Second Law of Thermodynamics Answer Key Discussion

Discussion #12 Things you should know when you leave Discussion today: 1. Second law of Thermodynamics: ΔSnet= ΔSsystem+ ΔSsurrounding i. ΔSnet = ΔSsystem + ΔSsurrounding > 0 process is spontaneous ii. ΔSnet = ΔSsystem + ΔSsurrounding < 0 process is not spontaneous iii. ΔSnet = 0 process is at Equilibrium T = Tequilibrium 2. ∆Ssurroundings = ∆Hsurr Tsurr =− ∆Hsystem Tsurr i. ΔSsur is thermal entropy due to distribution of quanta (q) of energy among fixed particles (m). Increase in number of quanta (q) of energy will increase the number of distinguishable arrangements. More quanta (q) of energy correspond to higher temperature. 3. ΔSsystem = ΔrS°= Σ [ni (S°)]product − Σ [nj (S°)]reactants= Sfinal−Sinitial i. ΔSsystem is positional entropy due to distribution of molecules. The more volume per molecule, the more distinguishable arrangements. ii. Where S° is Standard Entropy of Formation. Standard Entropy of Formation for elements in the standard state does not equal zero iii. Sgas > Ssolution > Ssolvent > Ssolid ∆Hsys iv. ∆Ssys = Tequilibrium 4. Entropy diagram for freezing point lowering of water ∆Sfr = ∆H൫H2 O(l)→H2O(s)൯ Tfp Ssolution (H2O+solute) ΔHfr – represents heat transfer when H2O(l)→H2O(s) it is the same in both cases. Ssolvent (H2O(l)) ∆S(l→s) = ∆S(solution→solid) = ∆H൫H2 O(l)→H2 O(s)൯ Tfp(solution) ∆H൫H2O(l)→H2 O(s)൯ Tfp(solvent) Ssolid(H2O(s)) Ssolid(H2O(s)) 1. Draw an entropy diagram that describes boiling point elevation. Tnew>T 1atm Solvent vapor above solution S Solution Solution ΔH Tnew Solvent Solvent vapor above Pure Liquid ΔH 0 Tboil Pure Liquid a. Compare the entropy change when a pure solvent boiling and when a solution boiling: ΔSsoln > ΔSpure ΔSsoln < ΔSpure ΔSsoln = ΔSpure b. Compare the magnitudes of entropy change when a pure solvent boiling and when a solution boiling: │ΔSsoln│ >│ΔSpure│ │ΔSsoln │< │ΔSpure│ ΔSsoln = ΔSpure c. The magnitudes of enthalpy change ΔHvap (water, l → water, g) compared to ΔHvap (solution, l → water, g) is… Greater smaller the same d. Tbp (solution) compared to Tbp (solvent) is ... Greater Smaller the same 2. A liquid boils at 400. K and its enthalpy of vaporization is ΔvapH = 40. kJ/mol. When a nonvolatile solute is added to the liquid, the entropy of the solution at 400 K is 10. J/(mol/K) higher than that if the pure liquid at 400 K. Calculate the boiling point of the solution. Hint: Draw the diagram. 1atm Solvent Vapor Above solution Solvent Boiling point Solvent vapor above Pure Liquid S(Solution) ∆S(solution) = 10J/molK S(Pure Liquid) ΔSsys (solvent-gas)= ΔSsys (solvent-gas)= H sys Teq H sys Teq = = H (H 2 O(l)  H 2 O(s)) T fp 40000 J / mol J  100 400 K molK ΔSsys(solution-gas) = ΔSsys (solvent-gas)− 10J/molK= 90J/molK; Teq= H sys S sys = 40000 J / mol  444 K J 90 molK 2 3. At 90 °C the vapor pressure of water is 0.9 atm and the vapor pressure of an aqueous glucose solution is 0.8 atm. Construct an entropy diagram for vapor pressure lowering and answer the questions below using 1 your drawing. Hint: have both entropy and pressure scales, and remember that S ∝ W ∝ V ∝ 𝑃 . Entropy Difference Solvent Vapor above Solution T=(90+273.15)K ΔH vap Solvent Vapor above Pure Liquid T=(90+273.15)K T Same S ΔH vap Solution P T  ΔSvap  0 Pure Liquid a. ΔHvap (water(l)→ water(g)) compared to ΔHvap (glucose solution(l)→ water(g)) is… Greater