CHEM 106 1nd Edition Lecture 9Outline of Last Lecture I. MicrostatesII. EntropyIII. Reactions + SpontaneityIV. Thermodynamics + Entropy of reactionsOutline of Current Lecture II. Absolute EntropyIII. 3rd law of thermodynamicsIV. Reactions + Spontaneity Current LectureChapter 14: Thermodynamics (continued)Remember from Chapter 5: Transferring heat between system + surroundingsExothermic: q < 0 -system lost heat to surroundingsEndothermic: q > 0 -system gained heat from surroundingsFig. 5.12 showing heat flow pg. 199Recall burning methane:NH4NO3(s) → NH4+(aq) + NO3-(aq) Reaction is spontaneous, even though it is endothermicThis is because ΔSsys is large + positive (Fig. 14.9 c)14.3 Absolute Entropy + 3rd Law of Thermodynamics3rd law- the entropy of a perfect crystal is zero at Temperature = absolute 0A perfect crystal has only 1 microstate so:S = k ln (1) = 0 Remember: Heat (q) is proportional to heat capacityq = n Cp ΔT = n ΔH enthalpy is relativeWe can determine absolute SStandard molar enthalpy is S°Table 14.2 pg. 673 S° not ΔS*don’t memorize values, look at the trends instead*Br2(g) 245.5 J/mol · k Br2(l) 152.2As phase changes from more condense to less condense, S° ↑Methane (g) 186.2 J/mol · k ethane (g) 229.5As molecule gets more complex, S° ↑As molecule get more flexible, S° ↑H2(g) 188.8 J/mol · k N2(g) 191.5 O2(g) 205.0As mass increases, S° ↑Entropy is a function of Temperature-entropy always ↑ with Temperature (for a given molecule) Fig. 14.11 pg. 674Note discontinuities → phase changes14.4 Calculating ΔS°rxnSimilar approach to calculating ΔH°rxn-need balanced chemical reaction- remember it is always products minus reactantsΔS°rxn = Σ n products S° products - Σ n reactants S° reactantsNumbers come off of table of values (appendix 4 in book)Ex: CH4(g) + 2O2(g) → CO2(g) + 2H2O(l)Pay attention to phasesValues: CO2(g) 213.8 H2O(g) 188.8 H2O(l) 69.9CH4(g) 186.2 O2(g) 205.0ΔS°rxn = [(213.8 + 2(69.9)] – [186.2 + 2(205.0)] = -242.6 J/kWhy is this answer negative?Because the products are more condensed than the reactants, therefore entropy decreased, so the answer makes sense.Process is spontaneous if ΔSuniverse ↑Assume constant Temperature and Pressure Gibbs Free Energy G-the energy available to do workRemember: ΔSuniverse = ΔSsystem + ΔSsurroundings And: -qsystem = + qsurroundings And: ΔSsystem = q revTAt a constant Pressure, qrev = ΔHsystem(equation 14.5)ΔSsurroundings = −q revT = −ΔH systemT Therefore, ΔSuniverse = ΔSsystem + (−ΔH rxnT) (all in terms of the system)-TΔSuniverse = -TΔSsystem + ΔHsystem = ΔHsystem – TΔSsystem ΔG → at constant Temperature and Pressure: -TΔSuniverseSo, ΔG = ΔH – TΔS Back to burning methane: (reaction is in previous notes)ΔS°rxn = ΔS°system = -242.6 J/kΔHrxn = ΔHsystem = -890.3 kJ/kNote the different units aboveΔSsurroundings = −ΔH systemT = +890.3 kJ298 K = 2.99 x 103 J/kSo, ΔSuniverse = -242.6 + 2.99 x 103 = 2.74 x 103 J/kΔSuniverse > 0 therefore the reaction is spontaneous-exothermic because ΔH overcomes the loss of entropyRemember: if ΔSuniverse > 0 = spontaneousΔG = -T ΔSuniverseIf ΔSuniverse > 0, then -T ΔSuniverse < 0 So ΔG < 0 = spontaneousKeep in mind: ΔG < 0 spontaneousΔG > 0 non spontaneousΔG is a state function, if the reaction is reversed then the sign on ΔG is reversed as wellWhat happens if ΔG is 0?Then -T ΔS = 0Therefore there is no thermodynamic driving forceTo be