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Homework 9 Key - CHEMISTRY 201X: Homework #9 Equilibrium...

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Unformatted text preview: CHEMISTRY 201X: Homework #9 Equilibrium Applications Due: Monday, December 3, 2007 (1) (6 pts) Snow and ice sublime spontaneously when the partial pressure of water vapor is below the equilibrium vapor pressure of ice. At 0C the vapor pressure of ice is 0.0060 atm (the triple point of water). Taking the enthalpy of sublimation of ice to be 50.0 kJ/mol, calculate the partial pressure of water vapor below which ice will sublime spontaneously at -15C. Use the Clausius-Clapeyron equation for equilibrium between H2O(s) and H2O(g): ln p1 H sub 1 1 p1 50, 000 1 1 = = - ; ln p2 R T2 T1 0.006 8.314 273.15 258.15 p1 = 0.0017 atm (2) (14 pts) At 1200 K in the presence of solid carbon, an equilibrium mixture of CO(g) and CO2(g) (called "producer gas") contains 98.3 mole percent CO and 1.69 mole percent CO2(g) when the total pressure is 1 atm. The reaction is CO2(g) + C(graphite) 2 CO(g) (a) (2 pts) Calculate p(CO) and p(CO2) (in atm). p(CO) = x(CO)(1 atm) = 0.983 atm p(CO2) = x(CO2)(1 atm) = 0.0169 atm (b) (2 pts) Calculate the equilibrium constant Kp at 1200 K from these data. p 2 (CO) (0.983)2 Kp = = = 57.2 p(CO 2 ) (0.0169) (c) (2 pts) Calculate G (in kJ) from these data. G = -(8.314 J/K)(1200 K)ln(57.2) = -40370 J = -40.4 kJ (d) (4 pts) Use thermodynamic data to calculate G (in kJ) and Kp at 1200 K. Compare with your results in parts (b) and (c). H = 2(-110.5 kJ) - [(-393.5 kJ) + (0)] = +172.5 kJ S = 2(197.9 J/K) - [(213.6 J/K) + (5.69 J/K)] = +176.5 J/K G = (+172.5 kJ) - (1200 K)(+0.1765 kJ/K) = -39.3 kJ Kp = exp[-(-39300 J)/(8.314 J/K)(1200 K)] = exp(3.939) = 51.4 This estimate relies on assuming that H and S remain constant with temperature, which is strictly incorrect. The equilibrium constant at 1200 K is CHEMISTRY 201X: Homework #9 Due: Monday, December 3, 2007 based on measuring amounts of gases, so the results in parts (b) and (c) should be more accurate. (e) (4 pts) Whenever CO(g) and CO2(g) exist in equilibrium, there must also exist O2(g). Estimate the equilibrium partial pressure of O2(g) in this mixture. The following equilibrium occurs simultaneously: 2 CO2(g) 2 CO(g) + O2(g) H = 2(-110.5 kJ) - [2(-393.5 kJ)] = +566 kJ S = [2(197.9 J/K) + (205.0 J/K)] - [2(213.6 J/K)] = +173.6 J/K G = (+566 kJ) - (1200 K)(+0.1736 kJ/K) = +357.7 kJ Kp = exp[-(357700 J)/(8.314 J/K)(1200 K)] = exp(-35.85) = 2.69 10-16 Kp = p 2 (CO)p(O 2 ) (0.983)2 p(O 2 ) = = 2.69 10-16 2 2 p (CO 2 ) (0.0169) p(O2) = 7.96 10-20 atm (3) (12 pts) Use these equilibria to estimate the solubilities (in g/mL) of the salts LiCl(s), NaCl(s), and NH4Cl(s) in water for temperatures 0C, 25C, 50C, 75C, and 100C. Plot these solubilities as a function of temperature. Discuss any trends you observe... LiCl(s) NaCl(s) NH4Cl(s) H2O Li+(aq) + Cl-(aq) Na+(aq) + Cl-(aq) NH4+(aq) + Cl-(aq) H2O H2O The general