Chm 152 Exam 4 Review – Ch. 18 19 Key

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  0.217 kJ/molK 5540 K At this temp and higher, the reaction is non-spontaneous. 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. Give the short-hand notation for the copper/silver cell. Cu(s) | Cu2+(aq) || Ag+(aq) | Ag(s) b. 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….). c. Write the balanced chemical reaction for the cell. Cu(s) + 2Ag+(aq) → Cu2+(aq) + 2Ag(s) d. 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 e. Based on your value and sign of E°cell is it spontaneous? Explain. Cu2++ 2e- → Cu E°red = 0.34V Ag++ e- → Ag E°red = 0.80V Yes. Since the E°cell is positive, it means the reaction will be spontaneous (∆G° < 0) f. Calculate the equilibrium constant (K) for the cell. E  cell  RT ln K nF J (289.15K )  8.314 mol K 0.46V  ln K J  (2mol e )96,500 V mol 35.82  ln K K = 3.6x1015 e 35.82  e (ln K ) g. Based on your value and sign of K, is the reaction spontaneous? Explain. Yes. K >> 1 is spontaneous and “product favored”. h. Which species is the oxidizing agent in the cell? i. Which species loses electrons? Cu(s) Cu2+(aq) Ag(s) Ag+(aq) Cu(s) Cu2+(aq) Ag(s) Ag+(aq) j. Based on your answers from c, g and h, determine if the following reaction is spontaneous. Explain. 2Ag(s) + Cu2+(aq) → Cu(s) + 2Ag+(aq) No, the reaction a written above is not spontaneous. Based on the spontaneous reaction written in part c, Cu(s) is more likely to give up electrons and Ag+(aq) is more likely to accept electrons. The reaction as written in part “j”, is in the reverse direction of the spontaneous reaction. Thus, it is non-spontaneous. k. Calculate ∆G° for the voltaic cell. G   nFE  cell   J (0.46V ) ∆G° = -89,000J 0 or -89kJ G   (2mol e - ) 96,500 V mol l. Calculate the Ecell if the [Cu2+] = 2.00M and [Ag+] = 0.0111M Ecell  E  cell  RT ln Q nF Q [Cu 2 ]  2 [ Ag ]  [2.00] [0.0111] 2 8.314 molJ K (289.15K ) E cell  0.46V  ln(16,232) J (2mol e )96,500 V mol   16000 Ecell = 0.45V 11. Based on reduction potentials, which of the following reactions should be spontaneous? Explain.   2Al(s) + 3Cl2(g) → 2Al3+(aq) + 6Cl-(aq) 2Al3+(aq) + 6Cl-(aq) → 2Al(s) + 3Cl2(g) Al3++ 3e- → Al E°red = -1.66V Cl2 + 2e- → 2ClE°red = 1.36V Since Aluminum has the largest negative reduction potential, we know that it will be the reducing agent (Al(s) is being oxidized). If we were calculating cell potential, we would “flip” this half-reaction to show oxidation potential. When we “flip” the Al half-reaction, Al(s) is on the reactants side. Thus, the reaction will Al(s) on the reactant side is the spontaneous reaction. Another way to think about this is if to compare Al(s) vs. Al3+(aq). The Al(s) form is more likely to give up electrons as the reducing agent (being oxidized), since you can think of Al3+ as already oxidized. (oxidation number for Al(s) = 0, Al3+ = +3) 12. Based on reduction potentials, which of the following reactants should give a spontaneous reaction? (HINT: Pay attend to states of matter…which should give up electrons which should accept electrons) Explain. a. b. c. d. F2 + 2e- → 2FE°red = 2.87V Br2 + 2e → 2Br E°red = 1.07V F2(g) + Br2(l) F2(g) + 2Br -(aq) 2F-(aq) + 2Br -(aq) 2F-(aq) + Br2(l) Based on reduction potentials, since Bromine has the smallest positive reduction potential, it will be the reducing agent (becoming oxidized) and the half-reaction that will be “flipped”. Once the bromine half-reaction is flipped, Br -(aq) is on the reactant side. The reaction with fluorine will not be flipped, and has F2(g) on the reactant side. Thus, those forms of Bromine and Fluorine will react spontaneously, F2(g) and Br -(aq). Another way to think about this is to compare the Br2(l) and Br-(aq) forms. Br- is more likely to give up electrons as the reducing agent (it becomes oxidized), since you can think of the Br2(l) form as already oxidized (oxidation number for Br - = -1, Br2 = 0). For Fluorine, the opposite is true: The F2(g) is more likely to accept electrons than the F-(aq), F-(aq) is already reduced.