Solutions to Thermodynamics Problems

Solutions to Thermodynamics Problems Chem03 Final Booklet Problem 1. Solution: Moles of AgNO3 = 0.050 L x 0.100 M = 0.05 moles AgNO3 = 0.05 moles Moles of HCl = 0.050 L x 0.100 M = 0.05 moles HCl = 0.05 moles Cl (aq) Since the reaction runs to completion, and all species are in a 1:1 ratio, 0.05 moles of AgCl are produced (you can prove this to yourself using an ICE table) Heat lost by reaction = heat gained by solution Heat gain = m s T = 100.0 g H O x 4.18 J 1 (3.40 C.60 C) g C = 330 J = Heat Loss, this is the heat evolved from the formation of 0.050 moles of AgCl Therefore, the heat produced per mol of AgCl = 330 J 1kJ 6.60 kj mol 0.050 mol AgCl 1000 J 1 Problem. Answer: -1370 kj mol-1 Heat capacity of calorimeter = C q=(360 s) (4000 J/s) = 1440 kj q = C T = C (50.3-5.0) rearrange and solve C = 56.9 kjc-1 Q rxn 1 (56.9kJ C )(7.4 5.0C) 1370 kj / mol (0.100mol) Problem 3. Solution: -14. J Absorbing heat is an endothermic process (positive) Work done by the system on the surroundings is negative by convention, therefore E = q + w = 1.0 J - 15. J= -14. J 1 of 36

Problem 4. = (0.8 J J mol C Solution: q = n s mol T ) 39.1 mol (38.0 0.0) C x Pressure-Volume work can be calculated by: w = -PΔV = -1.00 atm x (998 876 L) = -1 L atm x 1kJ 1000J 101.3J Latm = 30.9 kj x 1kJ 1000J Chem03 Final Booklet = -1.4 kj So, the total internal energy, ΔE = q + w = 30.9 kj + ( 1.4 kj) = 18.5 kj Problem 5. Solution: NO (g) + O 3(g) NO (g) + O (g) ΔH = 199 kj 3 O (g) O 3(g) ΔH = ½ ( 47 kj) O (g) ½ O (g) ΔH = ½ (+ 495 kj) NO (g) + O (g) NO (g) ΔH = 33 kj Problem 6. Solution: 1367 kj C H 5 OH (l) C H 4(g) + H O (l) ΔH = +44 kj C H 4(g) + 3 O (g) CO (g) + H O (l) ΔH = 1411 kj C H 5 OH (l) + 3 O (g) 3 H O (l) + CO (g) ΔH = 1367 kj Problem 7. Answer: ΔH = (33.8) (90.9) 0 = 11.94 kj mol of 36

Chem03 Final Booklet Problem 8. Solution: CH 3 CH OH (g) + 3 O (g) CO (g) + 3 H O (g) The energy lost from breaking bonds and energy gained in bond formation must be determined. Bonds Broken Bonds Formed 5 C-H Bonds (413 kj mol-1) 4 C=O bonds (CO) (799 kj mol-1) 1 C-C Bond (347 kj mol-1) 6 O-H bonds (467 kj mol-1) 1 C-O Bond (358 kj mol-1) 1 O-H Bond (467 kj mol-1) 3 O=O Bonds (498 kj mol-1) Total = 4731 kj Total = 5998 kj H = BE(Broken) BE(Formed) H = 4731 kj 5998 kj = -167 kj Problem 9. Answer: a) Water at 105 o C + b) Na (aq) + Cl (aq) c) HO/MeOH mixture d) Reactants e) F (g) 3 of 36

Problem 10. Solution: = S + ½ CH3Cl S H - ( CH4 S reaction coeff p S ( products) coeff p S (reactants) S + ½ S Cl ) Chem03 Final Booklet = (45. J mol-1k-1) + ½(130.6 J mol-1k-1) [(186. J mol-1k-1) + ½ (3 J mol-1k-1)] = +1.8 J mol-1 K-1 H reaction coeff p H ( products) coeff p H (reactants) = H f CH3Cl + ½ H f H - ( H f CH4 + ½ H f Cl = (+90 kj mol-1) + ½(0 kj mol-1) [(-74.85 kj mol-1) + ½(0 kj mol-1)] = +164.8 kj mol-1 ) At 98 K, ΔG o = ΔH o -TΔS o = 164.8 kj (98K)(1.8 J mol-1 K) = 161 kj mol-1 so non-spontaneaous Since both ΔH o and ΔS o are positive, the reaction will be spontaneous at high temperatures. To determine at which temperature the reaction becomes spontaneous, find when ΔG < 0 Solve for ΔG = 0 = ΔH o -TΔS o 16480 o 1 J mol T = o = = 1875 K 1 1 S 1.8J K mol 4 of 36