smaller the same b. S (water (l)) compared to S (glucose solution(l)) is… Greater smaller the same c. S (water(g)) compared to S (glucose solution(g)) is… Greater smaller the same d. ΔSvap (water(l) → water(g)) compared to ΔSvap (glucose solution(l) →water(g)) is… Greater smaller the same 4. An ionic solid M2X3(s) is highly soluble in water at 300 K and when it dissolves in water to form M3+ (aq) and X2−(aq), the entropy change of the system is −450 J/(K mol). Compared to its solubility at 300 K, the solubility of M2X3(s) in boiling water will be … Circle the correct answer and explain why: lower the same higher ΔSsystem <0 for the process to be spontaneous ΔSsurrounding has to be positive and H sys │ΔSsur│>│ΔSsys│ since ΔSsur= − >0 this means that process is exothermic ΔHsys < 0 T Increasing the T of an exothermic reaction makes it less spontaneous and solid less soluble. 5. To melt a certain liquid ΔH°fus = 15.0 kJ/mol. And the freezing point is at 45°C. Calculate ΔSsys of the substance going from liquid to solid. ΔH°sys= −ΔH°fus H sys ΔSsys= =-15000/(273+45)=-47.2 J/mol·K T 3 6. A liquid freezes at 215. K and its ΔfussH = 25. kJ/mol. When a nonvolatile solute is added to the liquid, the entropy of the solution is 15. J/(mol/K) higher than that of the pure liquid at 215K. Calculate the freezing point of the solution. Hint: use the diagram on page 1 for help 7. Compare the entropy change when a pure solvent freezes and when a solution freezes: ΔSsoln > ΔSpure ΔSsoln < ΔSpure ΔSsoln = ΔSpure 8. Compare the magnitudes of entropy change when a pure solvent freezes and a solution freezes: │ΔSsoln│ >│ΔSpure│ │ΔSsoln │< │ΔSpure│ ΔSsoln = ΔSpure 9. Compare the entropy change when vapor forms over pure water and over a solution, both at 25 C: ΔSsoln > ΔSpure ΔSsoln < ΔSpure ΔSsoln = ΔSpure 10. Calculate the temperature (in K) at which melting 36.0 grams of water (molar mass of water is 18.0 g/mol) changes the entropy of the universe by –16.0 J/K. For water, ∆fusH = 6.00 kJ/mol and ∆fusS = 22.0 J/(mol K). ΔSuniv = ΔSsystem + ΔSsurrounding= ΔSsystem - H system Tsur Since we have 2 moles of water: 6.00 –16.0 J/K= 22.0 J/(mol K)*2moles - kJ J  2moles 1000 molK kJ . Tsur T=200. K 4 In preparation for next week, not part of exam 3: 1. ΔG° = −T∙ΔSnet = ΔH°sys − TΔS°sys ° ∆Greaction = nj ൫∆Gf° ൯ − ∑ products nj ൫∆Gf° ൯ ∑ @25°C reactants a. ΔG° = − R∙T∙ln(K) =− 2.3· R∙T·log(K) −( ∆G° ) b. K = 10 R∙T∙2.3 c. ΔG°= −ne·F·E° = −q·E° Q Q 2. ΔG = R∙T∙ln (K) = 2.3 ∙ R ∙ T ∙ log (K) = ∆G° + R ∙ T ∙ ln(Q) ( ∆G−∆G° ) a. Q = 10 R∙T∙2.3 b. ΔG = −T ΔSnet = ΔHsys − TΔSsys c. ΔG= −ne·FE = − q·E i. ΔG<0 process is Spontaneous ii. ΔG>0 process is Not Spontaneous iii. ΔG=0 process is at Equilibrium 11. Remember that ΔGº = − RT ln K = ΔHºsys − TΔSºsys and so we can derive ln K=  ln Keq  H sys RT +  S sys R Reaction 1 Reaction 2 0 High T region Low T region 𝟏 →0 T→ ∞ 𝟏 →0 T→ ∞ 𝐓 𝐓 Reaction 3 𝟏 𝐓 Reaction 4 a. When ln K is positive. What does it mean? Why? b. When ln K is negative. What does it mean? Why? c. When ln K is zero. What does it mean? Why? ∆H° ∆S° sys ***Remember that ln 𝐾 = − R∙T + Rsys and that equations of the line is y = mx + b*** d. What is the y-intercept of the expression for lnK? e. What is slope of the expression for lnK? f. The slope is positive. What does it mean? Why? 5 g. The slope is negative. What does it mean? Why? For Reactions 1, 2, 3 