Chemistry II Midterm Exam 20 April, 2012

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1 Chemistry II Midterm Exam 0 April, 01 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. A 0.5-g sample of a polymer, dissolved in L of CCl 4, has an osmotic of 4.0 Torr at 7.C. (a) What is the molar mass of the polymer? (b) Use your answer of question (a) and the given data, for CCl 4, T b = 76.5C, k b = 4.95 K kg mol -1, density = 1.59 g cm -3. Why is boiling-point elevation method not very suitable for determining the molar mass of a polymer? (4+3%). The mixture of benzene and toluene is an ideal solution. What is the (a) vapor pressure and (b) mole fraction in the vapor of each component of a mixture of benzene and toluene in which x benzene (l) = 0.33 and x toluene (l) = 0.67 at 5C. The vapor pressures of benzene and toluene at 5C are 94.6 and 9.1 Torr, respectively. (c) The boiling points of benzene and toluene are 80.1 and 110C, respectively. Is the boiling point of this mixture lower 80.1C, higher than 110C, or between 80.1 and 110C? (4+4+%) 1

2 3. The vapro pressur of acetone ( 丙酮 ) at 7.7C is 13.3 kpa, and its enthalpy of vaporization is 9.1 kj mol -1. Estimate (a) the normal boiling point and (b) the entropy of vaporization of acetone. (5+5%) 4. Explain the followings briefly: (a) Henry s law; (b) hydrophilic. (3+3%) 5. The decomposition reaction of calcium sulfate CaSO 4 (s) in vacuum will produce calcium oxide CaO(s), sulfur dioxide SO (g), and oxygen gas. (a) Please write the balanced chemical equation of the decomposition reaction. (%) (b) Please calculate two equilibrium constants K at 600K and 1600K by using the thermodynamic data listed below, where all of the data are independent on the temperature. (6%) Compound H f (kj/mol) S m (J/ K mol) CaSO CaO SO Oxygen gas (c) What are the total equilibrium pressures if the reaction starts from pure CaSO 4 (s) at 600K? (3%) (d) What are the total equilibrium pressures if the reaction starts from pure CaSO 4 (s) at 1600K? (3%) 6. The reaction HHH(g) H (g) + CC (g) has K = at 98K. (a) The initial partial pressure are HCl, bar; H, 1 bar; and Cl, 3 bar. At equilibrium there is 1.0 mol H (g). What is the volume of the container? (3%) (b) If the total pressure at equilibrium is found to be 6.0 bar and the Cl:H atom ration is 1:5. What are the partial pressures of the three gases? (6%) 7. In the cells of the normal stomach of the human at 37, the ph values are 7 inside the cells and 1 outside the cells. In order to maintain ph =1 outside the stomach cells which is the healthy condition for the digestion process, the H + ions are required to be pumped from the inside of the cell to the outside. The pumping process of the H + ions can be written as the chemical equation: H + (inside the cell) H + (outside) (a) What is the standard free energy G r of the pumping process? Explain it. (4%) (b) Please calculate the chemical quotient Q of the pumping process in the normal stomach of the human. (3%)

3 (c) Please calculate the free energy G r required by the pumping process in the normal stomach of the human at 37. (4%) 8. Silver from a mL solution of silver(i) sulfate is plated on to the cathode of an electrolytic cell. (a) Hydronium ions are generated at one of the electrodes. Is this electrode the anode or cathode? (3%) (b) How many moles of H 3 O + are generated if a current of 0.10 A is applied to the cell for 4.0 h? (4%) (c) What will be the concentration of the H 3 O + after the electrolysis? Assume no change in the volume of solution. (3%) 9. Describe the following conditions for corrosion process of iron. (a) The iron nails do not rust when it is exposed to pure water (ph=7). Support your answer with calculation and data found in the standard potential table. (5%) (b) When an iron pipe is exposed to acid rain with [H + ] = 10-4 M and oxygen dissolving in the water. Assume the partial pressure of O in air is 0. bar. What is the reduction potential for the half reaction: (5%) O (g) + 4H + (aa) + 4e H O(l) (c) Based on the result in (b), will the oxidation of iron from Fe to Fe + be thermodynamically spontaneous? Describe your reasoning. (5%) (d) Which of the following metals, if coated onto iron, would prevent the corrosion of iron? Describe your reasoning. (5%) a. Zn b. Sn c. Co 10. Write a proper cell diagram for the following reaction based on the reaction in the alkaline battery cell: (3%) Zn(s) + OH (aq) ZnO(s) + H O(l) + e - MnO (s) + H O(l) + e - Mn(OH) (s) + OH (aq) 3

4 Standard Potential Table (at 5 ) 4

5 Answers Ans 1: (a) = irtc ; 4.0/760 = c ; c = (M) ; molar mass of the polymer = ( ) = g mol -1 4 分 (b) T b = 4.95 (0.5 ( ) ( )) = K T b is very small and very difficult to be measured precisely and accurately. 3 分 Ans : (a) P benzene = = 31. Torr, P toluene = = 19.5 Torr 各分 (b) x benzene (g) = 31. ( ) = 0.6, x toluene (g) = 19.5 ( ) = 0.38 各分 (c) between 80.1 and 110C 分 Ans 3: (a) ln( / ) = ( ) (1/T 1/80.8) T = K (or 6.3C) 5 分或是以 Touton s Law 算出 S vap = H vap T = 85 J/K mmm T = H vap /85 = /85 = K 5 分 (b) ln(p/1 bar) = H vap /RT + S vap /R ln( ) = ( )/( ) + S vap /8.314 S vap = 86.9 J/K mol 5 分或是以 Touton s Law 敘述如下 : 由於 aceton 丙酮是非極性液體其物理特性適合以 Touton s Law 敘述, 因此 S vap 85 J/K mol 5 分 Ans 4: (a) Henry s Law: the solubility of a gas is directly proportional to its partial pressure. 3 分 (b) water attracting 3 分 5