Problem 11. Calculate ΔH f o (C H 4(g) ) at 5 0C. Solution: Chem03 Final Booklet ΔH 0 = ΔH f 0 (C H 6(g) ) ΔH f 0 (C H 4(g) ) ΔH f 0 (H (g) ) 137.0 = 84.7 ΔH f 0 (C H 4(g) ) 0 ΔH f 0 (C H 4(g) ) = +5.3 kj mol Calculate ΔS 0 for the reaction at 5 0 C. Solution: ΔG 0 = ΔH 0 -TΔS 0 Calculate S 0 (C H 4(g) ) at 5 0 C. Solution: S S S 0 0 0 prod react 0 10.5 9.5 S ( CH4( g)) 130.6 S ( C H ( g)) 19.4JK 0 1 4 101.1 = 137.0 98 ΔS 0 ΔS 0 = 10.5J K Give a brief rationalization in physical terms for the algebraic sign of ΔS 0 calculated in part (ii) above. Solution: The product side contains fewer gas molecules than the reactant side more ordered. ΔS 0 is negative 5 of 36

Problem 1. Answer: ΔG 0 =-69.9kJ, ΔS 0 =-94.9kJ, ΔH 0 =98.4kJ, ΔE 0 =-97.1kJ Solution: G G ( SO ) G ( SO ) 370.3 300.4 69.9kJ 0 0 0 f 3( g) f ( g) 1 S S ( SO ) S ( SO ) S ( O ) 56.1 48.5 05.0 94.9JK 0 0 0 G H TS 0 1 3( g) ( g) ( g) 94.9 300K 1000 0 0 0 H G T S 69.9kJ 98.4kJ 0 0 0 H E P V E ngrt where ng 8.314 1 1000 0 0 E H ngrt 98.4kJ 300kJ 97.1kJ 1 Chem03 Final Booklet Problem 13. Solution: (a) G reaction coeff p G coeff f ( products) pg f (reactants) = (51 kj mol-1) [(½(0 kj mol-1) + 86.60 kj mol-1) = -35.6 kj mol-1 G RT ln K eq 1 G 35.6kJmol ln Keq 1 1 RT (8.314J mol K )(98K) ln K 14.4 K eq 1.8 6 eq 10 14 (b) Same method. Answer: K 4.5 eq 10 6 of 36

Problem 14. Step 1: Solve for Keq at 73 K: 1 1 ln K H K1 R T T1 K K e 73K 98K H 1 1 R 73K 98K K73 7.0 10 K K73 1.1 10 K Chem03 Final Booklet 1 ( 198.4 kjmol ) 1 1 4 1 8.314 JKmol 73K 98K 8 e Step : Solve for G o 73K : G G o 73K o 73K RT ln K eq = -(8.314 J mol-1 K-1) (98 K) ln(1.1 x 108) = -146 kj mol-1 Step 3: Solve for G 73K with given concentrations Q = [ SO3] SO O = [ ] [ ] (0.100) (0.500) (0.0100) = 4.0 G G RT Q o o 73K 73K ln G kj mol J mol K K G o 1 1 1 73K 146 (8.314 )(98 )ln 4.0 o 73K = -143 kj mol-1 7 of 36

Problem 15. (a) Calculate ΔH for the above reaction at 5 C. Answer: ΔH = 467.9kJ H 0 n p H 0 f ( prod) n r H 0 f (react) = (4x0+3(-393.5))-((-84.)+3(0)) = +467.9kJ (b) Calculate ΔS for the above reaction at 5 C. Answer: ΔS = 558.4 JK-1 0 0 ΔS =np S ( prod ) nrs ( react ) = (4(7.3)+3(13.7))-((87.4)+3(5.7)) = 558.4 JK-1 (c) Account for the algebraic sign of ΔS for the above reaction. Chem03 Final Booklet Answer: Entropy (disorder) increases (ie. ΔS is positive) because 5 moles of solid are connected to 4 moles of solid and 3 moles of gas. (d) Calculate ΔG for the above reaction at 5 C. Answer: 301.5 kj Solution: ΔG = ΔH - TΔS =467.9-98x558.4/1000=301.5kJ (e) Calculate the equilibrium constant, K, for the above reaction at 5 C. Answer: K=1.4x10-53 G ln K RT K 1.4 10 0 53 301,500 11.7 8.314 98 (f) Determine the temperature above which this reaction would become spontaneous under standard state conditions. Answer: 838K Solution: ie. We require ΔG 0 H ie ΔG =0 when T= S 0 0 467,900J 1 558.4JK 838K 8 of 36

Solutions to Equilibrium Practice Problems Chem03 Final Booklet Problem 16. Solution: PO 4 10 eq The expression for K 3 5 P O 4 eq eq PO 4 10 init 1 M In (a) Q 1 3, the reaction proceeds to the right. 5 5 P O 1 M 1 M 4 init init PO 4 10 init 5.0 M In (b) Q 3.7 3, the reaction proceeds to the left. 5 5 P O 8.0 M 0.70 M 4 init init 9 of 36

