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Unformatted text preview: CHEM 1B 4/12/2006 NAME__________________________ (LAST) (First) Show all calculations, and include units where applicable. (Total points = 100) 1. 2 points. Which of the following processes are spontaneous? (a) a house is built (b) a satellite is launched into orbit (c) a satellite falls back to earth (d) the kitchen gets cluttered Answer: c and d. Thermodynamics says nothing about how fast a spontaneous process will occur; that is the subject of Kinetics. It only says that the initial and final states differ in free energy, the change being negative. 2. 3 points. Will the following change be spontaneous at constant T and P ? Explain in three or four words. + = -10. kJ, 6 = +5.0 J/K, T = 298 K Answer: One measure of spontaneity is an overall decrease in free energy, *DFFRPSDQ\LQJWKH process. * +-767KHUHDUHWZRFRQWULEXWLRQVWRWKHIUHHHQHUJ\FKDQJH The +RIWKH system, (which, by the way, impacts the entropy of the surroundings)DQGWKH6RIWKHV\VWHP In this problem, both +DQG6DUHIDYRUDEOHVLQFHWKH\HDFKFRQWULEXWHWRDQLQFUHDVHLQWKHHQWURS\RI the universe. Alternately, they both contribute to a decrease in *PDNLQJWKHSURFHVVVSRQWDQHRus at all tempertures (T in Kelvin is always positive). 3. (a) 2 points. Which has the higher positional entropy, N2 gas at 1 atm or N2 gas at 1.0 10-2 atm? Answer: Because of the inverse relationship between volume and pressure at constant temperature for a given number of moles, a larger volume is available to molecules at lower pressure. This allows for many more ways of arranging them. 4. 4 points. Calculate 6o for the reduction of aluminum oxide by hydrogen gas: Al2O3 (s) + 3 H2 (g) 2 Al (s) + 3 H2O (g) Al2O3 (s) So = 51, H2 (g) So = 131, Al(s) So = 28, H2O (g) So = 189 (J/Kmol) Answer: ( 3189 + 228 ) - (51 + 3131 ) J/K = 179 J/K This value is for the change from Initial state to Final state. The Initial state consists of 1 mole pure Al2O3 (s) in its most stable form at 298 K, and 3 moles H2 (g) gas at 1 atm and 298 K. The Final state consists of 2 moles Al (s) in its most stable form at 298 K and 3 moles H2O (g) at 1 atm and 298 K. If the number of moles is doubled for all the components in the balanced equation, so is the value of 6 Note that superscript zero (nought) designates the standard state of 1 atm and the pure elements/compounds in their most stable form at, usually, 298 K. 5. 4 points. A chemical engineer wants to determine the feasibility of making ethanol by reacting water with ethylene according to the equation C2H4 (g) + H2O (l) C2H5OH (l) Is this reaction spontaneous under standard conditions? Show your calculation. *of values are: C2H5OH (l) = -175 kJ/mol, H2O (l) = -237 kJ/mol, C2H4 (g) = 68 kJ/mol Answer: *o = *of (ethanol liquid) - (*of (liquid water) + *of (ethane gas) ) = -6 kJ per mol rxn. 6. For ammonia (NH3 ), the enthalpy of fusion is 5.65 kJ/mol and the entropy of fusion is 28.9 J/Kmol. (a) 3 points. Calculate the approximate melting point of Ammonia. Answer: At the melting point the solid and liquid phases are at equilibrium. Either melting or freezing could take place by adding or removing heat. For a system at equilibrium, * 6LQFH* +-7S, and * UHDUUDQJHPHQWJLYHVH 76RU7 H6 . The values for freezing would have the same magnitude but opposite sign for the enthalpy and entropy changes. The fusion values are given, and substitution gives: T = (5.65 kJ/mol * 1000 J/kJ ) / ( 28.9 J/Kmol ) = 196 K (b) 1 point. Will NH3 (s) spontaneously melt at 250. K? Answer: Yes, because 250 K is above the melting point calculated in (a). Above the melting point, the favorable entropy term (positive) overcomes the unfavorable enthalpy term (positive) resulting in net increase in the entropy of the universe. (A positive enthalpy term for the system means a negative entropy term for the surroundings. At higher temperatures, +VXUU7EHFRPHVOHVVVLJQLILFDQWUHODWLYH WR6V\VDQGWKHVXPRIWKHWZREHFRPHVSRVLWLYH 4. 