ph = pk a + log 10{[base]/[acid]}

Transcription

1 FRONT PAGE FORMULA SHEET - TEAR OFF N A = x C = ( 5 / 9) ( F - 32) F = ( 9 / 5)( C) amu = x kg C = K K = C atm = 760 torr = 760 mm Hg 1 atm = bar pv = nrt R = L atm/mol K 1 L atm = J R = J/mol K 1 J= 1 kg m 2 /s 2 F = C/mol ln(p) = - H vap + C ln(p 2/p 1) = - ( H vap/r) { (1/T 2) - (1/T 1) } T p A = X A p A [B] = k p B p A = X Bp A T b = K b m B T f = K f m B = [B]RT H = U + pv G = H - TS G rxn = G rxn + RT ln Q ln K = - G rxn/rt K p = K C (RT) n If ax 2 + bx + c = 0, then x = ( - b [b 2-4ac] 1/2 ) 2a K a K b = K w = 1.0 x (at T = 25 C) ph = pk a + log 10{[base]/[acid]} G = - nfe cell E cell = E cell - (RT/nF) ln Q ln K = nfe cell/rt [A] t = [A] 0 e -kt ln[a] t = ln[a] 0 - kt t 1/2 = ln2/k [A] t = [A] 0/(1 + kt[a] 0) (1/[A] t) = (1/[A] 0) + kt t 1/2 = 1/(k[A] 0) k = A e -Ea/RT ln k = ln A - (E a/r)(1/t) ln(k 2/k 1) = - (E a/r) [ (1/T 2) - (1/T 1) ]

2 GENERAL CHEMISTRY 2 FINAL EXAM Name Panthersoft ID Signature Part 1 (48 points) Part 2 (64 points) Part 3 (88 points) TOTAL (200 points) Do all of the following problems. Show your work. 2

3 Part 1. Multiple choice. Circle the letter corresponding to the correct answer. There is one and only one correct answer per problem. [4 points each] 1) Which of the following methods for expressing concentration have no units? a) molality b) molarity C c) mole fraction d) Both a and b e) All of the above methods for expressing concentration have units 2) A solution is formed by adding moles of a nonvolatile and nonionizing solute to g of a volatile solvent. Compared to the pure solvent a) the boiling point of the solution will be higher than the boiling point of the pure solvent b) the freezing point of the solution will be higher than the freezing point of the pure solvent A c) the vapor pressure of the solution will be higher than the vapor pressure of the pure solvent d) Both a and b e) Both b and c 3) For a spontaneous reaction which of the following must be true? a) G rxn < 0 b) S rxn < 0 A c) S rxn > 0 d) Both a and b e) Both a and c 4) For which of the following reactions would we expect S rxn to be large and negative? a) MgCO 3(s) MgO(s) + CO 2(g) b) C 6H 6( ) C 6H 6(g) C c) C 3H 8(g) + 5 O 2(g) 3 CO 2(g) + 4 H 2O( ) d) both a and b e) both a and c 5) Consider the reaction I 2(g) + Cl 2(g) 2 ICl(g) A system containing I 2, Cl 2, and ICl is initially at equilibrium at some temperature T. The volume of the system is decreased to half of its initial value while keeping temperature constant. Which of the following will occur as the system re-establishes equilibrium? a) The moles of I 2 in the system will increase b) The moles of Cl 2 in the system will increase E c) The moles of ICl in the system will increase d) Both a and b e) None of the above 6) Benzoic acid (C 6H 5COOH) is a weak acid, with K a = 6.5 x 10-5 (at T = 25. C). The C 6H 5COO - ion is a) a strong acid b) a weak acid D c) a strong base d) a weak base e) None of the above, as ions have no acid or base properties 3

4 7) A Bronsted acid is a) a proton acceptor b) a proton donor B c) an electron pair acceptor d) an electron pair donor e) both b and c 8) Consider the following three substances: HBr, H 2S, and H 2Se. Of these substances a) HBr is the strongest acid and H 2S is the weakest acid b) HBr is the strongest acid and H 2Se is the weakest acid A c) H 2S is the strongest acid and HBr is the weakest acid d) H 2Se is the strongest acid and HBr is the weakest acid e) H 2Se is the strongest acid and H 2S is the weakest acid 9) Which of the following aqueous solutions will be a buffer solution? a) A solution containing M HBr (an acid) and M NaBr. b) A solution containing M HNO 2 (an acid) and M NaNO 2. E c) A solution containing M HNO 2 (an acid) and M NaOH. d) Both a and b e) Both b and c 10) The standard reduction potential for the process Cu 2+ (aq) + 2 e - Cu(s) is E = v. Based on this, we can say that the half-cell oxidation potential for the process 2 Cu(s) 2 Cu 2+ (aq) + 4 e - is a) E = v b) E = v D c) E = 0.00 v d) E = v e) E = v 11) A chemical reaction obeys the rate law rate = k [A] [B] 2 The overall order of the reaction is a) first order b) second order C c) third order d) both a and b e) both a and b and c 12) A catalyst a) changes the rate of reaction and also changes the equilibrium constant for a reaction b) changes the rate of reaction but does not change the equilibrium constant for a reaction B c) does not change the rate of reaction but changes the equilibrium constant for a reaction d) does not change the rate of reaction and also does not change the equilibrium constant for a reaction e) cannot be present in a system at equilibrium 4