Problem 17. Solution: Chem03 Final Booklet To determine the final concentrations, the first thing needed are the initial reactant concentrations and an expression for the reaction coefficient Q. [A (g) ] = and Q 1.00mol 0.50 L ABinit A B = 4.00 M, [AB (g) ] = [B (g) ] = (8.00).0 (4.00) (8.00) init init.00 mol 0.50 L = 8.00 M, The direction of the reaction needs to be determined. To do this, we compare Q vs K. Since Q =.0 > K = 0.5, the reaction proceeds to the left. To determine what final concentrations will be from initial concentrations, a handy tool the Initial, Change, Equilibrium (ICE) table can be used. All compounds involved in the reaction are included in an ICE table as follows, with the species that will be consumed on the left side, and the species that will be produced on the right side. Since the reaction proceeds to the left, A and B will be formed and AB will be consumed. Concentration (M) AB (g) A (g) B (g) Initial 8.00 M 4.00 M 8.00 M Change -x +x +x Equilibrium 8.00 x 4.00 + x 8.00 + x As the reaction proceeds, x moles of A (g) and B (g) are formed as x moles of AB (g) are consumed. Make sure to consider stoichiometric coefficients appropriately. The equilibrium values are simply the sum of the initial + change concentrations. Substitute the equilibrium concentrations into K, and solve for x (remember that K is defined for the reaction in the way that it was initially described): K ABeq A B (8.00 x) 0.5 (4.00 x) (8.00 x) eq eq so, 10 of 36

(8.00 x) 0.5(4.00 x)(8.00 x) 64 3x 4x 0.5(3 1 x x ) 64 3x 4x 16 6x 0.5x 3.5x 38x 48 0 Chem03 Final Booklet To solve for x, the quadratic formula must be used, 38 ( 38) 4(3.5)(48) b b 4ac x 1.46 or 9.39, a (3.5) Since x 9.39 = 18.78 > 8.00 (which would leave the equilibrium concentration of AB at equilibrium negative), then the x = 1.46 So, final concentrations are: [A] = 4.00 + 1.46 = 5.46 M [B] = 8.00 + 1.46 = 9.46 M [AB] = 8.00 (1.46) = 5.08 M To check, substitute these concentrations into the Equilibrium constant expression, ABeq A B (5.08) K 0.5, matches up. (5.46) (9.46) eq eq 11 of 36

Problem 18. Answer: K c 4 [ NO] [ H ] [ NO ] 4 4 3 Chem03 Final Booklet Solution: Pure solids and pure liquids and solvents are not included in the expression. Products go over Reactants, and coefficients in the equation are written as superscripts. Problem 19. Solution: K c = [CO (g) ] Remember: Solids and liquids are not included in the expression. Problem 0. Solution: Equation is equal the double and reverse of equation1 therefore K p = K p1 = ( 1 0.157 ) = 40.6 Problem 1. Answer: Equilibrium Constant = K 1 1/ = 1 Solution: The second equation if reversed (1/K) and halved (K1/). Combining these two gives 1/K1/. K 1 Problem. Answer: The reaction equation is: The relationship between K c and K p is: CO (g) + O (g) CO (g) K c = 3.3 10 91 K p = K c (RT) n gas In this case there are 3 moles of gas in the reactants and moles of gas in the products, so n = -1 So solving for K p : K p = K c (RT) n gas = (3.3 10 91 ) [(0.081 L atm mol K ) (98 K)] 1 = 1.35 10 90 1 of 36

Problem 3. Solution: Chem03 Final Booklet N (g) + C H (g) HCN (g) i: 1.00 1.00 1.00 c: +x +x -x e: 1.00+ x 1.00 + x 1.00-x Q=1>K so then rxn goes to the left (1.00 x) (1.00 x) K c (1.00 x) (1.00 x) K c 1 x K K c c 0.488 Problem 4. Solution: COCl (g) CO (g) +Cl (g) initial: 0.04 0 0 change: -x +x +x equil: 0.04-x x x x Kc = 0.04 x x + Kcx 0.04Kc = 0 Kc Kc (4)(0.04)( Kc) x 5.9 10 3 13 of 36