4 points. Predict the sign of 6o for the following reaction. (a) H2 (g) + 1/2 O2 (g) H2O (l) (b) Fe2O3 (s) + 3H2 (g) 2 Fe(s) + 3 H2O (g) page 2 Answer: (a) gaseous molecules (standard state) are being converted to a liquid => Decrease or negative (b) Same number of gaseous molecules on both sides, but water molecules are more complex and have more vibrational and rotational modes of motion. This suggests an increase in entropy (positive sign), but since it is difficult to make predictions for solids, one would have to consult tables of standard entropy values. 5. 5 points. For the following reaction, O (g) + O (g) O2 (g) (a) Predict the signs of + ________and 6 _________. (b) Would the reaction be more spontaneous at high or low temperatures? (Circle one) Explain in 10 to 15 words. Answer: Bonding results in a decrease in enthalpy (release of energy). But there is a concomitant decrease in entropy for this reaction (decrease in number of gaseous molecules). Since the enthalpy change is favorable, and the entropy change, unfavorable, the reaction would be spontaneous at lower temperatures. 6. 4 points. Hydrogen cyanide is produced industrially by the following exothermic reaction: 1000oC Pt-Rh 2 NH3 (g) + 3 O2 (g) + 2 CH4 (g) 2 HCN (g) + 6 H2O (g) Is the high temperature needed for thermodynamic or kinetic reasons? (circle one) Explain in 10 to 15 words. Answer: Kinetic reasons. Favorable enthalpy (exothermic) and favorable entropy (more gas molecules on product side) mean the reaction is spontaneous at all temperatures. The high temperature and catalyst are needed to speed up the reaction. 8. 6 points. For the following reaction, 2 NO2 (g) N2O4 (g) if the reactant and product are mixed at the following partial pressures at 25oC, predict the direction in which the reaction will shift to reach equilibrium. Show your calculation. R = 8.314 J/Kmol *of (NO2 ) = 52 kJ/mol and *of (N2O4 ) = 98 kJ/mol (a) P (NO2 ) = P (N2O4 ) = 1.0 atm Answer: *o rxn = 1 mol ( 98 kJ/mol ) - 2mol ( 52 kJ/mol) = -6 kJ per mole of reaction (i.e. N2O4 formed) At 1 atm, the reactant and product are in the standard state so the calculated *o value applies. It is negative, therefore reaction will take place spontaneously in the forward direction to reach equilibrium. (b) P (NO2 ) = 0.29 atm, P (N2O4 ) = 1.6 atm * *o + RT ln Q Q = 1.6 / (0.29)2 = 19 and ln Q = 2.94 * -6 kJ/mol + (8.314 10-3 kJ/Kmol) (298 K) (2.94) = 1 kJ/mol This value is positive, therefore reaction will proceed to the left to reach equilibrium. (*IRUWKH reverse reaction will be -1 kJ/mol.) page 3 9. 5 points. Consider the following reaction at 800. K: N2 (g) + 3F2 (g) 2 NF3 (g) An equilibrium mixture contains the following partial pressures: PN2 = 0.021 atm, PF2 = 0.063 atm, PNF3 = 0.48 atm. Calculate *o for the reaction at 800. K . If the equilibrium pressures were all 1 atm instead, what would *o be then? ________ Answer: A system at equilibrium has a *YDOXHRI]HUo. If, at equilibrium, all of the partial pressures are 1 atm, the system is in the standard state, its *LVGHVLJQDWHG*o , and in this case it has a value of zero. For pressures other than 1 atm, * *o + RT ln Q , and since * DWHTXLOLEULXP, 0 = *o + RT ln Keq and *o = - RT ln Keq = - (8.314 10-3 kJ/Kmol) (800 K) ln [0.482/0.021/0.0633] = -71 kJ/mol o This is * at 800 K. 11. Sketch the galvanic cell based on the following overall