Exam 4B, Spring 2001
Exam 4B, Spring 20011. Complete and balance the following equations.
a. ammonia plus nitric acid
b. Cu2+(aq) + NH3(aq) {NH3 is in excess}
c. As2O3(s) + NO3–(aq) → H3AsO4(aq) + NO(g)
d. Bi(OH)3(s) + Sn(OH)3–(aq) → Bi(s) + Sn(OH)62–(aq)
a. NH3(aq) + HNO3(aq) → NH4+(aq) + NO3–(aq)
b. Cu2+(aq) + 4 NH3(aq) → ← [Cu(NH3)4]2+(aq)
c. As2O3(s) + NO3–(aq) → H3AsO4(aq) + NO(g)
H3AsO4 indicates that the reaction is run in acid.
Oxidation: As2O3(s) + 5 H2O(l) → 2 H3AsO4(aq) + 4 H+(aq) + 4 e–
Reduction: NO3–(aq) + 4 H+(aq) + 3 e– → NO(g) + 2 H2O(l)
Net: 3 As2O3(s) + 4 NO3–(aq) + 7 H2O(l) + 4 H+(aq) → 6 H3AsO4(aq) + 4 NO(g)
d. Bi(OH)3(s) + Sn(OH)3–(aq) → Bi(s) + Sn(OH)62–(aq)
The three hydroxide species indicate that the reaction is run in base.
Oxidation: Sn(OH)3–(aq) + 3 H2O(l) → Sn(OH)62–(aq) + 3 H+(aq) + 2 e–
Reduction: Bi(OH)3(s) + 3 H+(aq) + 3 e– → Bi(s) + 3 H2O(l)
Net in acid: 2 Bi(OH)3(s) + 3 Sn(OH)3–(aq) + 3 H2O(l) → 2 Bi(s) + 3 Sn(OH)62–(aq) + 3 H+(aq)
Neutralization: 3 H+(aq) + 3 OH–(aq) → 3 H2O(l)
Net in base: 2 Bi(OH)3(s) + 3 Sn(OH)3–(aq) + 3 OH–(aq) → 2 Bi(s) + 3 Sn(OH)62–(aq)
2. Consider the reaction between solid calcium chloride and aqueous sodium sulfate.
a. Write the balanced reaction.
b. Find ΔH° for the reaction in units of kJ.
c. Find ΔS° for the reaction in units of J/K.
d. Find ΔG° for the reaction at 25 °C in units of kJ.
e. Find ΔG° for the reaction at 75 °C in units of kJ.
a. CaCl2(s) + Na2SO4(aq) → CaSO4(s) + 2 Na+(aq) + 2 Cl–(aq)
b. ΔH° = [(–1434) + 2(–240.1) + 2(–167.2)] – [(–795.8) + (–1390)] = –63. kJ/mol
c. ΔS° = [(106.7) + 2(59.0) + 2(56.5)] – [(105) + (138.1)] = 95. J/mol·K
d. 25 °C = 298 K, so ΔG° = –63. – (298)(0.095) = –91. kJ/mol.
d. 75 °C = 348 K, so ΔG° = –63. – (348)(0.095) = –96 kJ/mol.
3. Barium carbonate has a molar solubility of 5.7×10–5 M at 25 °C and 6.1×10–5 M at 85 °C. Find Ksp at each temperature and ΔH° and ΔS° for the solubilization reaction.
BaCO3(s) → ← Ba2+(aq) + CO32–(aq)
Ksp = x2, where x is the molar solubility of barium carbonate.
At 25 °C Ksp = (5.7×10–5)2 = 3.2×10–9
At 85 °C Ksp = (6.1×10–5)2 = 3.7×10–9
Use the van t'Hoff equation to find ΔH° and ΔS°.
At 25 °C (= 298 K): ln(3.2×10–9) = –ΔH°/(8.314×298) + ΔS°/(8.314)
At 85 °C (= 358 K): ln(3.7×10–9) = –ΔH°/(8.314×358) + ΔS°/(8.314)
Solve two equations in two unknowns:
–19.560 = –4.036×10–4ΔH° + 0.1203ΔS°
–19.415 = –3.356×10–4ΔH° + 0.1203ΔS°
Subtracting the two equations gives:
–0.145 = –0.680×10–4ΔH° so that ΔH° = 2130 J/mol = 2.13 kJ/mol
Then, ΔS° = (–19.560 + 4.036×10–4ΔH°)/0.1203 = (–19.560 + 4.036×10–4×(2130))/0.1203 = –155 J/mol·K
4. Consider the cell: Na(s) | Na+(aq) || Co2+(aq) | Co(s)
a. What is the balanced oxidation half-reaction?
b. What is the balanced reduction half-reaction?
c. What is the balanced net reaction?
d. What is the standard potential for the cell?
e. Is the reaction spontaneous or nonspontaneous? Why?
a. Na(s) → Na+(aq) + e–
b. Co2+(aq) + 2 e– → Co(s)
c. 2 Na(s) + Co2+(aq) → 2 Na+(aq) + Co(s)
d. E° = E°ox + E°red = 2.713 + –0.277 = +2.436 V
e. Spontaneous because E° > 0.