Problem 5. Solution: Kc [ CO ][ H ] [ CO][ H O] 5.10 Chem03 Final Booklet CO + H O CO +H I 0.1M 0.1 0.1 0.1 C -x -x +x +x E 0.1-x 0.1-x 0.1+x 0.1+x (0.1 x) 0.01 0.x x Kc 5.10 (0.1 x) 0.01 0.x x 0.01 0.x x 0.0511.0x x 0 0.041 1.x 1.x 0.041 x 0.034M Therefore, at equilibrium [H] = 0.1 + x = 0.1 + 0.034M = 0.134M Problem 6. Solution: Only C is true An increase in volume will shift the equilibrium to the right thus causing an increase in the total moles of CO at equilibrium. Problem 7. Solution: a) shift to the left b) no effect c) shift to the right d) shift to the right Problem 8. Solution: The reaction will shift left forming Ni(CO) 4(g) to reach equilibrium Problem 9. Answer: C Solution: Q = [NO] [Cl ] [NOCl] = (1.) (0.56) (1.3) = 0.51. Q=K, therefore we are already at equilibrium. 14 of 36

Chem03 Final Booklet Solutions to Acids and Bases Practice Problems Problem 30. Solution: Equal volumes of 0.1 M NaF and 0.1 M HF Adding an acid and the salt of its conjugate base can form a buffer, this is the case in (b). Problem 31. Solution: 1 and 3. A buffer normally consists of a weak acid and its conjugate base in roughly equal amounts. The ratio should be no greater than 0.1 to 10 of weak base to weak acid. In 1, the ratio of acid to base is 3:1. In 3 the ratio is 4:1. In, there is a greater amount of strong base than acid, so all of the acetic acid is consumed so it is not a buffer. 4 is not a buffer solution since all of the weak acid is consumed with strong base, 5 is not a buffer because sodium acetate is a weak base, so you have a weak base with a strong base in solution. Problem 3. Solution: B This is a buffer solution, therefore: [ A ] 0.400 ph pka [ HA] eq 0.600 eq log log(1.0 10 ) log 1.74 Problem 33. Solution: This is a buffer solution, therefore: [ A ] eq ph pka log [ HA ] eq [ CH COONa ] 0.100 =4.75 3 = 4.75 log [ CH3COONa ] Therefore log 0.100 =0 and [ CH3COONa ] 0.100 =1 [CH3COONa] = 0.10M # moles CH3COONa = 0.10 moles Problem 34. Solution: For this equilibrium [H+] = [In-] = 10-8 = 1 x 10-8 [ H ][ In ] Ka = 1 x 10-6 = = [ HIn] -8 (1 x 10 ) [ HIn] [HIn] = 1 x 10-10, therefore [HIn]/[In- ] = 1 x 10-10/1 x 10-8 =0.01 = 1/100 15 of 36

Problem 35. Solution: ph = 8.91 Chem03 Final Booklet ph = pka + = 9.0 + = 8.91 CN log HCN 1.0 log.0 Problem 36. Solution: C This is a buffer system as there are equal amounts of conjugate acid and base, but the addition of H+ can change the ph slightly. The original ph of the buffer is approximately equal to pka. When H+ ions are added from the strong acid HCl, A- is converted into HA. Therefore the addition of 0.01 moles H+ produces 0.01 moles HA and consumes 0.01 moles A-. nha= 0.5 + 0.01 = 0.51 mol na- = 0.5 0.01 = 0.49 mol [H+]=Ka x nha/na- = (1.8 x 10-4) (0.51/0.49) = 1.873 x 10-4 ph = -log (1.873 x 10-4) = 3.77 Problem 37. Solution: C There are more moles of carbonate buffer in solution c than any of the other solutions Problem 38. Solution: E The solution is a buffer with ph above 7. A buffer is resistant to both addition of strong acid and strong base and the concentration of the hydronium ion is not more than the hydroxide ion (ph >7). Problem 39. Solution: E HCN + H O CN + H 3 O + I 0.5 y 0 C 1 10 7 +1 10 7 +1 10 7 E 0.5 y + 1 10 7 +1 10 7 Ka = 6. x 10-10 = [(1x10-7)( Y+1x10-7)]/ 0.5 Y=0.003M x 1L = 0.003moles Mass = 49.0075g/molNaCN x 0.003 moles = 0.15g 16 of 36