strategy is as follows: Kc = [A+][Cl-] = [A+]2 because [A+] = [Cl-]. When these equilibria are established, the solution is saturated, and this represents the solubility of the salt. We calculate Kc from free energies: G(T) = -RT ln Kc = H - TS. Also, since the units of [A+] = mole/L, to convert to g/mL, we must take [A+](FWACl / 1000). LiCl T (K) 273 298 323 348 373 H = -37.4 kJ G (J) -39967.61 -40202.61 -40437.61 -40672.61 -40907.61 Kc 43985174.72 11055654.44 3440770.45 1266281.18 532813.21 S = +9.4 J/K [A+] (M) 6632.13 3325.00 1854.93 1125.29 729.94 FW = 42.394 g/mol Solubility (g/mL) 281.16 140.96 78.64 47.71 30.95 CHEMISTRY 201X: Homework #9 Due: Monday, December 3, 2007 300 Solubility of LiCl (g/mL) 250 200 150 100 50 0 260 280 300 320 340 360 380 Temperature (K) NaCl 273 298 323 348 373 -8200.08 -9280.08 -10360.08 -11440.08 -12520.08 H = +3.6 kJ 37.00 42.26 47.28 52.06 56.58 0.46 0.44 S = +43.2 J/K 6.08 6.50 6.88 7.22 7.52 FW = 58.443 g/mol 0.36 0.38 0.40 0.42 0.44 Solubility of NaCl (g/mL) 0.42 0.40 0.38 0.36 0.34 260 280 300 320 340 360 380 Temperature (K) NH4Cl 273 298 323 348 373 -5868.20 -7750.70 -9633.20 -11515.70 -13398.20 H = +14.7 kJ 13.25 22.80 36.07 53.43 75.09 0.50 0.45 0.40 0.35 0.30 0.25 0.20 0.15 260 280 300 S = +75.3 J/K 3.64 4.77 6.01 7.31 8.67 FW = 53.503 g/mol 0.19 0.26 0.32 0.39 0.46 Solubility of NH4Cl (g/mL) 320 340 360 380 Temperature (K) The solubilities of NaCl and NH4Cl increase with temperature; the solubility of LiCl decreases with temperature. The solubility equilibrium for LiCl is exothermic; that for NaCl and NH4Cl is endothermic, and these account for the trends in solubility. Also, LiCl is much more soluble in water than either NaCl and NH4Cl, which comes mainly from the enthalpy of the equilibrium reaction and must reflect how the ions interact with water molecules in the solution. The interaction is quite favorable for Li+(aq); much less favorable for Na+(aq) and NH4+(aq). CHEMISTRY 201X: Homework #9 Due: Monday, December 3, 2007 (4) (47 pts) The chemistry of nitrogen is marked by nitrogen's ability to exist in oxidation states ranging from +5 (nitric acid) to -3 (ammonia). Thermodynamic information regarding the oxidation or reduction of nitrogen-containing species in aqueous solutions is nicely summarized in a reduction-potential diagram. Here are the diagrams for nitrogen species in acidic and basic media: 0.96 V Acidic: NO3 0.79 V NO2 1.09 V HNO2 1.00 V 1.25 V -0.15 V Basic: NO3 0.01 V NO2 -0.46 V 0.08 V NO 0.76 V N2O 0.94 V N -0.73 V NH 2 3 NO 1.59 V N2 O 1.77 V N 0.27 V 2 NH4+ In these diagrams only the nitrogen-containing species are indicated. An arrow represents the properly balanced half-reaction, which is a reduction, and the standard reduction potential associated with that half-reaction is written above the arrow. Also, acidic media means pH = 0 ([H+] = 1.00 M) and basic media means pH = 14 ([H+] = 1.00 10-14 M; [OH-] = 1.00 M). (a) (8 pts) Write down the 8 half-reactions in acidic media