5 Part 2. Short answer. 1) Define the following terms [6 points each] a) amphoteric A substance that can act as either a Bronsted acid or a Bronsted base in an acid/base reaction. Example: HF(aq) + H 2O( ) H 3O + (aq) + F - (aq) water is a Bronsted base NH 3(aq) + H 2O( ) NH + 4 (aq) + OH - (aq) water is a Bronsted acid b) catalyst A substance that when added to a system speeds up a chemical reaction without itself being produced or consumed. Enzymes are one example of biological catalysts. 2) 0.45 g of a nonvolatile pure chemical substance (a polymer) are dissolved in liquid benzene (C 6H 6, MW = 78.1 g/mol), to form a solution with final volume V = ml. The osmotic pressure of the solution, measured at T = 20.0 C, was 33.2 torr. Based on this information find the molecular weight of the polymer. [10 points] = [B]RT [B] = = 33.2 torr (1 atm/760 torr) = 1.82 x 10-3 mol/l RT ( L atm/mol K)(293. K) # moles = ( L) (1.82 x 10-3 mol/l) = 3.63 x 10-4 mol MW = mass = 0.45 g = g/mol moles 3.63 x 10-4 mol 3) For each of the following questions circle the correct answer. There is one and only one correct answer per problem. [4 points each] a) The substance with G f = 0.0 kj/mol at T = 25. C CuCl 2(s) C 6H 12( ) O2(g) O 3(g) b) The acid that is a polyprotic acid HBr HBrO 2 HNO 3 H3PO4 c) The hydroxide compound that is a strong soluble base AgOH Ba(OH)2 Cu(OH) 2 Fe(OH) 3 d) The best indicator to use in the titration of a strong acid by a strong base alizarin yellow bromothymol blue bromophenol blue thymol blue pk ind = 11.0 pkind = 6.8 pk ind = 3.8 pk ind = 2.0 5

6 4) Equilibrium between sulfur oxides in the presence of molecular oxygen is difficult to study experimentally. Consider the following reaction 2 SO 2(g) + O 2(g) 2 SO 3(g) (K C = 1.9 x at T = 25.0 C) a) Give the expression for K C, the equilibrium constant, in terms of reactant and product concentrations. [4 points] K C = [SO 3] 2 [SO 2] 2 [O 2] b) For a particular system the initial concentrations of SO 3 and O 2 are both mol/l. No SO 2 is initially present. Find the value for [SO 2], the concentration of SO 2, when equilibrium is reached. [10 points] Initial Change Equilibrium SO x 2x O x x SO x x So ( x) 2 = 1.9 x If we assume x << , then (2x) 2 ( x) (0.0400) 2 = 1.9 x x 2 = 5.26 x x = 7.25 x (4x 2 )(0.0400) So at equilibrium, [SO 2] = 2(7.25 x ) = 1.45 x mol/l 5) Balance the following oxidation-reduction reaction for the indicated condition. [12 points each] H 3AsO 3(aq) + Cr 2O 7 2- (aq) H 3AsO 4(aq) + Cr 3+ (aq) (in acid solution) ox H 3AsO 3(aq) + H 2O( ) H 3AsO 4(aq) + 2 e H + (aq) x 3 red Cr 2O 7 2- (aq) + 6 e H + (aq) 2 Cr 3+ (aq) + 7 H 2O( ) net 3 H 3AsO 3(aq) + Cr 2O 7 2- (aq) + 8 H + (aq) 3 H 3AsO 4(aq) + 2 Cr 3+ (aq) + 4 H 2O( ) 6

7 Part 3. Problems. 1) Thermochemical data (at T = 25. C) for several substances is given below, and may be of use in doing the following problems, which also take place at T = 25. C. Substance H f (kj/mol) G f (kj/mol) S (J/mol K) C(s) CO 2(g) CO(NH 2) 2(s) H 2O(g) NH 3(g) O 2(g) a) What is the value for S rxn for the reaction [8 points] C(s) + O 2(g) CO 2(g) S rxn = [ S (CO 2(g)) ] [ S (C(s)) + S (O 2(g)) ] = [ ] [ (5.74) + (205.14) ] = 2.86 J/mol K b) What are the values for G rxn and K for the reaction [16 points] CO(NH 2) 2(s) + H 2O(g) CO 2(g) + 2 NH 3(g) G rxn = [ G f(co 2(g) + 2 G f(nh 3(g)) ] [ G f(co(nh 2) 2(s)) + G f(h 2O(g)) ] = [ ( ) + 2 ( ) ] [ ( ) + ( ) ] = kj/mol ln K = - G rxn = - ( J/mol) = 0.55 RT (8.314 J/mol K) (298. K) K = e 0.55 =

