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

Transcription

1 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 F = C/mol R = J/mol K 1 J= 1 kg m 2 /s 2 p A = X A p A [B] = k p B p A = X B p A T b = K b m B T f = K f m B = [B]RT K p = K C (RT) n H = E + pv G rxn = G rxn + RT ln Q G = H - TS ln K = - G rxn /RT 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]} [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 ) ] G = - nfe cell E cell = E cell - (RT/nF) ln Q ln K = nfe cell /RT

2 GENERAL CHEMISTRY 2 FINAL EXAM JUNE 16, 2017 Name Panthersoft ID Signature Part 1 (50 points) Part 2 (78 points) Part 3 (112 points) TOTAL (240 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 is not a colligative property? a) boiling point elevation b) freezing point depression D c) vapor pressure lowering d) molecular diffusion e) osmotic pressure 2) 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 6 H 6 ( ) C 6 H 6 (g) C c) C 3 H 8 (g) + 5 O 2 (g) 3 CO 2 (g) + 4 H 2 O( ) d) both a and b e) both a and c 3) For a particular chemical reaction H rxn < 0 and S rxn > 0. Assuming that both H rxn and S rxn are approximately independent of temperature, which of the following is expected to be true? a) The reaction is always spontaneous b) The reaction is never spontaneous A c) The reaction is spontaneous at low temperatures but not at high temperatures d) The reaction is spontaneous at high temperatures but not at low temperatures e) Any of the above might be true for the reaction 4) Consider the following chemical reaction CO(g) + H 2 O(g) CO 2 (g) + H 2 (g) A closed system at a constant temperature contains all of the above gases and is initially at equilibrium. For which of the following changes will the number of moles of H 2 (g) in the system increase when equilibrium is reestablished? a) Addition of moles of CO(g) into the system b) Addition of moles of CO 2 (g) into the system A c) Increasing the volume of the system d) Both a and c e) Both b and c 5) A Bronsted acid is a) a proton donor b) an electron pair donor A c) a proton acceptor d) an electron pair acceptor e) both a and b 6) Consider the following three substances: HBr, H 2 S, and H 2 Se. Of these substances a) HBr is the strongest acid and H 2 S is the weakest acid b) HBr is the strongest acid and H 2 Se is the weakest acid A c) H 2 S is the strongest acid and HBr is the weakest acid d) H 2 Se is the strongest acid and HBr is the weakest acid e) H 2 Se is the strongest acid and H 2 S is the weakest acid 3

4 7) For a galvanic cell reaction to be spontaneous for standard conditions which of the following must be true? a) G rxn > 0 and E cell > 0 b) G rxn > 0 and E cell < 0 C c) G rxn < 0 and E cell > 0 d) G rxn < 0 and E cell < 0 e) both c and d 8) 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 9) 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 10) A catalyst a) changes the rate of reaction but does not change the equilibrium constant for a reaction b) changes the rate of reaction and also changes the equilibrium constant for a reaction A c) does not change the rate of reaction and also does not change the equilibrium constant for a reaction d) does not change the rate of reaction but changes the equilibrium constant for a reaction e) cannot be present in a system at equilibrium Part 2. Short answer. 1) For the reaction 2 FeCl 3 (s) + 3 H 2 O(g) Fe 2 O 3 (s) + 6 HCl(g) write the expression for K C. [6 points] K C = [HCl] 6 [H 2 O] 3 4

5 2) 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 6 H 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 aqueous solution with the highest normal boiling point a M solution a M solution a M solution a M solution of NaCl of Ba(OH) 2 of Fe(NO3)3 of K 2 SO 4 e) The best indicator to use in the titration of a strong acid by a strong base alizarin yellow bromothymol blue bromophenol blue thymol blue K ind = 11.0 Kind = 6.8 K ind = 3.8 K ind = 2.0 3) 0.45 g of a nonvolatile pure chemical substance (a polymer) are dissolved in liquid benzene (C 6 H 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, and so [B] = = (33.2 torr) (1 atm/760 torr) = x 10-3 mol/l RT ( L atm/mol K) (293.2 K) So the moles of solute is moles = (0.200 L) (1.816 x 10-3 mol/l) = x 10-4 mol And so MW = 0.45 g =1240 g/mol (3.631 x 10-4 mol) 4) Hydrochloric acid (HCl) is a strong acid. What are the values for ph and poh for a 6.0 x 10-4 mol/l solution of hydrochloric acid, at T = 25. C? [10 points] Since HCl is a strong acid, it completely dissociates HCl(aq) H + (aq) + Cl - (aq) [H + ] = [HCl] = 6.0 x 10-4 mol/l ph = - log 10 (6.0 x 10-4 ) = 3.22 poh = ph =

