Chm152 Exam 4 Review

CHM 152 Exam 4 Review – Ch. 18 19 KEY 1. Predict whether the entropy change will be positive or negative for the following: a. H2O (g)  H2O (l) S__-__ b. C6H12O6(s)  2C2H5OH(l) + 2CO2(g) S__+_ c. 2NH3(g) + CO2(g)  H2O(l) + NH2CONH2(aq) S__-__ d. NaCl(s)  NaCl(aq) S__+__ e. Cu(s) (100oC)  Cu(s) (25oC) S__-__ f. 2NH3(g) N2(g) + 3H2(g) S__+__ 2. Calculate ∆H°rxn for C6H12O6(s)  2C2H5OH(l) + 2CO2(g) using ∆H°f values. ∆H°rxn =[(2 mol)(-277.7kJ/mol)+(2 mol)(-393.5kJ/mol)]-(1mol)(-1273.02 kJ/mol) =-69.38kJ C6H12O6 C2H5OH CO2 ∆Hf° (kJ/mol) -1273.02 -277.7 -393.5 S° (J/mol∙K) 212.1 160.7 213.6 3. Calculate ∆S°rxn and ∆S°surr. ∆S°rxn =[(2 mol)(160.7 J/molK)+(2 mol)(213.6 J/molK)]-(1mol)(212.1 J/molK)=536.5 J/K Δ Ssurr   Δ Hsys ΔSsurr  T  ( 69380 J)  232.7 J/K 298.15 K 4. Based on your values for ∆S°rxn and ∆S°surr, is the reaction spontaneous under standard thermodynamic conditions? Yes, ∆S°rxn/sys + ∆S°surr = 536.5 J/K + 232.7 J/K = 769.2 J/K. Since this value is >0, it is spontaneous. 5. Is this reaction always spontaneous? If not, determine at what temperatures it changes from spontaneous to non-spontaneous. With a -∆H°, and +∆S°, this reaction will be spontaneous under all conditions. 6. For the following reaction: 2Mg(s) + O2(g)  MgO(s) , ΔHorxn = -1202 kJ/mol; ΔSorxn = -217 J/mol∙K  Calculate ∆G° using ∆G=∆H-T∆S ∆G=-1202 kJ/mol – (298.15K)(-0.217 kJ/mol K) = -1137 kJ/mol  Is the reaction spontaneous under standard conditions? Yes, since ΔH and ΔS (along with 298.15K) are all under standard conditions (the “°” tells us standard conditions).  Based on the signs of ΔH and ΔS, comment on the spontaneity of this reaction at different temperatures. Spontaneous at “low” temperatures, non-spontaneous at “high” temps.  At what temperature does the reaction become spontaneous/non-spontaneous? 0 =-1202 kJ/mol –T(-0.212 kJ/mol K) 1202 kJ/mol  5670 K At this temp and higher, the reaction is non-spontaneous. 0.212 kJ/mol K 7. a) Caclculate Grxn using G=S (assuming 298.15K) 3C2H2(g)  C6H6(g) ΔH°f (kJ/mol) ΔG°f (kJ/mol) S° (J/mol K) C2H2(g) 227.4 209.9 200.9 C6H6(g) 82.9 129.7 269.2 ΔH°rxn = (1 mol C6H6)(82.9 kJ/mol) - (3 mol C2H2)(227.4 kJ/mol) = -599.3 kJ ΔS°rxn = (1 mol C6H6)(269.2 J/mol) – (3 mol C2H2)(200.0 J/mol) = -333.5 J G=S ΔG°rxn = -599.3 kJ – (298.15 K)(-0.3335 kJ) = -499.9 kJ b) Calculate Grxn for the following reaction using ΔG°f only. ΔGrxn = (1 mol C6H6)(129.7 kJ/mol) – (3 mol)(209.9 kJ/mol) = -500.0 kJ  Are the values similar? Should they be? Yes they are. And they should be. You are calculating the same thing, ΔGrxn.  