8 2) Titration is a common method for finding the concentration of stock solutions of acids or bases. The following problem concerns a titration carried out at T = 25. C. A student titrates a ml sample of a M solution of hypochlorous acid (HClO, K a = 2.9 x 10-8 ) with a stock solution of sodium hydroxide (KOH), a strong soluble base. a) What is the initial ph of the sample HClO solution? [8 points] HClO(aq) + H 2O( ) H 3O + (aq) + ClO - (aq) K a = [H 3O + ][ClO - ] [HClO] Initial Change Equilibrium H 3O + 0 x x (x) (x) = 2.9 x 10-8 If we assume x << ClO - 0 x x ( x) HClO x x x 2 = (0.1814)(2.9 x 10-8 ) = 5.26 x 10-9 x = (5.26 x 10-9 ) 1/2 = 7.25 x 10-5 ph = - log 10(7.25 x 10-5 ) = 4.14 b) After ml of sodium hydroxide solution is added the equivalence point for the titration is reached. Based on this result, what is the concentration of the sodium hydroxide stock solution? [8 points] Neutralization reaction is HClO(aq) + NaOH(aq) Na + (aq) + ClO - (aq) + H 2O( ) Since equal amounts of acid and base react, then at equivalence point moles acid = moles base moles acid = moles base, and so M acid V acid = M base V base M base = M acid (V acid/v base) = ( M) (25.00/38.17) = M c) In the above titration, what is the ph at the equivalence point (when ml of stock HClO solution has been mixed with ml of stock NaOH solution)? [12 points] The volume at this point is V = ml ml = ml At this point, all the weak acid has been converted into weak base. So this is finding the ph of a weak base solution. The moles of ClO - is n = ( M) ( L) = 4.54 x 10-3 mol The initial ClO - concentration is [ClO - ] = (4.54 x 10-3 mol)/( L) = M K b = 1.0 x /K a = (1.0 x )/(2.9 x 10-8 ) = 3.45 x 10-7 ClO - (aq) + H 2O( ) HClO(aq) + OH - (aq) K b = [HClO] [OH - ] [ClO - ] Initial Change Equilibrium HClO 0 x x (x)(x) = 3.45 x 10-7 If we assume x << OH - 0 x x ( x) ClO x x x 2 = (0.0718)(3.45 x 10-7 ) = 2.48 x 10-8 x = 1.57 x 10-4 poh = 3.80 ph =

9 3) Consider the following galvanic cell (at T = 25. C) Cu 2+ ( M) Cu(s) Ag(s) Ag + ( M) Find the following. You may need to use some of the reduction data given below in doing part of this problem Ag + (aq) + e - Ag(s) Cu + (aq) + e - Cu(s) Cu 2+ (aq) + 2e - Cu(s) Cu 2+ (aq) + e - Cu + (aq) E = 0.80 v E = 0.52 v E = 0.34 v E = 0.16 v a) The half cell oxidation reaction, the half cell reduction, and the net cell reaction [5 points] ox Cu(s) Cu 2+ (aq) + 2 e - E = v red Ag + (aq) + e - Ag(s) x 2 E = v cell Cu(s) + 2 Ag + (aq) Cu 2+ (aq) + 2 Ag(s) E cell = v b) E cell [5 points] E cell = 0.46 v (see above) c) E cell [6 points] From Nernst E cell = E cell (RT/nF) ln(q) Q = [Cu 2+ ] = (0.5000) = 1.25 x 10 5 [Ag + ] 2 ( ) 2 E cell = (0.46 v) - (8.314 J/mol. K)(298. K) ln(1.25 x 10 5 ) = 0.31 v (2)(96485 C/mol) 9

10 4) MPAN (peroxymethacryloyl nitrate) is one of a family of related compounds that are atmospheric pollutants. The reaction, given below, is for most conditions an irreversible first order reaction MPAN products The rate constant for the reaction is k = 5.6 x 10-6 s -1 at T = 0.0 C, and k = 1.2 x 10-2 s -1 at T = 50.0 C Find the following a) The value for t 1/2 for MPAN at T = 0.0 C. [5 points] t 1/2 = ln(2) = 1.24 x 10 5 s (5.6 x 10-6 s -1 ) b) The value for A and E a, the Arrhenius parameters, for the above reaction (including correct units) [15 points] ln(k 2/k 1) = - (E a/r) { (1/T 2) (1/T 1) } E a = - R ln(k 2/k 1) = - (8.314 J/mol. K) ln(1.2 x 10-2 /5.6 x 10-6 ) = kj/mol { (1/T 2) (1/T 1) } { (1/323. K) - (1/273 K) } We may use the rate constant at 0.0 C to find A k = A exp(-e a/rt) A = k exp(e a/rt) = (5.6 x 10-6 s -1 ) exp[( J/mol)/(8.314 J/mol. K)(273. K)] = (5.6 x 10-6 s -1 ) e = 1.9 x s -1 10