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Chem03 Final Booklet Problem 40. Solution: E A buffer normally consists of a weak acid and its conjugate base in roughly equal amounts. The ratio should be no greater than 0.1 to 10 of weak base to weak acid. In 1, the ratio of acid to base is 3:1. In 3 the ratio is 4:1. In, there is a greater amount of strong base than acid. Problem 41. Solution: C or E In order to determine which solutions are able to act as buffer solutions, determine what ions will be found in the solution and whether those ions are acidic, basic or spectator ions. If acidic and/or basic ions are found, calculate the amount of ions that are present. For solution A) all ions present (H+, NO3-, and Na+) are spectator ions. No buffer abilities possible. For solution B) all ions present (Na+, OH- and Cl-) are spectator ions. No buffer abilities possible. For solution C) This is a 1:1 ratio of conjugate acid to its conjugate base. This IS a buffer. For solution D) the HCl completely dissociates to form H+ and Cl- ions. NH3 can form an equilibrium where NH3 + H+ NH4+. However, HCl and its dissociated ions are present in 0.01mol amounts the same as the amount of NH3. Therefore, all the NH3 is used up in reacting with the HCl, so no buffering abilities are possible. For solution E) the NaOH completely dissociates to form Na+ and OH- ions. The 0.004mol of OH- will react with the 0.01mol of HF to form 0.004mol of F- and have 0.006mol left of HF since the OH- is the limiting reagent. This means we have HA and A- both present in a ratio of 3:. This is a buffer. Problem 4. Solution: C B + HO BH+ + OH- Calculate from the given ph, the concentration of OH- ions that dissociate form from the reaction of the base with water. poh = 14 ph = 14 (8.88) = 5.1 [OH-] = 10-pOH = 10-5.1 = 7.585 x 10-6M Assume 1 L of solution, therefore the [B] = 0.40mol/L and [BH+] = 0.50mol/L. K = [BH +][OH ] [B] (0.50)( 7.585 x 10 6) = = 4.74 x 10 6 0.40 18 of 36

Problem 43. Solution: D Chem03 Final Booklet Because equal volumes of the acid and weak base are being mixed, all concentrations (M) can be treated as moles (mol). HNO3 is a strong acid so it completely dissociates HNO3 H+ + NO3-0.1 0.1 0.1 NH3 will react with the H+ released by the HNO3 to form NH4+. Initially there is 0.3M or 0.3moles of NH3. Upon addition of 0.1moles H+ (from the HNO3), 0.1mol of NH3 will react to form 0.1mol of NH4+, leaving 0.mol NH3 unreacted. Therefore, NH3 + H+ NH4+ Using the following equation, solve for [H+]. [H+] = Ka x [HA] [A-] Kw = Ka x Kb so that Ka = Kw = 1.0 x 10-14 = 5.56 x 10-10 Kb 1.8 x 10-5 [H+] = 5.56 x 10-10 x [0.1] =.78 x 10-10M [0.] ph = -log[h+] = -log[.78 x 10-10] = 9.56 19 of 36

Problem 44. Solution: B Chem03 Final Booklet Write out the equations that are occurring in the solution described above. Because both solutes are being added to 1L of water, all concentrations (M) can be treated as moles (mol). CH5COONa is a soluble salt so it completely dissociates CH5COONa CH5COO- + Na+ 0.1 0.1 0.1 HCl is a strong acid so it completely dissociates HCl H+ + Cl- 0.01 0.01 0.01 CH5COO- will react will all the H+ to form CH5COOH. Initially there is 0.1 mole of CH5COO-. When 0.01mol of H+ is added, the CH5COO- reacts leaving 0.09mol. There is also 0.1mol of CH5COOH to start, but when the CH5COO- reacts with the H+, it forms 0.01mol more CH5COOH so that the total amount of CH5COOH is 0.11mol. CH5COOH CH5COO- + H+ Before HCl is added 0.1 0.1 After HCl is added 0.11 0.09 Calculate [H+] using the following equation. (n = moles) [H+] = Ka x nha = 1.41 x 10-5 x (0.11) = 1.7 x 10-5M na- (0.09) Calculate ph from the [H+]. ph = -log[h+] = -log[1.7 x 10-5] = 4.76 0 of 36

Chem03 Final Booklet Solutions to Electrochemistry Practice Problems Problem 45. Solution: List the oxidation and reduction steps: Reduction: Oxidation: Br (g) + e Br (aq) + Ag (aq) + e Ag (s) E anode E cathode = 1.080 V = 0.799 V E cell = E cathode - E anode = 1.080 V 0.799 V = 0.81 V E 0.057V ln n cell K eq or E 0.059V log n cell K eq log K log K eq eq K eq necell 0.059V (0.81V ) 9.49 0.059V 3.110 9 Problem 46. Solution: The half-cell reactions are as follows (note: these are not at standard state or E cell = 0 V): Reduction: Oxidation: Pb + (aq) + e Pb (s) E cathode =? Pb + (aq) + e Pb (s) E anode =? E cell E cathode = E anode = = Ecathode - Eanode 0.059V 1 0.15 log = - 0.1546 V 0.100M 0.059V 1.15 log [ Pb 0 0.059V 1 E cell 0.0700 V = 0.1546 0.15 log [ Pb ] ] [Pb + (aq) ] = 4.3 10 4 + Solubility PbSO 4 Pb (aq) SO 4 Initial Some 0 0 Change -s +s +s Equilibrium Some-s S S K sp = [Pb + (aq) ][SO 4 ] K sp = (4.3 10 4 ) = 1.9 10 7 1 of 36