reaction which is incomplete as written. Assume acidic conditions. Answer: Anode and cathode compartments, electrodes immersed in respective solutions, connecting wire, direction of electron flow, voltmeter or load, salt bridge or porous disk, solution components. Cr3+ (aq) + Cl2 (g) Cr2O72- (aq) + Cl- (aq) (a) 3 points. First write the half-reaction for reduction and balance it. Answer: Cl2 (g) + 2 e- 2 Cl- (aq) Change in oxidation state of Cl is from 0 to -1 (b) 3 points. Next write the half-reaction for oxidation and balance it. Answer: 2 Cr3+ (aq) + 7 H2O Cr2O72- (aq) + 6 e- + 14 H+ Change in oxidation state of Cr is from (III) to (VI) (c) 3 points. Now combine the half-reactions in the proper way and write the overall balanced equation. Answer: Multiply the reduction half by 3 to balance the number of electrons. 3 Cl2 (g) + 6 e- + 2 Cr3+ (aq) + 7 H2O 6 Cl- (aq) + Cr2O72- (aq) + 6e- + 14 H+ The six electrons cancel out giving the balanced equation. (d) 3 points. Calculate the standard cell potential, Eo , for this galvanic cell. Answer: Eo cell = Eoox + Eo red = Do not multiply Eo values by the number of electrons or number of moles. Eo is an intensive property and does not depend on quantities of reactants. It is defined as energy or work per unit charge. (e) 2 points. What is the direction of electron flow? (Use an arrow in your sketch above.) Answer: From anode to cathode (f) 2 points. Identify the Cathode and the Anode in your sketch above. Answer: Oxidation takes place at the anode, reduction at the cathode (match vowels and consonants). (g) 2 points. What material is used for the electrodes? Answer: A Platinum wire or other inert conductor. (h) 2 points. For this standard cell, what are the concentrations or partial pressures of the components in each compartment? Anode: Cathode: All of the components (reactant and product) in (b) above. GAses at 1 atm, all other solutes at 1 M . All of the components (reactant and product) in (a) above. GAses at 1 atm, all others at 1 M page 4 12. 5 points. Answer the following questions using the data in the table of standard reduction potentials (all under standard conditions). Answer: Electrons flow from half-reactions with more NEGATIVE reduction potentials TO halfreactions with more POSITIVE reduction potentials. In addition, electrons flow from the reduced species (reducing agent) in one half-reaction to the oxidized species (oxidizing agent) in the other halfreaction. Here is another way of looking at it: The half-reaction with the more negative reduction potential is written as an oxidation (i.e. electron(s) on right side of arrow), and the half-reaction with the more positive reduction potential is written as a reduction (i.e. electron(s) on left side of reaction arrow). Also, reactions with the largest potentials (reduction, oxidation, or redox, as written from left to right) are the most favored. The foregoing analysis tells us only if a redox reaction is thermodynamically favored. There may be other factors (kinetics, adhering oxide layer, overvoltage) which prevent it from occurring. (a) Is H+ (aq) capable of oxidizing Cu(s) to Cu2+ (aq) ? NO + (b) Is H (aq) capable of oxidizing the metals, Al, Fe, Zn, Ni, Cr, and Sn ? (all or none) (c) Is Fe3+ (aq) capable of oxidizing I- (aq) ? NO + (d) Is H2 (g) capable of reducing Ag (aq) ? YES (e) Is Fe2+ (aq) capable of reducing Cr3+ (aq) to Cr2+ (aq) ? NO 13. 4 points. Consider the following species (at standard conditions): Na+ , Cl- , Ag+ , Ag, Zn2+ , Zn, Pb (a) Which is the strongest oxidizing agent? (b) Which is the strongest reducing agent? Na+ Zn YES 14. 