whose reduction potentials are given above. NO3-(aq) + e- + 2 H+(aq) NO2(g) + H2O(l) NO2(g) + e- + H+(aq) HNO2(aq) HNO2(aq) + e- + H+(aq) NO(g) + H2O(l) NO3-(aq) + 3 e- + 4 H+(aq) NO(g) + 2 H2O(l) 2 NO(g) + 2 e- + 2 H+(aq) N2O(g) + H2O(l) N2O(g) + 2 e- + 2 H+(aq) N2 (g) + H2O(l) N2 (g) + 6 e- + 8 H+(aq) 2 NH4+(aq) 2 NO3-(aq) + 10 e- + 12 H+(aq) N2 (g) + 6 H2O(l) Ered = +0.79 V Ered = +1.09 V Ered = +1.00 V Ered = +0.96 V Ered = +1.59 V Ered = +1.77 V Ered = +0.27 V Ered = +1.25 V (b) (7 pts) Write down the 7 half-reactions in basic media whose reduction potentials are given above. NO3-(aq) + 2 e- + H2O(l) NO2-(aq) + 2 OH-(aq) NO2-(aq) + e- + H2O(l) NO(g) + 2 OH-(aq) Ered = +0.01 V Ered = -0.46 V CHEMISTRY 201X: Homework #9 NO3-(aq) + 3 e- + 2 H2O(l) NO(g) + 4 OH-(aq) 2 NO(g) + 2 e- + H2O(l) N2O(g) + 2 OH-(aq) Due: Monday, December 3, 2007 Ered = -0.15 V Ered = +0.76 V 2 NO3-(aq) + 8 e- + 5 H2O(l) N2O(g) + 10 OH-(aq) Ered = +0.08 V N2O(g) + 2 e- + H2O(l) N2 (g) + 2 OH-(aq) N2 (g) + 6 e- + 6 H2O(l) 2 NH3(aq) + 6 OH-(aq) Ered = +0.94 V Ered = -0.73 V (c) (2 pts) Explain why the N(V) species is written as NO3- and not HNO3. Draw Lewis structures to help. HNO3(aq) is a strong acid, i.e., a strong electrolyte in water. Nitric acid completely dissociates in water; the nitrate ion is stabilized by 3 resonance structures. H O N O O O O N O H In the Lewis structure for "molecular" nitric acid, there would be 2 resonance structures. The formal charge at N is +1 in both structures; the formal charge for the O atom involved in the N-O bond is -1; the formal charges at all other atoms are 0. O N O O O O N O O O N O The Lewis structures of the nitrate ion involves 3 resonance structures. Here, the formal charge at N is +1. The additional resonance structure provides additional stability to the nitrate ion over the "molecular" nitric acid in an environment of water, which is a sufficiently strong base to "deprotonate" nitric acid. (d) (2 pts) Explain why the N(III) species is written as HNO2 in acidic media, but as NO2- in basic media. Draw Lewis structures to help. HNO2(aq) is a weak acid, Ka = 4.5 10-4. The following equilibrium holds: HNO2(aq) H+(aq) + NO2-(aq); Ka = [H + ][NO-2 ] . [HNO 2 ] In acidic media (pH = 0; [H+] = 1.0 M), [NO2-] << [HNO2]; In basic media (pH = 14; [H+] = 10-14 M), [NO2-] >> [HNO2]. CHEMISTRY 201X: Homework #9 Due: Monday, December 3, 2007 H O N O O N O O N O In the Lewis structure of HNO2, there is no resonance stabilization. The formal charge at the N atom is 0. The nitrite ion is stabilized by 2 resonance structures; the formal charge at the N atom is also 0. HNO2 is a weaker acid than HNO3 because there exists more covalent bonding throughout the molecule than in HNO3. When the formal charge on N is positive, the electron density in the H-O bond is polarized with more electron density attracted toward the O atom. (e) (2 pts) Explain why the N(-3) species is written as NH4+ in acidic media, but as NH3 in basic media. Draw Lewis structures to help. NH3(aq) is a weak base, Kb = 1.8 10-5. The following equilibrium holds: NH3(aq) + H2O(l) NH4+(aq) + OH-(aq); Kb = + [NH 4 ][OH - ] . [NH 3 ] In acidic media (pOH = 14; [OH-] = 10-14 M), [NH3] << [NH4+]; In basic media (pOH = 0; [OH-] = 1.0 M), [NH3] >> [NH4+]. H H N H H H N H H In acidic media, the lone pair at the N atom in ammonia would be a good Lewis base toward the proton in water, and it forms the ammonium ion. In basic media, the hydroxide is a stronger base than ammonia, so the ammonia molecule persists. (f) (2 pts) What is the standard reduction potential in acidic media for the reduction of nitric acid to nitrous acid? NO3-(aq) + e- + 2 H+(aq) NO2(g) + H2O(l) Ered = +0.79 V NO2(g) + e- + H+(aq) HNO2(aq) Ered = +1.09 V -------------------------------------------------------------------------------------------NO3-(aq) + 2e- + 3 H+(aq) HNO2(aq) + H2O(l) Ered = (+0.79 + 1.09)V / 2 Ered = +0.94 V We can also use Gred = -nF Ered. CHEMISTRY 201X: Homework #9 NO3-(aq) + e- + 2 H+(aq) NO2(g) + H2O(l) Due: Monday, December 3, 2007 Gred = -76.24 kJ NO2(g) + e- + H+(aq) HNO2(aq) Gred = -105.19 kJ -------------------------------------------------------------------------------------------Gred = -181.43 kJ NO3-(aq) + 2e- + 3 H+(aq) HNO2(aq) + H2O(l) Gred = -181.43 kJ = -(2 mol)(96,500 coul/mol)Ered Ered = +0.94 V (g) (4 pts) Is nitrous acid in acidic media stable with respect to disproportionation to nitric acid and nitric oxide at standard conditions? Explain by writing a balanced reaction and calculating G. HNO2(aq) NO3-(aq) + NO(g) HNO2(aq) + e- + H+(aq) NO(g) + H2O(l) Ered = +1.00 V HNO2(aq) + H2O(l) NO3-(aq) + 2e- + 3 H+(aq) Ered = +0.94 V -------------------------------------------------------------------------------------------3 HNO2(aq) HNO3(aq) + 2 NO(g) + H2O(l) E = (1.00 V) - (0.94 V) E = +0.06 V > 0 No, HNO2(aq) is unstable with respect to disproportionation; G = -nFE = -(2 mole)(96,500 coul/mol)(0.06 V) = -11.6 kJ (h) (4 pts) Is the nitrite ion in basic media stable with respect to disproportionation to the nitrate ion and nitric oxide at standard conditions? Explain by writing a balanced reaction and calculating G. NO2-(aq) NO3-(aq) + NO(g) NO2-(aq) + e- + H2O(l) NO(g) + 2 OH-(aq) Ered = -0.46 V Ered = +0.01 V NO2-(aq) + 2 OH-(aq) NO3-(aq) + 2 e- + H2O(l) ---------------------------------------------------------------------------------------------------3 NO2-(aq) + H2O(l) NO3-(aq) + 2 NO(g) + 2 OH-(aq) E= (-0.46 V) - (+0.01 V) E = -0.47 V < 0 No, NO2-(aq) is stable with respect to disproportionation; G = -nFE = -(2 mole)(96,500 coul/mol)( -0.47 V) = +90.7 kJ (i) (4 pts) Active metals like zinc will react with concentrated nitric acid to produce the ammonium ion. Write the balanced chemical reaction for this process and calculate E and G for the reaction. CHEMISTRY 201X: Homework #9 Zn(s) + NO3-(aq) Zn2+(aq) + NH4+(aq) Due: Monday, December 3, 2007 NO3-(aq) + 8 e- + 10 H+(aq) NH4+(aq) + 3 H2O(l) Ered = ??? We can