6 Ans5 : (a) CaSO 4 (s) CaO(s) + SO (g) + O (g) (%) (b) H r = ( )*+(-96.83)*+0-( )* = kj/mol S r = (39.75)*+(48.)* (106.7)* = J/K/mol At 600K At 1600K G r = * * = J/mol K = e G r RT = exp(-9609/8.314/600) = (3%) G r = * * = J/mol K =e G r RT = exp(-9609/8.314/1600) = (3%) (c) Because K<10-4, Total equilibrium pressure = 0 bar (3%) 或是以下的計算及答案也算正確 Assume that P(O ) is x bar at equilibrium, P(SO ) will be x bar. K = 4x 3 = x = Total equilibrium pressure = 3x = bar 0 bar (3%) (d) Assume that P(O ) is x bar at equilibrium, P(SO ) will be x bar. K = 4x 3 = x = Total equilibrium pressure = 3x = 0.17 bar (3%) Ans6: (a) Since reactants are strongly favored, it is easier to push the reaction as far to the left as possible then start from new initial conditions. Pressures (bar) HCl(g) H (g) + Cl (g) original new initial change x + x + x final 4.0 x + x.0 + x 6

7 P P K = H Cl PHCl 34 = = x = bar nrt V = = p = L ( x)(.0 + x) ( x)(.0) x (4.0 x) (4.0) (1 mole)( L bar K mol )(98K) bar (b) HCl(g) H (g) + Cl (g) P P K = = H Cl PHCl (Eq. 1) At equilibrium,.p Total = P H + P Cl + P HCl = 6 bar (Eq. ) (4%) And, since the atomic ratio Cl : H is 1:5, Moles Cl : Moles H = 1:5 = n HCl + n Cl : n HCl + n H 5 n HCl + n Cl = n HCl + n H n H = n HCl + 5n Cl P gas n gas at constant T and V, so P H = P HCl + 5P Cl (Eq. 3) (1%) We now have three equations in three unknowns, so we can rearrange and substitute to find each partial pressure. Substituting equation 3 into equation gives P Total = P HCl + 5P Cl + P Cl + P HCl = 3P HCl + 6P Cl = 6 bar P HCl = P Cl (1%) Substitution back into equation 3 gives P H = 4 + P Cl (1%) Using these two expressions for partial pressures in equation 1 gives K = 4+P Cl P Cl P Cl = Since the equilibrium constant is so small, we would expect the partial pressure of chlorine to be very low at equilibrium, so we can neglect it to simplify this expression. 7

8 K = 4 P Cl 4 = P Cl = bar P HCl = bar P H = 4 bar (3%) Ans7: (a) G r = 0 J/mol (or G r = 0 kj/mol )(%) 因為在 standard condtion,[h + ] outside = [H + ] inside =1M 因此內外的氫離子濃度相同, 處於平衡狀態, 此時的 G r 應該等於 0 (%) (b) Q = [H+ ] outside [H + ] inside = ph=1 ph=7 = = 106 (3%) (c) G r = G r + RTlnQ = *310*ln(10 6 ) = J/mol = kj/mol (4%) Ans8: (a) 4Ag (aq) e 4Ag(s) E = 0.80 V H + (aq) or hydronium ions are generated at the anode. (3%) (b) Q = nf and Q = It = 8.95 x 10 - mol of electrons Since 4 moles of electrons are needed in the balanced reaction to generate 4 moles of H 3 O + there is a 1:1 ratio. Since 8.95 x 10 - mol of electrons were supplied a maximum of 8.95 x 10 - mol H 3 O + can be generated. (4%) (c) (3%) 8

9 Ans9: (a). Standard condition [OH - ] = 1 pure water, ph = 7, [OH-] = 10-7 E = E E = E E = 0 0 ( 0.83) E = 0.4V log n log n Q [ OH ] ( [ OH ] (10 ) log (1) ph = 7) (1M ) Fe+(aq) e- + Fe(s) E = V Iron has only a very slight tendency to be oxidized by water at ph = 7. For this reason, iron can be used for pipes in water supply systems and can be stored in oxygen-free water without rusting. (5%) (b). [H + ] = 10-4 M and p(o ) = 0. bar E = E logq n PO (1atm) H E 1.3 log = n PO (0.atm) H E = = 0.968V + 4 [,1M ] + 4 [,10 M ] 4 (5%) (c). The E = V is higher than the value for Fe + /Fe. iron can reduce oxygen in aqueous solution at ph = 4. (5%) (d) a. Zn b. Sn c. Co Fe Zn+/Zn Sn+/Sn Co+/Co Fe+/Fe E Zinc lies below iron in the electrochemical series; so, if a scratch exposes the metal beneath, the more strongly reducing zinc releases electrons to the iron. (5%) Ans10: (3%) Zn(s) ZnO(s) OH (aq) Mn(OH) (s) MnO (s) graphite Zn(s) ZnO(s) OH (aq) MnO (s) Mn(OH) (s) graphite 9 或是皆正確

Chemistry II Midterm Exam 22 April, 2011

Chemistry II Midterm Exam 22 April, 2011 Constants R = 8.314 J/mol K = 0.08314 L bar/k mol = 0.0821 L atm/k mol = 8.314 L kpa/k mol 1 bar = 750.06 torr = 0.9869 atm F = 9.6485 10 4 C/mol 1. You have a