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Problem 47. Answers: Cr = +3, O = - Ca = +, C = +4, H = +1, O = - Fe = +3, C = +4, O = - Chem03 Final Booklet Problem 48. a) Answer: 0.15 V Pb + + e Pb E o = 0.13 V Co Co + + e E o = +0.8 V o E cell = 0.8 V 0.13V = 0.15 V (b) Answer: B Reduction occurs at the cathode; therefore, the lead electrode is the cathode. Problem 49. Answer: MnO 4 (aq) > Zn (s) > I (aq) > I (aq) > Zn + (aq) > MnO (aq) An oxidizing agent gets reduced therefore MnO 4 (aq) is the strongest oxidizing agent as it has the largest E o. Problem 50. Solution: Br (aq) will be reduced and I (aq) Br (aq) + e Br (aq) I (aq) E = 1.09 V I (aq) + e E = 0.54 V will be oxidized. 3 of 36

Problem 51. What is the value of E cell? Chem03 Final Booklet Answer: 0.55 V Br (aq) + e Br (aq) I (aq) Br (aq) + I (aq) Then put the Ecell into the Nerst: E = 1.09 V I (aq) + e E = 0.54 V Br (aq) + I (aq) E cell = 0. 55V E cell = 0.55 ( 0.059 ) (log ( [0.] [0.1] [0.][10] ) = 0.66V Problem 5. Answer: D>B>A>C Solution: Reducing agent undergoes oxidation (most easily oxidized = strongest reducing agent) From A + + B A + B + B can oxidize to B+, B is a stronger reducing agent than A. From A + + C no reaction C cannot oxidize to C+ A is a stronger reducing agent than C. From B + + D B + D + D can oxidize to D + D is a stronger reducing agent than B. The decreasing order of reactivity (most easily oxidized to least easily oxidized) is: D>B>A>C, where D is the strongest reducing agent. 4 of 36

Problem 53. Answer: B + Given: Hg (aq) + Br (aq) To Find: anode reaction Hg Br (s) ANODE allows oxidation to take place, which is the loss of electrons. - Answers A, C and E are incorrect as they are reduction reactions. - Answer D is incorrect as the charges on both sides are NOT balanced. Hg (l) acts as an intermediate ANODE Hg (l) + Br (aq) + CATHODE: Hg (aq) Chem03 Final Booklet Hg Br (s) + e + e Hg (l) Problem 54. Answer: Ag + only Cu Cu + + e E = 0.34V So addition of this to the cathode reaction should be a positive number for the reaction to proceed. The only one that this works for is Ag + Problem 55. Answer: 0.5 V 3 (I (s) + 6 H O (l) HIO 3(aq) + 10 H + + 10 e ) E = 1.0V 5 (ClO 3(aq) 3 I (s) + 5 ClO 3(aq) + 6 H + + 6 e Cl aq + 3 H O (l) ) E = +1.45V + 3 H O (l) 6 HIO 3(aq) + 5 Cl (aq) E = 0.5V Problem 56. Answer: 0.6 V In the reaction above, Sn + (aq) is being oxidized to Sn 4+ (aq) (it lost e ) and Fe 3+ (aq) is being reduced to Fe + (aq) (gaining 1 e ). E cell = E reduction + E oxidation. Since all E values are given as E reduction values, the E red value for Fe in the reaction is +0.77 V. The E value for Sn in the reaction must be reversed because Sn is undergoing oxidation, therefore E ox = 0.15V. E cell = E reduction + E oxidation = (+0.77V) + ( 0.15V) = 0.6V 5 of 36

Problem 57. Answer: A only Chem03 Final Booklet Calculate the E cell for all three reactions. In statement I), E cell = -1.86V (Cd + (aq) is being reduced and Cl (g) is being oxidized). In statement II), E cell = -0.9V (Sn + (aq) is being oxidized and reduced into Sn 4+ (aq) and Sn respectively). In statement II) E cell = 0.91V (Sn is being oxidized and Fe3+ is being reduced). E cell values that are positive means that the redox reactions will occur spontaneously, while a negative E cell value means the redox reaction is not spontaneous. Problem 58. Answer: I, II, and III Solution: Statement I) is true becausecu + has a more positive E reduction value than Cr 3+, which means that Cu + is a better oxidizing agent than Cr 3+ (remember that an oxidizing agent oxidizes other substances and becomes reduced in the process). Problem 59. Answer: -0.6 + In the reaction given above, Ag (aq) o overall E cell for the reaction is calculated as is being reduced to Ag while Ni is being oxidized to Ni + (aq). The o E cell = E reduction + E oxidation The reduction half of the reaction is given as + 0.80V. In order to calculate the E value for the oxidation half, rearrange the E o cell equation to solve for E oxidation. o E oxidation = E cell E reduction = (+1.06V) (+0.80V) = +0.6V. The E oxidation value is the opposite sign from the E reduction value, so in order to solve for the E reduction n value as asked, the sign on the E oxidation value must be reversed. Therefore, E reduction for Ni + (aq) = 0.6V 6 of 36