2 points. (a) If you had to purchase only one of the following metals and its salt to make a concentration cell, which would it be, Silver or Nickel? Why? (A quantitative evaluation might be useful, as in (b).) Answer: Silver is oxidized to Ag+ , n=1 -- Only one electron in the denominator of Nernst equation (E = Eo - 0.0591/n logQ at 25oC ) gives a larger cell potential. Nickel is oxidized to Ni2+ with loss of 2 electrons. n=2 makes the cell potential smaller. (b) How would you make a concentration cell, using one of the metals in (a)? You could either sketch such a cell, or answer the following questions: 1 point. Which is the cathode? The compartment with the higher concentration of Ag+ 1 point. Which is the anode? The compartment with the lower concentration of Ag+ 1 point. What is the electrode material? Silver metal 1 point. What is the direction of electron flow? From Anode to Cathode 3 points. Suggest the concentrations for this cell and calculate the cell potential at 25oC . Answer: Suppose Low conc. = 0.10 M AgNO3 (aq) and High conc. = 1.0 M AgNO3 (aq) The silver electrode dissolves in the anode compartment, increasing the conc. of Ag+, but in the cathode compartment, the silver electrode gains mass and Ag+ (aq) is reduced, decreasing the concentration of silver ion in that compartment. Anode: Cathode: Ag (s) Ag+ (aq) + Ag+ + e eAg (s) Eo ox = -0.80 V Eo red = + 0.80 V Net: Ag+(aq) cathode (1 M ) Ag+ (aq) anode ( 0.10 M ) Eo = Eo ox + Eo red = 0 since the same standard reaction But need to use the Nernst equation since the concentrations are not both 1 M (standard conditions) Ecell = Eocell - RT/nF lnQ = Eocell - 0.0591/n log Q 0 - 0.591 log (0.10 M / 1.0 M ) = - 0.591 log 0.10 = 0.591 V If not sure about Q, choose the one that gives a positive potential for the cell. 15. The overall reaction in the lead storage battery is Pb(s) + PbO2(s) + 2H+ (aq) + 2HSO4- (aq) 2 PbSO4 (s) + 2H2O (l) (a) 3 points. Calculate Ecell at 25oC for this battery when [H2SO4] = 4.5 M , that is, [H+ ] = [HSO4-] = 4.5 M . At 25oC , Eocell = 2.04 V for the lead storage battery. Answer: Use the Nernst equation with n= 2 and Q = 1/ ( [H+]2 [HSO4-]2 ) Ecell = 2.04 V - 0.0591/2 log(1/[(4.52)(4.52)]) = 2.12 V page 5 (b) 2 points. As the battery discharges, how does Ecell change? (two or three words) Answer: It decreases to zero as the reactants are depleted and Q above becomes bigger and bigger. Once the reaction reaches equilibrium, * DQGWKHEDWWHU\LVGHDG (* -nFE ) 16. 5 points. Aluminum is produced commercially by the electrolysis of Al2O3 in the presence of a molten salt. If a plant has a continuous capacity of 1.00 million amp, what mass of aluminum can be produced in 2.00 h ? (You first need to determine the oxidation state of aluminum in Al2O3 . Also, note that you are only being asked the mass of aluminum produced, not the mass of Al2O3 required.) Answer: 2.00 hours 60 min/h 60 s/min 1 106 Coulomb/s 1mol e- /96,485 C 1 mol Al/3 mol e- 27 g Al/mol Al = 6.71 105 grams 17. (a) 6 points. What are the possible reactions that can take place at the cathode and the anode when a 1.0 M solution of NiBr2 is electrolyzed? Oxidation of bromide ion to Br2 Eoox = -1.09 V ; oxidation of water to oxygen and H+ Eoox = -1.23 V The first one is less negative and therefore more favored => Bromide is easier to oxidize, assuming standard conditions. Anode: Cathode: Reduction of Ni2+ , Eored = -0.23 V Reduction of water to H2 Eo red = -0.83 V The first one is less negative and therefore favored under standard conditions. (b) 3 points. Which reactions actually take place? Why? The charge on one mole of electrons = 96,485 Coulombs (also known as 1 Faraday) ...
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