6 5) For the chemical reaction Cl 2 (g) + 2 NO(g) 2 NOCl(g) the value for the equilibrium constant at a particular temperature is K C = 3.7 x A system initially contains M NO and M NOCl. No Cl 2 is initially present in the system. Find the concentration of Cl 2 in the system at equilibrium. [10 points] K C = [NOCl] 2 = 3.7 x 10 8 Initial Change Equilibrium [Cl 2 ] [NO] 2 Cl 2 0 x x NO x x NOCl x x So ( x) 2 = 3.7 x 10 8 Assume x << , then x ( x) 2 (0.0030) 2 = 3.7 x 10 8 x = (0.0030) 2 = 3.9 x x (0.0250) 2 (0.0250) 2 (3.7 x 10 8 ) Our assumption that x is small is good. So at equilibrium, [Cl 2 ] = x = 3.9 x M 6) The solubility product for aluminum hydroxide (Al(OH) 3 ) is K sp = 1.3 x at T = 25. C. a) Give the balanced solubility reaction for Al(OH) 3. [4 points] Al(OH) 3 (s) Al 3+ (aq) + 3 OH - (aq) K sp = [Al 3+ ] [OH - ] 3 b) What is the molar solubility of aluminum hydroxide in a ph = 4.20 buffer solution? [8 points] Since we have a buffer system, we know ph = 4.20 will be constant. So poh = ph = = 9.80 [OH - ] = 10 -poh = = 1.58 x M K sp = [Al 3+ ] [OH - ] 3 and so [Al 3+ ] = K sp = (1.3 x ) = 3.3 x 10-4 M [OH - ] 3 (1.58 x ) 3 Since for every mole of Al 3+ formed a mole of Al(OH) 3 dissociates, the molar solubility is equal to 3.3 x 10-4 M. 7) Balance the following oxidation reduction reaction for acid conditions [10 points] Cu + (aq) + IO 3- (aq) Cu 2+ (aq) + I 2 (aq) oxidation Cu + (aq) Cu 2+ (aq) + e - x 10 reduction 2 IO 3- (aq) + 12 H + (aq)+ 10 e - (aq) I 2 (aq) + 6 H 2 O( ) net 10 Cu + (aq) + 2 IO 3- (aq) + 12 H + (aq) 10 Cu 2+ (aq) + I 2 (aq) + 6 H 2 O( ) 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 problem. Substance H f (kj/mol) G f (kj/mol) S (J/mol K) C(s) CO 2 (g) CO(NH 2 ) 2 (s) H 2 O(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 2 O(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 2 O(g)) ] = [ ( ) + 2 ( ) ] [ ( ) + ( ) ] = kj/mol ln K = - G rxn = - ( J/mol) = 0.55 K = e 0.55 = 1.73 RT (8.314 J/mol K) (298.2 K) 2) For each of the following solutions find the value for ph at T = 25. C. 7

8 a) A M solution of hypochlorous acid (HOCl, K a = 3.5 x 10-8 ). [10 points] HOCl(aq) + H 2 O( ) H 3 O + (aq) + OCl - (aq) K a = [H 3 O + ] [OCl - ] = 3.5 x 10-8 Initial Change Equilibrium HOCl x x H 3 O + 0 x x OCl - 0 x x So (x) (x) = 3.5 x 10-8 Assume x << 0.042, then x 2 = 3.5 x 10-8 ( x) (0.0420) x 2 = (0.0420) (3.5 x 10-8 ) = 1.47 x 10-9 x = (1.47 x 10-9 ) 1/2 = 3.83 x 10-5 The assumption that x was small was good. So [H 3 O + ] = x = 3.83 x 10-5 M ph = - log 10 (3.83 x 10-5 ) = 4.42 b) L of a solution containing moles of nitrous acid (HNO 2, K a = 4.5 x 10-4 ) and moles of sodium hydroxide (NaOH), a strong soluble base. [12 points] Nitrous acid and sodium hydroxide will react by the process HNO 2 (aq) + NaOH(aq) Na + (aq) + NO 2- (aq) + H 2 O( ) and so The reaction involves a strong soluble base and so will go to completion. NaOH is the limiting reactant, initial HNO 2 = moles initial NaOH = moles - initial NO 2 = 0.0 moles final HNO 2 = = moles final NaOH = 0.0 moles - final NO 2 = moles So after the neutralization reaction (since there is L solution) [HNO 2 ] = M [NO 2- ] = M Since we have significant amounts of a weak acid and its conjugate base we can use the Henderson equation (though doing this problem using the ICE table will also work). ph = pk a + log 10 {(base)/(acid)} = - log 10 (4.5 x 10-4 ) + log 10 (0.060/0.140) = = = ) A portion of the electrochemical series is given below and may be of use in doing the following problem. 8

