(s) → H2(g) + CO(g)ΔH°rxn = 131.3 kJ ΔS°rxn = 133.8 J/k T = 981.3KIf ΔS < 0, ΔG ↑ as T ↑ spontaneous at lower T, can become non-spontaneous @ elevated T14.6 Coupled ReactionsCan couple a non-spontaneous reaction with a spontaneous reaction to make overall reaction spontaneousWorks like Hess’s Law, Chapter 5Very common in metabolismSimpler example to write out:2Fe2O3(s) → 4Fe(s) + 3O2(g)Reverse of iron rustingΔG = +742.3 kJCan make spontaneous by coupling with another reaction that is spontaneous4 Al(s) + 3O2(g) → 2Al2O3 (s)ΔG = -1,582.3 kJ/mol2Fe2O3(s) → 2Fe(s) + 3O2(g)4Al (s) + 3O2 → 2Al2O3 (s)2Fe2O3 + 4Al(s) → 2Al2O3 (s) + 2Fe(s)According to Hess’s Law:ΔG = 742.3 + (-1,582.3) = -839.9 kJFor simplest stoichiometryFe2O3(s) + 2Al(s) → Al2O3(s) + Fe(s)ΔG = -420 kJLimitations of ThermodynamicsTells us nothing about the rate of the reaction!Also tells us nothing about the mechanism, e.g. how the reaction