CH 233H First Midterm Exam Friday, April 28, 217 Name KEY Please show your work for partial credit. If you need more spacefor an answer, use the back of the page and indicate where we should look. You may not use notes or other materials with chemical information without the instructor s approval; necessary information is provided on pages at the back of the exam. Please do not use ipods or other music players.
1. (15 points) Provide a concise definition in terms of energy for each of the three laws of thermodynamics. First Law: Energy is neither created nor destroyed. Second Law: In any spontaneous process, the entropy of the universe must increase. Third Law: The entropy of a perfect crystal at K is. 2. (2 points) Consider the following reaction at 298 K, where isopropanol decomposes to acetone and hydrogen in the gas phase: A. What is ΔHr? ΔHf : -272.8 kj/mol -218.5 kj/mol so ΔHr = (-218.5 + ) - (-272.8 kj/mol) = +54.3 kj/mol B. What is ΔSr? S : 328 J/(mol-K) 34 J/(mol-K) so ΔSr = (34 + 13.7 kj/mol) - (328 kj/mol) = 16.7 J/(mol-K) 13.7 J/(mol-K)
2. (continued) C. At what temperature does a total pressure of 1. atm show an equilibrium partial pressure of.172 atm isopropanol? If the total pressure is 1 atm, set X = Pacetone = Phydrogen, and 2X +.172 = 1. X =.414 Keq = PacetonePH2 Pisopropanol = (.414)(.414) =.9964 (round to 1.).172 ΔGr = -RTlnKeq = -[8.314 J/(mol-K)](T)ln(1.) = J/mol since ln 1 = (ln.9964 = -.36, actually) ΔGr = ΔHr - TΔSr = 54,3 J/mol T(16.7 J/mol-K) Solve for T: T = 543 J/mol / 16.7 J/(mol-K) = 59 K ( = 236 C).
3. (24 points) One of the experimental next-generation batteries mentioned in the text is the aluminum-air battery, that operates with the net reaction: 4 Al (s) + 3 O2 (g) + 6 H2O (l) + 4 OH- (aq) 4 [Al(OH)4-] (aq) A. Write the two half-cell reactions responsible for generating electrons in this battery. Al (s) + 4 OH- (aq) Al(OH)4- (aq) + 3 e- E = +2.31 V O2 (g) + 2H2O (l) + 4 e- 4OH- (aq) E = +.41 V Net: 4 Al (s) + 4 OH- (aq) + 3O2 (g) + 6 H2O (l) 4[Al(OH)4-] (aq) (l) E = + 2.711 V B. Specify which half reaction will represent the anode, and which the cathode. Explain your reasoning. Anode: Al (s) going to Al(OH)4- (aq); this is the source of electrons in the circuit where oxidation occurs.. By process of elimination, O2 (g) going to water is there the reduction occurs (consumption of electrons) and therefore that is the cathode. C. What is the maximum voltage this cell can produce? The full cell potential is +3.539 V.