9 2 H + (aq) + 2 e - H 2 (g) E = 0.00 v Co 2+ (aq) + 2 e - Co(s) E = v Pb 2+ (aq) + 2 e - Pb(s) E = v Cd 2+ (aq) + 2 e - Cd(s) E = v Sn 2+ (aq) + 2 e - Sn(s) E = v Ni 2+ (aq) + 2 e - Ni(s) E = v Cr 3+ (aq) + 3 e - Cr(s) E = v Mn 2+ (aq) + 2 e - Mn(s) E = v a) Give the half cell oxidation reaction, the half cell reduction reaction, the net cell reaction, and the value for E cell and E cell for the galvanic cell whose cell diagram is given below. [24 points] Cr(s) Cr 3+ (4.5 x 10-2 mol/l) Sn 2+ (6.0 x 10-4 M) Sn(s) oxidation 2 Cr(s) 2 Cr 3+ (aq) + 6 e - E = v reduction 3 Sn 2+ (aq) + 6 e - 3 Sn(s) E = v net 2 Cr(s) + 3 Sn 2+ (aq) 2 Cr 3+ (aq) + 3 Sn(s) E cell = v From Nernst E cell = E - (RT/nF) lnq n = 6 Q = [Cr 3+ ] 2 = (4.5 x 10-2 ) 2 = 9.4 x 10 6 [Sn 2+ ] 3 (6.0 x 10-4 ) 3 So E cell = 0.60 v [ (8.314 J/mol K)(298.2 K)/6(96485 C/mol) ] ln(9.4 x 10 6 ) = 0.60 v 0.07 v = v b) Manganese metal can be obtained from the electrolysis of a molten solution of manganese II chloride (MnCl 2 ). How many grams of manganese metal can be produced when a sample of molten MnCl 2 undergoes electrolysis for 4.00 hours using a current i = 25.0 amps (1 amp = 1 C/s)? [10 points] The electrolysis reaction is Mn 2+ Mn So moles e - = 25.0 C/s (4.00 hr) (60 min) (60 s) 1 mol e - = mol e - (1 hr) (1 min) C grams Mn = mol e- 1 mol Mn g Mn = g Mn 2 mol e - 1 mol Mn 4) The chemical reaction 9

10 A + B product is an irreversible first order reaction, and follows the rate law rate = k [A] The value for k is 4.3 x 10-4 s -1 at T = 25.0 C, and 3.6 x 10-2 s -1 at T = 60.0 C. a) In an experiment carried out at T = 25.0 C the initial concentration of A is [A] 0 = mol/l. What will the concentration of A be after seconds? [8 points] For a first order reaction [A] t = [A] 0 e -kt And so after 500. s, [A] t = (0.468 mol/l) exp[ - (4.3 x 10-4 s -1 ) (500.0 s) ] = mol/l b) What is the value for t 1/2, the half-life for the reaction, at t = 60.0 C? [8 points] t 1/2 = ln 2 = ln 2 = 19.3 s k (3.6 x 10-2 s -1 ) c) Assuming the reaction obeys the Arrhenius equation, what are the values for E a, the activation energy, and A, the pre-exponential factor, for the reaction? [16 points] ln(k 2 /k 1 ) = - (E a /R) { (1/T 2 ) - (1/T 1 ) } So E a = - R ln(k 2 /k 1 ) = - (8.314 J/mol K) ln(3.6 x 10-2 /4.3 x 10-4 ) = x 10 5 J/mol { (1/T 2 ) - (1/T 1 ) ] [ (1/333.2 K) - (1/298.2 K) ] = kj/mol Once we know E a, we can find A from the rate constant. k = A e -Ea/RT so A = k e Ea/RT Using the value for k at T = 25.0 C (298.2 K) gives A = (4.3 x 10-4 s -1 ) exp [( J/mol)/(8.314 J/mol K)(298.2 K) ] = (4.3 x 10-4 s -1 ) exp(42.15) = 8.7 x s -1 10