Chem03 Final Booklet Problem 60. Answer: n = 4 H + O H O can be broken down into its half-cell reactions: H 4 H + + 4 e O + 4 H + + 4 e H O Since there is an exchange of four electrons, n = 4 in the Nernst Equation. Problem 61. Answer: 0.1 V Solution: E = E o 0.057V n lnq = 0.15V 0.057V [ Co ln [ Pb ] = 0.1V ] 7 of 36

Chem03 Final Booklet Problem 6. (a) When current is allowed to flow, which species is oxidized? Solution: MnO 4 + 8 H + + 5 e Mn + + 4 H O E o = 1.51 V Cr 3+ + 7 H O Cr O 7 + 14 H + + 5 e E o = 1.33 V Oxidation is a loss of electrons. Cr 3+ is being oxidized to Cr O 7 (b) When current is allowed to flow, which species is reduced? Solution: MnO 4 is being reduced to Mn + (c) What is the value of E cell? Solution: MnO 4 + 8 H + + 5 e Mn + + 4 H O E o = 1.51 V Cr 3+ + 7 H O Cr O 7 + 14 H + + 5 e E o = 1.33 V MnO 4 + Cr 3+ + 3 H O Cr O 7 + 6 H + + Mn + E o = 0.18V (d) What is the oxidation state of Cr in Cr O 7? Solution: oxid. # = [7(-) + ]/ = +6 8 of 36

Chem03 Final Booklet Solutions to Kinetics Practice Problems Problem 63. Solution: v k[ I ] [ S O ] v k I S O m n 1 1 8 1 m n [ ] [ 8 ] 5 1.510 m n k(0.080) (0.040) 6 6.510 m n k(0.040) (0.040) v v 1 3 m m 1 5 1.510 m n k(0.080) (0.040) 6 6.510 m n k(0.080) (0.00) n n 1 rate k[ I ] [ S O ] k rate [ I ][ S O 8 1 1 8 ] 5 1.510 k1 3.9110 M s (0.080) (0.040) 3 1 1 Problem 64. Solution: Rate = k[h O] x [CH 3 Cl] y x Rate1 k(0.0100) 3.6 10 x Rate k(0.000) 1.44 10 4 4 0.5x =.5, therefore x = and the reaction is second-order in H O x Rate1 k(0.0100) 3.6 10 x Rate k(0.000) 1.44 10 4 4 5.54 ( y 3 ) = 3.69 ( 3 )y =, therefore y = 1 and the reaction is first-order in CH 3 3Cl Rate = k[h O] [CH 3 Cl] 1 9 of 36

Problem 65. Solution [ A] 0 ln[ A] t ln[ A] kt, ln kt 1/ [ A ] t1/ [ A] 0 for t1/ [ A] t 1/ [ A] 0 [ A] 0 ln ln ln kt1/ [ A] t [ A] 1/ 0 ln t1/ k 0 1/ 1/ Chem03 Final Booklet Problem 66. Solution: If 0% decomposes, then 80% of the sample remains. ln[ A] ln[ A] kt rearrange 0 ln 0.8[A] o k(50s) [A] o k 4.46310 3 sec 1 t 1/ ln k 155 sec Problem 67. Solution: (a) v v k[ CH NNCH ] n 1 3 3 1 n k[ CH3NNCH 3] 6.810 n (5.1310 ) 5 1.110 1 n (.0510 ) n 0.5 0.5 n 1 Therefore, the reaction is the first order. Rate = k [CH 3 NNCH 3(g) ], therefore k = (b) ln[a] t = ln[a] 0 kt 6.810 M / 5.1310 s = 5.46 x 10-5 s-1 M ln[0.1 5.1310 ] = ln[5.1310 ] (5.4610 5 ) t -5.7 = -.97 - (5.4610-5)t (5.4610-5)t =.30 t = 4.1104 s = 11.7 hours (c) [A]10% = 0.15.1310 = 5.110-3M Rate = k[a] = (5.4610-5 s-1)( 5.110-3 M) =.8010-7 M/s 30 of 36

Chem03 Final Booklet 1 1 Problem 68. Solution: Because this reaction is second order kt if 1/[A] was plotted vs. t [ A] [ A] o then the y-intercept would be 1/[A]0 and the slope would be k 1/[A] slope = +k 1/[A] Problem 69. Answer t Problem 70. Solution: Since Hrxn > 0, then Hproducts > Hreactants 31 of 36