4. (21 points) One of the demonstrations we saw showed that burning magnesium metal was not extinguished by dry ice (solid CO2) but instead continued to react quite violently. A. Write a balanced equation for reaction of Mg with atmospheric O2, and calculate ΔG for the reaction at the combustion temperature of 3 C. Mg (s) + ½O2 (g) MgO (s) ΔHf : S : 32.68 25.1-61.7 kj/mol 57.24 J/(mol-K) ΔHr = -61.7 kj/mol ΔSr = -78. J/mol-K ΔGr = ΔHr =-TΔSr = -617 J/mol 3273K(-78. J/mol-K) = -34646 J/mol = -346.4 kj/mol B. Write a balanced equation for reaction of Mg with gaseous CO 2, and calculate ΔG for the reaction at the same temperature. Mg (s) + ½CO2 (g) MgO (s) + ½C (s) ΔHf : S : 32.68-393.5 213.7-61.7 57.24 5.74 J/(mol-K) ΔHr = -44.9 kj/mol ΔSr = -79.42 J/mol-K ΔGr = ΔHr =-TΔSr = -449 J/mol 3273K(-79.42 J/mol-K) = -1449 J/mol = -144.9 kj/mol C. Would CO2 be an effective extinguisher for a sodium metal fire? Explain. Assuming the same combustion temperature: 2Na (s) + ½O2 (g) Na2O (s) ΔHf : S : 51.21 25.1-418 kj/mol 75.4 J/(mol-K) ΔHr = -418 kj/mol ΔSr = -129.9 J/mol-K ΔGr = ΔHr =-TΔSr = -418 J/mol 3273K(-129.9 J/mol-K) = +7162 J/mol = +7.162 kj/mol BUT: we recognize that a lower temperature would make this entropically more disfavored reaction possible. 2Na (s) + ½CO2 (g) Na2O (s) + ½C (s) ΔHf : S : 51.21-393.5 213.7-418 75.4 5.74 J/(mol-K) ΔHr = -221.3 kj/mol ΔSr = -8.15 J/mol-K ΔGr = ΔHr =-TΔSr = -2213 J/mol 3273K(-8.15 J/mol-K) = +412 J/mol = +41.2 kj/mol We d have to decide on a temperature to solve exactly, but we can say that compared to the Mg combustion reaction in CO2, Na + CO2 is less exothermic (ΔHr ) but less entropically disfavored and thus less likely to be favorable. We might be able to use CO 2 to distinguish a sodium fire but it would be risky, depending on the temperature. In general CO 2 is never used to extinguish metal fires.
5. (2 points) Electroplating is a common technique used to apply a thin layer of metal, usually an expensive one, on an object made from a cheaper metal. One example used today is nickel plating, often used in home plumbing fixtures because of the aesthetic qualities of nickel. A. Show the half cell reaction responsible for the deposition of Ni (s) from an aqueous solution of NiSO4, NiCl2 and H3BO4. Ni+2 (aq) + 2 e- Ni (s) E = -.257 V B. What is the minimum voltage required to achieve deposition of Ni (s)? The -.257 V, but that would presume the oxidation of H2 at the cathode (see (C)). C. What is the most likely cathode half-cell reaction that occurs in nickel plating? Show the half cell reaction and its potential. Does this affect your answer for (B)? Explain. The candidates for oxidation are water (1.229 V), chloride (1.548 V), sulfate (2.1 V) and borate (not listed, but likely similar to sulfate or higher). Water is the easiest oxidation: 2H2O (l) O2 (g) + 4 H+(aq) + 4 e- E = -1.229 V This would add to the nickel reduction voltage unless a better reducing agent was added at the cathode.
Selected data that may be of use: Physical constants: g = 9.8 m/s2 ε = 8.85419 1-12 C2/(Nm2) c = 2.99792458 11 cm/s R =.826 L-atm/(mol-K) = 8.314 J/(mol-K) N = 6.22 123 k = 1.381 1-23 m2kg/(k-s2) h = 6.626 1-34 m2kg/s F = 96485 C/mol π = 3.14159 e = 2.71828 Gravitational Constant Electric susceptibility of a vacuum Speed of light Gas constant Avogadro s Number Boltzmann constant Planck s constant Faraday s constant Properties of State Species ΔH f S H2 (g) 13.7 J/(mol-K) N2 (g) 191.6 J/(mol-K) O2 (g) 25.1 J/(mol-K) C (s) (graphite) 5.74 J/(mol-K) (CH3)2CHOH (g) -272.8 kj/mol 328 J/(mol-K) (CH3)2CO (g) -218.5 kj/mol 34 J/(mol-K) CO2 (g) -393.5 kj/mol 213.7 J/(mol-K) H2O (l) -285.8 kj/mol 69.91 J/(mol-K) OH- (aq) -23. -1.75 J/(mol-K) Al (s) 28.33 J/(mol-K) Al(OH)4- (aq) -153 kj/mol -99.5 J/(mol-K) Na (s) 51.21 J/(mol-K) Na2O (s) -417.98kJ/mol 75.4 J/(mol-K) Li (s) 29.12 J/(mol-K) Li2O (s) -595.8 kj/mol 37.89 J/(mol-K) Mg (s) 32.68 J/(mol-K) MgO (s) -61.7 kj/mol 57.24 J/(mol-K)
Electromotive series: