Chemistry II Midterm Exam 22 April, 2011

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1 Chemistry II Midterm Exam 22 April, 2011 Constants R = J/mol K = L bar/k mol = L atm/k mol = L kpa/k mol 1 bar = torr = atm F = C/mol 1. You have a g mixture of table sugar (C 12 H 22 O 11 ) and table salt (NaCl). When this mixture is dissolved in 150. g of water, the freezing point is found to be C. K f of water is 1.89 K kg/mol, K b of water is 0.51 K kg/mol. Assume that the density of the solution is g/ml. (a). What is the molality of all solutes in this solution? (2%) (b). What is the boiling point of the solution? (2%) (c). Calculate the percent by mass of sugar in the original mixture. (4%) (d). What is the molarity of all solutes in this solution? (4%) (e). What is the osmotic pressure of the solution at 25? (4%) 1

2 2. The vapor pressure of H 2 O doubles when the temperature is raised from 50 to Assume Δ H o vap and Δ S o vap of water are constant. (a). Please calculateδ H o vap of water. (4%) (b). The vapor pressure of water at 25 is torr. Please calculateδ S o vap of water. (4%) (c). The pressure of a pressure cooker is 2.5 bar at cooking. Please find the boiling temperature of pure H 2 O in the pressure cooker. (4%) 3. A solution contains two volatile liquids A and B. Please complete the following table, in which the symbol indicates attractive intermolecular forces Relationship of Attractive Forces Deviation from Raoult s Law (none, negative or positive) Δ H sol (>0, <0, = 0) (A A, B B ) >> A B a (1%) b (1%) (A A, B B ) << A B c (1%) d (1%) (A A, B B ) = A B e (1%) f (1%) 4. The equilibrium constant of reaction I 2 (g) 2I(g) is 6.8 at 1200K. The partial pressure of I 2 (g) and I(g) in a reactor contains are 0.33 bar and 1.2 bar respectively at 1200K. (a). What is the reaction quotient? (2%) (b). What is the spontaneous direction of the reaction? (3%) (c). What is the standard reaction Gibbs free energy at 1200K? (4%) (d). What is the reaction Gibbs free energy at 1200K? (4%) (e). What is the total pressure at equilibrium? (4%) 5. An important reaction step used in the production of sulfuric acid is the oxidation of sulfur dioxide to sulfur trioxide: 2SO 2 (g) + O 2 (g) 2SO 3 (g). (a). The equilibrium constant K of reaction is at 500K and at 298 K. Is this a exothermic reaction or endothermic reaction? (2%) Please explain it. (2%) (b). If the reaction reaches equilibrium, what happens to the partial pressure of SO 3 when the partial pressure of SO 2 is decreased? (2%) (c). If the reaction reaches equilibrium, what happens to the partial pressure of O 2 when the partial pressure of SO 3 is increased? (2%) 6. A reaction used in the production of gaseous fuels from coal, which is mainly carbon, is C(s) + H 2 O(g) CO(g) + H 2 (g). (a) Evaluate K and K C at 900K, given that the standard Gibbs free energies of 2

3 formation of H 2 O(g) and CO(g) at 900K are kj/mol and kj/mol, respectively. (4%) (b) A 5.2 kg sample of graphite and 125 g of water were placed into a 10L container and heated to 900K. What are the equilibrium concentrations? (4%) 7. (8%) When iodide ions (I ) react with iodate ions (IO 3 ) in basic aqueous triiodide ions (I 3 ) are formed. Write the oxidation, reduction, and net ionic equations for the reaction. (Hint: the same product is obtained in each half-reaction.) 8. (8%) Acidified aqueous permanganate (MnO 4 ) solutions and acidified aqueous dichromate (Cr 2 O 7 ) solutions are powerful oxidizing agents. Suppose solutions of the two reagents are prepared and serve the two half-cells in a galvanic cell with platinum electrodes that generates a current in an external circuit. (a) Determine the standard potential of the cell constructed from these half-cells. (b) Write the net ionic equation for the cell having a positive standard potential. (c) Write the cell diagram of the reaction. 9. (7%) A tin electrode in 0.015M Sn(NO 3 ) 2(aq) is connected to a hydrogen electrode in which the pressure of H 2 is 1.0 bar. (a) Write down the cell diagram of the reaction and determine its standard cell potential. (b) If the potential is 0.061V at 25 C, what is the ph of the electrolyte at the hydrogen electrode? 10. (10%) A fuel cell generates electricity directly from a chemical reaction, as in a battery, but uses reactants that are supplied continuously, as in an engine. (a) Write the oxidation (anodic) and reduction (cathodic) half-reactions of a H 2 /O 2 fuel cell under basic condition (electrolyte: KOH (aq) ) and determine its standard cell potential. (b) What is the maximum electrical work that it can perform at the standard state condition by oxidation of 1 mol H 2? 3

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5 Ans 1: (a) The molality of all solutes = 2.24/1.89 =1.185 (2%) (b) The boiling point of the solution = *0.51 = (2%) (c) Assume that there is y g of sucrose y=5.7 g the percent by mass of sugar in the original mixture = 5.7/10.4*100% = 54.8% (4%) (d) The molarity of all solutes = (4%) (e) the osmotic pressure of the solution = 1.267*0.0821*298 = atm (or = 1.297* = bar) (4%) Ans 2: (a) Using Clausius-Clapeyron Equation, we can get Δ H o vap = J/mol 44 kj/mol (4%) (b) Torr J K mol ln S vap Torr 298 K Δ S o vap = J/K/mol (4%) or Torr J K mol ln Svap 760 Torr 298 K Δ S o vap = J/K/mol (4%) (c) Assume the boiling temperature of pure H 2 O in the pressure cooker is T 2. Clausius-Clapeyron Equation, we can get Using T 2. = K = (4%) 5

6 Or T 2. = K = (4%) Ans 3: (a) Positive (1%) (b) Δ H sol >0 (1%) (c) Negative (1%) (d) Δ H sol <0 (1%) (e) none (1%) (f) Δ H sol =0 (1%) Ans 4: (c) Q=1.2 2 /0.33 = 4.36 (2%, 如果將 Q 寫出單位以 0 分計算 ) (d) Because Q(=4.36) < K(=6.8), forward (3%) (e) ΔG o r = *1200*ln6.8 = J/mol kj/mol (4%) (f) ΔG r =ΔG o r *1200*ln4.36 = J/mol -4.4 kj/mol (4%) (g) I 2 (g) 2I(g) x +2x 0.93-x +2x K = (2x) 2 /(0.93-x)=6.8 X = or (meaningless) Partial pressure of I 2 is =0.262 Partial pressure of I is 2*0.668=1.336 Total pressure = = bar (4%) Ans 5: (a) Exothermic reaction (2%) because K at 500 is smaller than K at 298K (2%) (b) decrease (2%) (c) increase (2%) Ans 6: (a) G r o = RT lnk; lnk = G o r RT G r o = ( kj mol -1 ) ( kj mol -1 ) = 6.8 kj mol -1 6

7 6800 J mol lnk = 1 (8.314 J K mol 1 1 )(900 K) = ; K = (2%, 如果將 K 寫出單位以 0 分計算 ) K C = K(RT) -1 = 0.403/( *900) = (2%, 如果將 K c 寫出單位以 0 分計算 ) (b) mol C g C = 433. mol C g C 1mol H2O 125 g H 2 O g H O 2 = 6.94 mol H 2 O H 2 O is limiting. Conc. of H 2 O = 6.94 mol/10.0 L = mol L -1 Concentration 1 (mol L ) H 2 O (g) CO(g) H 2 (g) initial change x x x final x x x K C = COH H 2 2 = ( x)( x) (0.694 x) x 2 = x; = x x = 0 x = or (not meaningful) [CO] = [H 2 ] = M [H 2 O] = = M (4%) Ans 7: Let s start with I + IO 3 I 3 (unbalanced): (1) Oxidation: 3I I 3 I are oxidized to I 2 + I 3I I 3 + 2e (2) Reduction: 3IO 3 I 3 I are reduced to I 2 + I 3IO 3 e I 3 3IO 3 H 2 O + e I 3 OH (3) Net equation = (1)x8 + (2) 24I + 3IO 3 H 2 O I 3 OH The stoichiometric numbers are divided by 3 and put the phase notation 7

8 for each species in the chemical equation. 8I aq + IO 3 aq H 2 O l I 3 aq OH aq Ans 8: Example 13.7 (a) MnO 4 aq + 8H aqe Mn 2+ aqh 2 O l E 0 = V Cr 2 O 7 aq + 14H aqe Cr 3+ aqh 2 O l E 0 = V E 0 (cell) = E 0 (R)- E 0 (L) = = V (3%) (b) 6MnO 4 aq + 48H aqe Mn 2+ aqh 2 O l Cr 3+ aqh 2 O l Cr 2 O 7 aq + 70H aqe 6MnO 4 aq + 11 H 2 O l Cr 3+ aqmn 2+ aqh aqcr 2 O 7 aq (c) Pt (s) Cr 2 O 7 aq Cr 3+ aqh aq H aqmno 4 aq Mn 2+ aqpt (s) (2%) Ans 9: Exercise (a) Sn (s) Sn 2+ (aq, H (aq,ph =?H 2 (g, 1.0 bar)pt (s) E 0 (cell) = E 0 (R)- E 0 (L) = 0 (-0.14) = V (3 %) (b) Nernst equation: E = E /2log([Sn 2+ ]/[H + ] 2 ) log([sn 2+ ]) 0.059pH ph = 2.25 (4%) Ans 10: Box 13.1 (a) Anode: 2H 2(g) + 4OH (aq) l e Cathode: O 2(g) e l 4OH (aq) (2%) Net reaction: 2H 2(g) + O 2(g) l E 0 (cell) = E 0 (R)- E 0 (L) = 0.4 (-0.83) = V (2% 直接寫出答案者不給分 ) (b) w e,max = G 0 = nfe 0 = = kj Note that this free energy is produced by oxidation of 2 moles of H 2. Let s rewrite the chemical equation as H 2(g) + 1/2O 2(g) l, the maximum work can be performed by the fuel cell is 237 kj per mol of hydrogen. (4%) 8

Chemistry II Midterm Exam April 24, 2009

Chemistry II Midterm Exam April 24, 2009 Constants R = 8.314 J / mol K = 0.08314 Lbar / K mol = 8.314 L kpa / K mol F = 9.6485 10 4 C/mol h = 6.63 10-34 J s h = 1.05 10-34 J s k = 1.3806504 10 23 J / K