and 4 in the plot above, answer the following questions. h. Is the reaction endothermic or exothermic and at what temperatures? Reaction 1: i. Reaction 3: Reaction 4: If the reaction above are redox reactions, which ones will have positive standard voltages and at what temperatures? Reaction 1: j. Reaction 2: Reaction 2: Reaction 3: Reaction 4: At what temperatures each of reactions will increase or decrease entropy of the system? Reaction 1: Reaction 2: Reaction 3: Reaction 4: k. For each reaction choose at what temperatures the ratio of product to reactants will be greater than 1 or smaller than 1? Reaction 1: Products/ Reactants >1: High T Low T never always Products/ Reactants <1: High T Low T never always Reaction 2: Products/ Reactants >1: High T Low T never always Products/ Reactants <1: High T Low T never always Reaction 3: Products/ Reactants >1: Products/ Reactants <1: Reaction 4: Products/ Reactants >1: Products/ Reactants <1: l. High T High T High T High T Low T never Low T Low T never never Low T never always always always always It is a well-known fact that mixing any group 1 metal and water leads to an explosion. The plot of which reaction corresponds to the reaction between Na and water: 2Na(s) + 2H2O(l)  2Na+(aq)+2 OH−(aq) + H2(g) Reaction 1: Reaction 2: Reaction 3: Reaction 4: m. The plot of which reaction corresponds to freezing of water? Reaction 1: Reaction 2: Reaction 3: Reaction 4: 12. For the following combustion reaction, calculate ΔG° at 25°C : C6H12O6 (s) + 6O2 (g) → 6 CO2 (g) + 6 H2O (l) a. Without doing a calculation, at what temperatures will this process cease to be spontaneous at standard states? O2(g) C6H12O6 (s) CO2 (g) H2O(l) ΔG°f(kJ/mol) (only at 25°C) 0. −900. −400. −240. 13. Circle the value of ΔG when both products and reactants are in their standard states. (Note: Q=1) ΔG = 1 ΔG = 0 ΔG = ΔG° 6 None of these 14. Circle the value of ΔG when Q = K. ΔG = 1 ΔG = 0 ΔG = ΔG° None of these 15. Circle the value of ΔG° when Q = K. ΔG° = 1 ΔG° = 0 ΔG° = -RT lnK None of these 16. Circle the value of ΔG° when Q > K. ΔG° = 1 ΔG° = 0 ΔG° = -RT lnK None of these 17. When would ΔG° = 0? 18. If ΔG= −5.0kJ/mol and ΔGº= 3.0kJ/mol circle everything that must be true: Q=1 Q>1 Q<1 QK a. Assuming temperature of 298 K, calculate the values of K and Q 19. Here are thermodynamic data for the gas phase dimerization. Calculate equilibrium constant at 37°C for the dimerization. 2 HC(O)OH(g) → (HC(O)OH)2 (g). ΔfH°( HC(O)OH(g))= -350.0 kJ/mol, Δ fH°( (HC(O)OH)2 (g))= -800.0 kJ/mol, S°(HC(O)OH(g))= 250.0 J/(K mol), S°( (HC(O)OH)2 (g))= 350.0 J/(K mol). a. Find ΔrxnH°= b. Find Δrxn S°= c. Find ΔGº (Hint: ΔGº = ΔH°sys − TΔS°sys) : d. Calculate equilibrium constant: Hint: ΔGº = − RT lnK : e. Calculate the ΔG of dimerization reaction when the pressure of HC(O)OH is at 2.0 bars and the pressure of (HC(O)OH)2 is at 3.5 bars at 37°C. Hint: ΔG =ΔG ° + RT ln Q i. Calculate Q: ii. ΔG= iii. Calculate the total energy available for useful work from the dimerization of 1.00 kg of the reactant used at 37°C. 20. For autoionization of water at 330. K, Kw=1.0x10−13 and at 300K, Kw=1.0x10−14. Calculate the value of ΔH in kJ/mol and ΔS in J . Do your calculations to 1 significant figure. molK 21. The standard free energy change for the process A (aq) + B (aq)  3C (aq) is ΔG° = +54 kJ/mol. 7 If the concentration of B is decreased from 1M to 0.50 M (but concentrations of A and C stay at standard states) circle the value of the free energy change, ΔG, relative to ΔG°. ΔG < ΔG° ΔG = ΔG° ΔG > ΔG° 22. If ΔG= 5.0 kJ/mol and ΔGº= 4.0 kJ/mol circle everything that must be true: Q=1 Q>1 Q<1 QK a. Assuming temperature of 298 K calculate the values of K and Q.Useful Hint: at T=298K value of 2.3RT = 2.3· 8.314(J.