CHEM 106 1nd Edition Lecture 22Outline of Last Lecture II. Using K to Calculate [Equilibrium]III. Le Chatelier’s PrincipleIV. Calculations based on Ka. ICE table with Quadratic FormulaOutline of Current Lecture V. Equilibrium & ThermodynamicsVI. Van Hoff PlotCurrent LectureChapters 16: Chemical Equilibria (continued)Note: This material will be on exam 316.9 Equilibrium & ThermodynamicsRemember from chapter 14:ΔGrxn is a measure of the energy available to do useful workNeed to relate Q, K, ΔG, ΔG°Back to Fig. 16.5 pg. 781Q < K equilibrium shifts to products, as written (right)And ΔG < 0, spontaneousQ > K equilibrium shifts to reactants, as written (left)And ΔG > 0, only spontaneous in reverse directionQ = K reaction is at equilibrium (by definition)And ΔG = 0; no energy driver in either directionNote: True only for specific given conditions, e.g. [concentration], T, PWe use Q to relate ΔGrxn back to ΔGf°, e.g. standard conditions (1 atm, 298 K, etc.)ΔG = ΔG° + RT ln Q+RT ln Q is the correction factor for not being at standard conditionsEx: N2O4 ↔ 2NO2ΔGrxn° = 4.8 kJWe can calculate ΔGrxn° from standard thermo data in tablesThe calculated value only applies under standard conditions, e.g. whenPNO2 = PN2O4 = 1 atmBut what is the value of ΔG under other conditions?Suppose all reactants: e.g. N2O4m = 1 atm (no NO2 present)Qp=PNO2PN2O4=01Qp < Kp, form products (shift left) so, ΔGrxn < 0, spontaneous(even though ΔGf° > 0)Suppose all products: e.g. PNO2 = 2 atm PN2O4 = 0 atmvery smallQ=2¿( 0)¿=+∞Q > K, form reactantsΔGrxn > 0; not spontaneous in forward directionIs spontaneous in reverse reaction, so shifts to reactants (left)Under equilibrium conditions:Q = K ΔGrxn = 0 = ΔG° + RT ln KΔG° = -RT ln KWe can solve for K:K = e-ΔG°/RT This is the mathematical expression for where we started:If ΔG° >> 0, K << 1; if ΔG° << 0, K >> 1Back to N2O4 ↔ 2NO2 ΔG° = 4.8 kJWhat is Kp?−ΔG °RT=−4.8kJmol8.314Jmol ∙ K298 K=−1.94Kp = e-1.94 = 0.144What happens if we change T?You need to know how this works in a qualitative sense for Exam 2- Heat as either a reactant or productBut now we can also treat it quantitatively Remember: ΔG° = ΔH° - T ΔS°And, ln K = - ΔG°/RTSo, ln K = - ΔG°/RT = - ΔH°/RT + ΔS°/RThis is the quantitative treatment of the temperature dependence on KIf T ↑, the smaller the influence of the - ΔH°/RT termIncreasing T shifts the reaction to opposite side of “heat” term as reactant or productVan Hoff Plot ln K=−Δ H °R(1T)+∆ S °RPlot ln K vs 1/T Slope=−Δ H °R;intercept =ΔS °RMeasuring K as a function of T gives an estimate of ΔH° & ΔS°Slope of line will be negative for the plot of ln K vs. 1/T when:ΔH°>0; ln K ↓ as 1/T ↑ (or as T ↓); endothermicSlope of line will be positive for the plot of ln K vs. 1/T when:ΔH° < 0; ln K ↑ as 1/T ↑ (or as T ↓);