calculate the standard reduction potential by combining two half-reactions: 2 NO3-(aq) + 10 e- + 12 H+(aq) N2 (g) + 6 H2O(l) Gred = -1206.25 kJ N2 (g) + 6 e- + 8 H+(aq) 2 NH4+(aq) Gred = -156.33 kJ -------------------------------------------------------------------------------------------------2 NO3-(aq) + 16 e- + 20 H+(aq) 2 NH4+(aq) + 6 H2O(l) Gred = -1362.58 kJ Gred = -1362.58 kJ = -(16 mol)(96,500 coul/mol)Ered Ered = +0.88 V Therefore, we have... NO3-(aq) + 8 e- + 10 H+(aq) NH4+(aq) + 3 H2O(l) Ered = +0.88 V Zn(s) Zn2+(aq) + 2 e- Ered = -0.76 V ----------------------------------------------------------------------------------------------------4 Zn(s) + NO3-(aq) + 10 H+(aq) 4 Zn2+(aq) + NH4+(aq) + 3 H2O(l) E = (+0.88 V) - (-0.76 V) = +1.64 V G = -(8 mol)(96,500 coul/mol)(+1.64 V) = -1266 kJ (j) (6 pts) In dilute nitric acid, zinc will react with the hydrogen ions to produce hydrogen gas. Write the balanced chemical reaction and plot E as a function of pH for pH values between 0 (1.00 M H+) and 14 (1.00 10-14 M H+). Set the pressure of H2 gas at 1.00 atm. For which pH values is this reaction predicted to be spontaneous? Zn(s) + 2 H+(aq) Zn2+(aq) + H2(g); E = (0.00 V) - (-0.76 V) = +0.76 V [Zn 2+ ] 0.0592 2+ E = (+0.76 V) - log + 2 = (+0.76 V) - (0.0296 log [Zn ]) - 0.0592 pH 2 [H ] the nitrate ion is a spectator ion here. Set [Zn2+] = 1.00 M, then E = (+0.76 V) - 0.0592 pH This reaction is spontaneous when E > 0, or for pH < 12.84. CHEMISTRY 201X: Homework #9 Due: Monday, December 3, 2007 0.8 E (V) 0.6 0.4 0.2 Spontaneous 0.0 Not spontaneous -0.2 0 2 4 6 8 10 12 14 pH (k) (4 pts) Metals like copper will react with concentrated nitric acid to produce nitric oxide. Write the balanced chemical reaction for this process and calculate E and G for the reaction. Cu(s) + NO3-(aq) Cu2+(aq) + NO(g) NO3-(aq) + 3 e- + 4 H+(aq) NO(g) + 2 H2O(l) Ered = +0.96 V Cu(s) Cu2+(aq) + 2 e- Ered = +0.34 V ----------------------------------------------------------------------------------------------------3 Cu(s) + 2 NO3-(aq) + 8 H+(aq) 3 Cu2+(aq) + 2 NO(g) + 4 H2O(l) E = (+0.88 V) - (+0.34 V) = +0.54 V G = -(6 mol)(96,500 coul/mol)(+0.54 V) = -312.7 kJ (l) (2 pts) Is concentrated or dilute nitric acid a better oxidizing agent? Explain. Use the following half-reaction, for example (any reduction of nitrate in acidic media will do...) NO3-(aq) + e- + 2 H+(aq) NO2(g) + H2O(l) Ered = +0.79 V p(NO 2 ) 0.0592 Ered = Ered - = (+0.79 V) + 0.0592 log [NO3-] - 0.1184pH log 1 [NO-3 ][H + ]2 In concentrated nitric acid, [NO3-] 1.00 M; pH 0, so Ered +0.79 V. In dilute nitric acid, [NO3-] < 1.00 M; pH > 0, so Ered < +0.79 V, and, therefore, nitric acid becomes a worse oxidizing agent. Concentrated nitric acid is a better oxidizing agent... ...
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This homework help was uploaded on 03/27/2008 for the course CHEM 201 taught by Professor Miller during the Fall '07 term at Iowa State.

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