Old Midterm Exam 2 - answer key # 9 units corrected

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1. (12 pts) Consider an energy diagram for an overall reaction A + C  D ( 1 ) that involves two elementary steps: a) Which of the rate constants, k 1 , k -1 or k 2 is the smallest and which is the largest? Explain briefly. Smallest: _ k 1 __ Largest: _ k 2 ___ The fastest reaction has the lowest activation energy and the largest rate constant b) What is the overall reaction order in [ C ]? Explain briefly. Order in [C]: _ 0 __ The reaction 1 st step is the slowest (rate-determining). This reaction step is zero order in [C], so is the overall reaction rate c) Which transition state, TS 1 or TS 2 , corresponds to the rate-determining step? Explain briefly. Answer: _ TS 1 __ The reaction 1 st step is rate-determining. The corresponding TS is TS 1 d) Is the overall reaction (1) spontaneous or not (circle one)? Explain briefly. yes no The reaction Gibbs energy,  G rxn , is negative. Hence, it is a spontaneous reaction

2 2. (8 pts) Use the < and/or = signs and rank the following. Assume otherwise identical conditions. a) H 2 O(l), 1M HCl(aq), 1M Na 2 HPO 4 (aq), according to their increasing ability to dissolve poorly water- soluble BaHPO 4 (s): 1M Na 2 HPO 4 (aq) (the common ion effect) < H 2 O(l) < 1M HCl(aq) (reacts with HPO 4 2- to form H 3 PO 4 ) b) 1M NH 3 (aq), H 2 O(l), 1M HCl(aq), according to their increasing ability to dissolve poorly water-soluble AgCl(s): 1M HCl(aq) (the common ion effect) < H 2 O(l) < 1M NH 3 (aq) (forms a soluble coordination compound with Ag + , Ag(NH 3 ) 2 + )

3 3. (12 pts) Consider two reactions below. For each of the reactions decide if the reaction is reactant-favored or product-favored. K a (HSO 4 - ) = 0.012, K a (H 3 PO 4 ) = 7.5×10 -3 , K a (H 2 PO 4 - ) = 6.2×10 -8 ; K a (HPO 4 2- ) = 4.2×10 -13 ; K a (H 2 CO 3 ) = 4.2×10 -7 , K a (HCO 3 - ) = 5.6×10 -11 Explain briefly. Show your work: Reaction 1 : NaHSO 4 ( aq ) + NaH 2 PO 4 ( aq )  Na 2 SO 4 ( aq ) + H 3 PO 4 ( aq ) Circle one: reactant-favored product-favored Reason: K a (HSO 4 - ) > K a (H 3 PO 4 ); acid-base reactions proceed from stronger acids to weaker acids. Reaction 2 : NaHCO 3 ( aq ) + NaH 2 PO 4 ( aq )  H 2 CO 3 ( aq ) + Na 2 HPO 4 ( aq ) Circle one: reactant-favored product-favored Reason: K a (H 2 CO 3 ) > K a (H 2 PO 4 - ); acid-base reactions proceed from stronger acids to weaker acids.

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4 4. (12 pts) Calculate the pH of 1.00 L of solution containing 0.55 mol NH 4 Cl ( pK a = 9.25) and 1.33 mol NH 3 : a) as prepared; b) after addition of 0.20 mol of NaOH( s ) to the solution. Show your work: We deal with an ammonia buffer solution The Henderson-Hasselbach equation: 𝒑𝑯 ൌ 𝒑𝑲 𝒂 ሺ𝒘𝒆𝒂𝒌 𝒂𝒄𝒊𝒅ሻ ൅ 𝒍𝒐𝒈 ቀ ሾ𝒄𝒐𝒏𝒋𝒖𝒈𝒂𝒕𝒆 𝒃𝒂𝒔𝒆ሿ 𝟎 ሾ𝒘𝒆𝒂𝒌 𝒂𝒄𝒊𝒅ሿ 𝟎 ቁ = 9.25 + log(1.33/0.55) = 9.25 + 0.38 = 9.63 a) pH of the solution, as prepared ____ 9.63 ______ NaOH will react with NH 4 + : NaOH(aq) + NH 4 Cl(aq)  NaCl(aq) + NH 3 (aq) + H 2 O(l) Assuming that the volume of the resulting solution will be 1.00L, the resulting solution will have 1.33+0.20=1.53 mol NH 3 and 0.55-0.20=0.35 mol NH 4 Cl The solution pH is 𝒑𝑯 ൌ 𝒑𝑲 𝒂 ሺ𝒘𝒆𝒂𝒌 𝒂𝒄𝒊𝒅ሻ ൅ 𝒍𝒐𝒈 ቀ ሾ𝒄𝒐𝒏𝒋𝒖𝒈𝒂𝒕𝒆 𝒃𝒂𝒔𝒆ሿ 𝟎 ሾ𝒘𝒆𝒂𝒌 𝒂𝒄𝒊𝒅ሿ 𝟎 ቁ = 9.25 + log(1.53/0.35) = 9.25 + 0.64 = 9.89 b) pH after addition of NaOH( s ) ___ 9.89 _____

