Chem 116 POGIL Worksheet - Week 12 - Solutions Second & Third Laws of Thermodynamics Balancing Redox Equations

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1 Chem 116 POGIL Worksheet - Week 12 - Solutions Second & Third Laws of Thermodynamics Balancing Redox Equations Key Questions 1. Does the entropy of the system increase or decrease for the following changes? Indicate whether ΔS system > 0 or ΔS system < 0. a. Water is boiled Producing a gas creates more disorder, so entropy increases. ΔS system > 0 b. CaCO 3 (s) CaO(s) + CO 2 (g) Two moles of product are formed for every one mole of reactant, and a gas is produced. Both factors indicate an increase in disorder, so entropy increases. ΔS system > 0 c. N 2 (g) + 3 H 2 (g) 2 NH 3 (g) Fewer gas molecules are produced, so entropy decreases. ΔS system < 0 2. Fill in the missing values for the following reactions occurring at 25 o C, and determine if the reaction is spontaneous or non-spontaneous. Reaction ΔH (kj/mol) ΔS (J/mol@K) ΔG (kj/mol) Spontaneous? H 2 (g) + Br 2 (g) 6 2 HBr(g) Yes 2 H 2 (g) + O 2 (g) 6 2 H 2 O(l) Yes 2 N 2 (g) + O 2 (g) 6 2 N 2 O(g) No N 2 O 4 (g) 6 2 NO 2 (g) No 3. Given the following ΔG o f values, calculate the standard free energy for the combustion of one mole of C 2 H 6 (g), and determine if the reaction is spontaneous or non-spontaneous: C 2 H 6 (g) 32.9 kj H 2 O(l) kj CO 2 (g) kj The balanced equation for the combustion is Therefore, we calculate ΔG o as follows: C 2 H 6 (g) + 7/2 O 2 (g) 6 2 CO 2 (g) + 3 H 2 O(l) ΔG o = GnΔG o f(products) GmΔG o f(reactants) ΔG o = (3)( kj) + (2)( kj) ( 32.9 kj) = Y spontaneous

2 Note that we ignored O 2 (g) in this calculation, because it is an element in its standard state, for which ΔG o f = Calculate ΔH o, ΔS o, and ΔG o for the following reaction at 25 o C. 2 H 2 O 2 (l) 6 2 H 2 O(l) + O 2 (g) Given the following data: Substance ΔH o f (kj/mol) S o (J/molAK) H 2 O(l) O 2 (g) H 2 O 2 (l) ΔH o = (2)( kj) (2)( kj) = kj = kj ΔS o = (2)(69.91 J/K) J/K (2)(109.6 J/K) = J/K = J/K We can calculate ΔG o from our values of ΔH o and ΔS o, using T = 298 K. ΔG o = ΔH o TΔS o = kj (298 K)(125.6 J/K) = kj J = kj = kj 5. Consider the reaction CaCO 3 (s) 6 CaO(s) + CO 2 (g) for which ΔH o = kj/mol, ΔG o = kj/mol, and ΔS o = J/KAmol. a. Is this reaction spontaneous at 25 o C? At 25 o C, ΔG o is positive, so the reaction is not spontaneous. The reverse reaction would be spontaneous with ΔG o = kj/mol.

3 b. Assuming that ΔH and ΔS do not change significantly with changing temperature, is this reaction spontaneous at 1200 K? At 1200K, ΔG. ΔH o TΔS o = kj/mol (1200 K)( kj/kamol) = kj/mol kj/mol = 14.5 kj Now ΔG is negative and the reaction becomes spontaneous. This is typical of most decomposition reactions, which tend to be non-spontaneous at low temperatures, but become spontaneous at higher temperatures. Note that in carrying out this calculation, we converted the given value of ΔS o from units of J/K@mol to kj/k@mol by moving the decimal three places to the left, so that the units of TΔS would be the same as ΔH o. 6. For the vaporization of cyclohexane, C 6 H 12 (l) º C 6 H 12 (g) ΔH o = kj/mol and ΔS o = J/K@mol. Assuming that these values do not change significantly with increasing temperature, estimate the boiling point temperature of cyclohexane. [Hint: Recall that boiling means that the liquid and vapor are in equilibrium, and therefore ΔG = 0.] At the normal boiling point, both the liquid and vapor phases are in equilibrium, and ΔG = 0. Therefore, we can solve ΔG. ΔH o TΔS o = 0 for T. ΔG = 0 = ΔH o TΔS o Y ΔH o = TΔS o Y T = ΔH o /ΔS o The handbook lists the boiling point of cyclohexane as EC. Our calculated value is in error by a small amount, because ΔH o and ΔS o actually do change slightly with temperature. 7. Under standard conditions, ΔG o = kj/mol for the reaction N 2 (g) + 3 H 2 (g) º 2 NH 3 (g) What is the value of ΔG for the reaction at 298 K when the partial pressures of a mixture are atm for N 2 (g), atm for H 2 (g), and 5.00 atm for NH 3 (g)? Is the reaction spontaneous or non-spontaneous under these conditions? First calculate the Q value.