(mol·K) ·298K =5698J/mol ≈ 6kJ/mol 23. Given the two reactions below, find the relationships between ΔG1° and ΔG2°. A(aq) + B(g) C(s) + D(aq) with Keq=K1 and ΔG°=ΔG1° 1 1 1 1 A(aq) + B(g)  C(s) + D(aq) with Keq=K2 and ΔG°=ΔG2° 3 3 3 3 24. For the following reaction at 25ºC using information below. H2(g) N2(g) NH3(g) 3H2(g) + N2(g)  2NH3(g) a. Calculate value of ΔG°. ΔGº kJ/mol 0 0 −16 P (atm) 0.010 0.040 0.20 b. Calculate value of K. c. Calculate value of Q. d. Calculate value of ΔG. e. Based on your calculation, did the reaction pass an equilibrium point? f. Based on your calculation, will the reverse reaction be spontaneous or not? 25. For the following reaction: Br2(l) Br2(g) ΔH°f For the reaction above at 343 K use the information (kJ/mol) given, to calculate ΔHsys, ΔSsys, ΔSsur, ΔSuniv, ΔG° and Br2(g) 30. the temperature at which process will cease to be Br2(l) 0 spontaneous. Do your calculations to 2 significant figures. Assume that ΔH°rxn and ΔS°rxn are independent of temperature. 8 S° (J/mol∙K) 250. 150. ΔG°f(kJ/mol) (only at 25°C) 3.0 0 26. You have a buffer with a pH = 5 at two different temperatures: T1 = 300 K and T2 = 330 K HA(aq) + H2O (l)  A− (aq) + H3O+ (aq) You need to have the following ratios of weak acid and conjugate base to make this happen: at T1 → [A− ]/[HA] = 1/2 at T2→ [A− ]/[HA] = 1/4 To answer the following questions you need to draw the plot of lnK verses 1/T. a. Will this reaction be exothermic, endothermic or neither. b. The system entropy change will be: Circle one: ΔS > 0 Answers: 1. a. ΔSsoln < ΔSpure b. │ΔSsoln │< │ΔSpure│ c. the same d. greater 2. 444K 3. a. The same b. Smaller c. smaller d. The same 4. lower 5. -47.2 J/mol·K 6. 7. 8. 9. 10. 200.K 190.K ΔSsoln < ΔSpure │ΔSsoln │> │ΔSpure│ ΔSsoln = ΔSpure Not part of exam 3 11. a. spontaneous; ΔG° < 0 b. non-spontaneous; ΔG° > 0 c. equilibrium; no more “potential” for the rxn to proceed d. ΔS°sys/R e. ΔH°sys/R f. exothermic; ΔH < 0 for -ΔH°sys/R to be (+) g. enodothermic; ΔH > 0 for -ΔH°sys/R to be (-) h. Reaction 1: exo, all T Reaction 2: exo, all T ΔS = 0 S<0 Reaction 3: endo, all T Reaction 4: endo, all T j. Reaction 1: E° always (+) Reaction 2: E° only (+) at low T Reaction 3: E° only (+) at high T Reaction 4: E° never (+) k. Reaction 1: increase all T Reaction 2: decrease all T Reaction 3: increase all T Reaction 4: decrease all T l. Reaction 1: R/P > 1 always; R/P < 1 never Reaction 2: R/P > 1 low T; R/P < 1 high T Reaction 3: R/P > 1 high T; R/P < 1 low T Reaction 4: R/P > 1 never; R/P < 1 always m. Reaction 1 n. Reaction 2 12. -2940kJ/mol a. ΔHºsys<0 (exothermic) ΔSºsys>0 13. ΔG = ΔG° 14. ΔG = 0 15. ΔG° = -RT lnK 16. ΔG° = -RT lnK 17. K=1 18. Q < 1 Q < K a. 0.3. 0.040 19. a. −100.0 kJ/mol 9 b. c. d. e. −150.0 J/(K mol) −53.48 kJ/mol 1.0399∙109 i. Q=0.875 ii. -53.84 kJ/mol iii. −585.2kJ 20. 60 kJ/mol; -60 J/mol∙K 21. ΔG > ΔG° 22. a. Q>1; Q>K b. 0.2, 1.5 1 23. 𝐾2 = 3√𝐾1 ∆𝐺2° = ∆𝐺1° 3 24. a. -32 kJ/mol b. 4.04∙105 c. 1∙106 d. 2.2kJ/mol e. Yes f. Spontaneous 25. 26. a. ΔH < 0 b. S < 0 10