5 5. (10 pts) Calculate pH of a 0.20 M NaCN( aq ) solution at 25 o C ( K b (CN - ) = 2.5×10 -5 ). Show your work: CN - (aq) is a weak Brønsted base ( K b (CN - ) = 2.5×10 -5 ): CN - (aq) + H 2 O(l)  HCN(aq) + OH - (aq) 𝐾 ௕ ൌ ሾு஼ேሿሾைு ష ሿ ሾ஼ே ష ሿ is small and the reaction equilibrium will lie on the reactants side [HCN] = [OH - ] = x is small, as compared to [CN - ] 0 = 0.20 M. Let’s put together an ICE Table: [CN - ], M [HCN], M [OH - ], M Initial 0.20 0 0.00 Change - x x x Equilibrium 0.20- x x x Next, substitute the equilibrium values into the expression for K b above: 𝐾 ௕ ൌ 𝑥 ଶ 0.20 െ 𝑥 ൌ 2.5 ൈ 10 -ହ Assuming that x << 0.20, we get: 𝐾 ௕ ൌ 𝑥 ଶ 0.20 ൌ 2.5 ൈ 10 -ହ By solving we get x = 0.0022 M It is evident that x << 0.20 M Hence, pOH = -log(0.0022) = 2.65 and pH = 14.00 – 2.65 = 11.35 pH = __ 11.35 __

6 6. (12 pts) Use the < and/or = signs and rank the following. Assume otherwise identical conditions. a) i) 35 Cl( g ) + 35 Cl( g ), ii) 36 Cl( g ) + 36 Cl( g ), iii) 37 Cl( g ) + 37 Cl( g ), according to increasing gas kinetic collision rate: lowest _ (iii) < (ii) < (i) highest The kinetic collision rate constant of particles 1 and 2 : 𝒌 ൌ 𝑷𝑵 𝑨 𝝅ሺ𝒓 𝟏 ൅ 𝒓 𝟐 ሻ 𝟐 𝒗 𝒓𝒆𝒍 തതതതത 𝐞𝐱𝐩 ቀ ି𝑬 𝒂 𝑹𝑻 ቁ ൈ 𝟏𝟎 𝟑 ሺ𝑳/𝒎 𝟑 ሻ ; 35 Cl atoms are the lightest, moving at a higher average relative speed. 37 Cl atoms are the heaviest, moving at a lower average relative speed b) H 2 S, H 2 SO 3 , H 2 SO 4 , according to increasing pK a in water: lowest H 2 SO 4 < H 2 SO 3 < H 2 S highest stronger acids have lower pK a c) Cl - , Br - , F - , according to increasing pK b in water: lowest F - < Cl - < Br - highest stronger bases have lower pK b d) Xe(g), Ar(g), Kr(g), according to increasing standard molar entropy: lowest Ar(g) < Kr(g) < Xe(g) highest standard molar entropy for similar substances increases with molar mass

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7 7. (12 pts) Consider the following reversible reaction: HSO 4 - ( aq ) + H 2 O( l ) SO 4 2- ( aq ) + H 3 O + ( aq )  H rxn > 0 Write the equilibrium constant expression for this reaction: K = ൣௌை ర మష ൧ሾு య ை శ ሿ ሾுௌை ర ష ሿ In which direction would the equilibrium above shift upon disturbing (circle the correct answer)? a) upon dilution with pure water: left no change right b) upon increase of [HSO 4 - ]: left no change right c) upon increase of [SO 4 2- ]: left no change right d) upon increase of temperature: left no change right