4 Now solve ΔG = ΔG o + RT lnq for the non-standard value of ΔG. ΔG = kj/mol + ( kj/k@mol)(298 K) ln( ) = kj/mol kj/mol = kj/mol = kj/mol Under standard conditions, with ΔG o = kj/mol, the reaction would proceed spontaneously in the forward direction. However, under the stated conditions, ΔG is positive, which is non-spontaneous in the forward direction. This suggests that the reaction must proceed to the left to form more reactants to reach equilibrium. 8. Under standard conditions, ΔG o = kj for the reaction N 2 (g) + 3 H 2 (g) º 2 NH 3 (g). What is the value of K, the thermodynamic equilibrium constant, at 25 EC? K = The calculated K corresponds to K p, because the standard state for all species is gas. 9. For each of the following, separate the skeletal (unbalanced) equation into two half reactions. For each half reaction, balance the elements, and then add electrons to the right or left side to make a net charge balance. Identify which half reaction is the oxidation and which is the. Then, multiply each half reaction by an appropriate factor so that the two multiplied half reactions add together to make a balanced redox equation. a. Hg S 2 O 3 Hg + S 4 O 6 2+ Hg 2 + 2e 2 Hg 2 S 2 O 3 S 4 O 6 + 2e oxidation Hg S 2 O 3 2 Hg + S 4 O 6 redox

5 b. Al + Cr 3+ Al 3+ + Cr 2+ Al Al e 3(Cr 3+ + e Cr 2+ ) Al + 3 Cr 3+ Al Cr 2+ oxidation redox c. Au 3+ + I Au + I 2 2(Au e Au) 3(2 I I 2 + 2e ) oxidation 2 Au I 2 Au + 3 I 2 redox 10. Use the ion-electron method to complete and balance the following skeletal redox equations, occurring in either acidic or basic aqueous solution, as indicated. Identify the oxidation and half reactions in each case. a. In acid, Cu + NO 3 Cu 2+ + N 2 O 4 Cu Cu e 2e + 4 H NO 3 N 2 O H 2 O 4 H NO 3 + Cu N 2 O 4 + Cu H 2 O oxidation b. In acid, XeO 3 + BrO 3 Xe + BrO 4 6e + 6 H + + XeO 3 Xe + 3 H 2 O 3(H 2 O + BrO 3 BrO 4 + 2H + + 2e ) oxidation XeO BrO 3 Xe + 3 BrO 4 c. In acid, MnO 4 + CH 3 OH Mn 2+ + HCO 2 H 5(H 2 O + CH 3 OH HCO 2 H + 4 H + + 4e ) oxidation 4(5e + 8 H + + MnO 4 Mn H 2 O) 12 H CH 3 OH + 4 MnO 4 5 HCO 2 H + 4 Mn H 2 O d. In acid, Cr 2 O 7 + I 2 Cr 3+ + IO 3 5(6e + 14 H + + Cr 2 O 7 2 Cr H 2 O) 3(I H 2 O 2 IO H e ) oxidation 5 Cr 2 O I H + 10 Cr IO H 2 O

6 e. In base, Pb(OH) 4 + ClO PbO 2 + Cl Pb(OH) 4 PbO H 2 O + 2e oxidation 2e + H 2 O + ClO Cl + 2 OH Pb(OH) 4 + ClO PbO 2 + Cl + H 2 O + 2 OH f. In base, SO 2 + MnO 4 SO 4 + MnO 2 3(4 OH + SO 2 SO H 2 O + 2e ) oxidation 2(3e + 2 H 2 O + MnO 4 MnO OH ) 4 OH + 3 SO MnO 4 3 SO MnO H 2 O