Thus, Ea rev = 66 41 = 5 kj/mol (draw an energy diagram to see this better) Chem03 Final Booklet Problem 71. Solution: Let T1 = 96 C = 369 K, and T = 5 C = 98 K. Using the Arrhenius Equation ln ( k 1 k ) = E a R ( 1 T 1 T 1 ) 10 1 3.55 x 10 s E 1 1 a 10 1 1 1 ln 1. x 10 s 8.314 JK mol 369 K 98 K and solve for Ea, Ea = 1.8 kj/mol Problem 7. Solution: from Arrhenius equation since vk, then ln k 1 E 1 R T T k a 1 ln k ln v E a 1 1 k 1 v 1 R T T 1 ln(3) E a 1 8.314 35 1 E a 77433.3 J / mol 77.4 kj / mol 313 1 Problem 73. Solution: Since we have an elementary process, then rate of the reaction is 1 1 This reaction is second order therefore kt [ A] [ A] 1 0.1M 1 0.M 5 M-1 = k(35. min) k(35. min) k = 1.4 x 10-1 M-1min-1 o rate k[a], 3 of 36

Chem03 Final Booklet Problem 74. The steps are elementary, and the first step is the rate determining step. So the overall rate law only depends on the first step: rate = k[o3][no] 33 of 36

Chem03 Final Booklet Solutions to Gases Practice Problems Problem 75. Solution PV PV nrt nr T PV 1 1 PV PV T T 73 7.00 C K 110K T T PV atm ml.00 atm 100mL 1 1.00 600 1 1 1 T 110 K 73 847 C Problem 76. Answer: C Increasing the temperature increases the kinetic energy of the molecules in the liquid causing the molecules to move more quickly and increase the vapor pressure. Problem 77. Solution 775 mmhg 1atm 760 mmhg ; Patm 775mmHg Patm 1 atm 1.0 atm 760 mmhg 73 310 583 T K C K P RT dp Mr x P Mr x P d x P x RT P 1 1 L atm mol K K 0.0806 583 MrP.64 g / L 13.8 g / mol x 1.0 atm MrP 13.8 g / mol x x 3.99 4 P4 Mr 31.0 g / mol P Problem 78. Solution 735 mmhg 1atm 760 torr; P atm 735torr P atm 1 atm 0.967 atm 760 mmhg 73 110 383 T K C K x 4.04 4 1 1 L atm mol K K 4 8 RT 0.0806 383 M m 0.970 g 56.5 g / mol CH CH x x PV 0.967 atm 0.559 L x Mr Mr CH CH 56.5 g / mol 14.0 g / mol C H 34 of 36

Chem03 Final Booklet Problem 79. Solutions C H O 6O 6CO 6H O 6 1 6 RT PV nrt VO n O P no 6 n 1 C H O 6 1 6 no 6 n 6 C6H1O 6 6 1 6 6 1 6 g 1 1 atm L K mol K m 1.50 0.0806 73 C H O RT VO Mr P 180.0 g / mol 1.00 atm 6 1 6 6 6 1.1 C H O 6 1 6 m Mr C H O C H O L Problem 80. Solution: TK 73 30.0C 303.0 K M X m X RT PV 0.5 g 8.3145 J mol1 K 1 303.0 K 101.3 kpa 70.4 10 3 L 79.5g/mol Problem 81. Answer: C 760mmHg Solution: 0.5atmO = 190 mmhg O 1atm Total pressure = 190mmHg O + 440mmHg N = 630mmHg Problem 8. Answer: B Solution: PV = nrt (1.00atm)(.4L) = n(0.081atml/molk)(303k) n = 0.90 moles helium gas Problem 83. Answer: B 10g no 0.315 3 Solution: nrt 0.315 8.314 373 P 3.3kPa V 30 35 of 36

Problem 84. Answer: A Chem03 Final Booklet Solution: To determine which gas has the highest density, consider the equation g molar mass ( Density = mol ) pressure(kpa) R(gas constant) temperature(k) If measured at the same pressure and temperature, the gas that will have the highest density will have the largest molar mass. The molar masses of the gases are as follows: F, 38g/mol; CH6, 30.08g/mol; HS, 34.086g/mol; NO, 30.01g/mol and SiH4, 3.13g/mol. Therefore, F will have the highest density because it has the largest molar mass. Problem 85. Answer: D PV 1 1 PV Solution: PV = nrt T T 1 (1.00 atm)(600 ml) (.00 atm)(100 ml) 80K T T = 847C Problem 86. Solution: We need to determine what mass is in the gas phase, and the rest must be in the liquid phase. We make the assumption that the liquid takes up a negligible volume of the container. The pressure of the water vapour will be 33.7mmHg P=1 atm / 760mmHg * 33.7 mmhg = 0.3075 atm Now calculate the moles of gas: n = PV RT = 0.3075 atm 0.5 L = 0.0055 mole 0.0806 L atm mol 1 K 1 (73.15 + 70)K mass of gaseous water is: 18 g mol 0.0055 mol = 0.098 g So the mass of liquid water is 0.5-0.098 = 0.40 g. 36 of 36