What is the value of the equilibrium constant at 298 K? - 500.0x10 3 J  ln K - (8.314J/molK)(298.15K)  ∆G=-RTlnK e 201.7  e ln K K=3.955x1087 Is this reaction spontaneous at 298 K? Yes, because ∆Grxn < 0. We could also look at K, which is very large. Each of these shows spontaneity of the reaction at standard comditions. Calculate Grxn with the product and reactant both starting at 35atm and 501K. (Hint, what is the value of Q?). Q 35  8.16 x10  4 3 (35) ΔG = ΔG° + RTlnQ ΔG = -500.0x103J + (8.314 J/mol K)(501K)ln(8.16x10-4) = -529.6x103 J or -529.6kJ It is “more spontaneous” under the new conditions. 8. Balance the following equations: Assume acidic media: Cr2O72-(aq) + Cl-(aq)  Cr3+(aq) + Cl2(g) Cr2O72-  2Cr3+ 2Cl-  Cl2 Cr2O72-  2Cr3+ + 7H2O 2Cl-  Cl2 + 2e- 14H+ + Cr2O72- 2Cr3+ + 7H2O 3(2Cl-  Cl2 + 2e-) 6e-+ 14H+ + Cr2O72- 2Cr3+ + 7H2O 6Cl-  3Cl2 + 6e- 14H+ + Cr2O72- + 6Cl-  2Cr3+ + 7H2O 3Cl2 Assume basic media: MnO42- + C2O42-  Mn2+ + CO2 2- 2+ C2O4  2CO2 2- 2+ C2O4  2CO2 + 2e MnO4  Mn 22- MnO4  Mn + 4H2O - + 2- 2+ 2(C2O4  2CO2 + 2e ) 2- + 2- 2+ 2C2O4  4CO2 + 4e 8H + MnO4  Mn + 4H2O 2- 8H + MnO4  Mn + 4H2O - + 2- - - 2+ 4e + 8H + MnO4  Mn + 4H2O - + 2- 2- 2+ 8OH + 8H + MnO4 + 2C2O4  Mn + 4H2O + 4CO2 + 8OH 2- 2- 2+ 8H2O + MnO4 + 2C2O4  Mn + 4H2O + 4CO2 + 8OH 2- 2- 4H2O + MnO4 + 2C2O4  Mn 2+ + 4CO2 + 8OH - - 4H2O + MnO42- + 2C2O42-  Mn2+ + 4CO2 + 8OH- - 9. Assign oxidation numbers of all elements in the following reaction:  Al(s) + O2(g) → Al2O3(s)  MnO2(s) + 4HCl(aq) → MnCl2(aq) + Cl2(g) + 2H2O(l)  HNO3(aq) + NaOH(aq) → H2O(l) + NaNO3(aq) Reactants: Al = 0, O = 0 Products: Al = +3, O = -2 Reactants: Mn = +4, O = -2, H = +1, Cl = -1 Products: Mn =+2, Cl in MnCl2 = -1 Cl in Cl2 = 0, H = +1, O = -2 Reactants: H in both reactants = +1, N = +5, Na = +1, O in both reactants = -2, Products: H = +1, O in both products = -2, Na = +1, N = +5  For each reaction determine what is reduced and what is oxidized. O: 0 → -2, reduced; Al: 0 → +3, oxidized Mn: +4 → +2, reduced; Cl: -1 → 0, oxidized Oxidation numbers for last reaction unchanged; no a REDOX reaction  For each reaction determine the oxidizing agent and reducing agent. Oxidizing agent: O2(g) Reducing agent: Al(s) Oxidizing agent: MnO2(s) Reducing agent: HCl When naming “agents”, chemists customarily give the entire formula not just the element 10. Draw a voltaic cell diagram using silver and copper (assume the Cu2+ ion). Place the anode on the left hand side. Also show the following: salt bridge; direction of electron travel; Indicate which electrode is being dissolved into solution and which electrode is being “plated” (show what the ions are doing at each electrode). a. Would CuCl2 make a good salt for the salt bridge in this cell? Explain. No. The Ag+ ion will react with Cl- from the salt bridge forming a precipitate. Also, Cu2+ is part of the chemical reaction, this might also complicate matters (Usually we use NaNO3, KNO3, etc….). b. Write the balanced chemical reaction for the cell. c. What is the E°cell for the cell above? Cu2++ 2e- → Cu E°red = 0.34V Ag++ e- → Ag E°red = 0.80V Since copper is oxidized, E°ox = -0.34V. Thus E°cell 0.80V+(-0.34V) = 0.46V