8 8. (8 pts) For the following reaction NaH 2 PO 4 ( aq ) + Na 2 S( aq ) Na 2 HPO 4 ( aq ) + NaHS( aq ) a) write a balanced net ionic equation : H 2 PO 4 - ( aq ) + S 2- ( aq ) HPO 4 2- ( aq ) + HS - ( aq ) b) identify two conjugate acid-base pairs involved: One conjugate acid-base pair, acid: H 2 PO 4 - , conjugate base: HPO 4 2- Another conjugate acid-base pair, base: S 2- , conjugate acid: HS -

9 9. (8 pts) Consider the following reaction used for the industrial production of H 2 (g): CH 4 (g) + H 2 O(g)  CO(g) + 3H 2 (g) Calculate the standard entropy change,  S o rxn , in J/K , and the standard enthalpy change for this reaction,  H o rxn , in kJ given the following information: ∆𝐻 ௙ ௢ ሺ𝐶𝐻 ସ , 𝑔ሻ = -74.6 kJ/mol ; 𝑆 ௢ ሺ𝐶𝐻 ସ , 𝑔ሻ = 186.3 J/(K mol) ; ∆𝐻 ௙ ௢ ሺ𝐻 ଶ 𝑂, 𝑔ሻ = -241.8 kJ/mol ; 𝑆 ௢ ሺ𝐻 ଶ 𝑂, 𝑔ሻ = 188.8 J/(K mol) ; ∆𝐻 ௙ ௢ ሺ𝐶𝑂, 𝑔ሻ = -110.5 kJ/mol ; 𝑆 ௢ ሺ𝐶𝑂, 𝑔ሻ = 197.7 J/(K mol) ; 𝑆 ௢ ሺ𝐻 ଶ , 𝑔ሻ = 130.7 J/(K mol) . Show our work: ∆𝑯 𝒓𝒙𝒏 𝒐 ൌ ∑ 𝒏 𝒊 ∆𝑯 𝒇 𝒐 ሺ𝒑𝒓𝒐𝒅𝒖𝒄𝒕𝒔ሻ െ ∑ 𝒏 𝒋 𝑯 𝒇 𝒐 ሺ𝒓𝒆𝒂𝒄𝒕𝒂𝒏𝒕𝒔ሻ = 1 mol×(-110.5) kJ + 3 mol×0 kJ – 1 mol×(-74.6) kJ – 1 mol×(-241.8) kJ = 205.9 kJ Reaction standard enthalpy,  H o rxn : _ + _ __ 205.9 ___ kJ sign magnitude ∆𝑺 𝒓𝒙𝒏 𝒐 ൌ ∑ 𝒏 𝒊 𝑺°ሺ𝒑𝒓𝒐𝒅𝒖𝒄𝒕𝒔ሻ െ ∑ 𝒏 𝒋 𝑺°ሺ𝒓𝒆𝒂𝒄𝒕𝒂𝒏𝒕𝒔ሻ = 1 mol×197.7 J/(K mol) + 3 mol×130.7 J/(K mol) - 1 mol×186.3 J/(K mol) - 1 mol×188.8 J/(K mol) = 214.7 J/ K Reaction standard entropy,  S o rxn : _ + _ __ 214.7 ___ J / K sign magnitude

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10 10. (6 pts) Consider another reaction used for the production of H 2 (g): CO(g) + H 2 O(g)  CO 2 (g) + H 2 (g) . Find the equilibrium constant K C for the reaction at 700.0 o C given the following information: ∆𝐻 ௥௫௡ ௢ = -41.2 kJ ; ∆𝑆 ௥௫௡ ௢ = -42.0 J/K . Show our work: We will consider a reaction producing 1 mol of H 2 : 𝒍𝒏ሺ𝑲ሻ ൌ െ ∆𝑯 𝒐 𝑹𝑻 ൅ ∆𝑺 𝒐 𝑹 ൌ ൅ 𝟒𝟏𝟐𝟎𝟎 𝑱 𝒎𝒐𝒍 𝟖.𝟑𝟏 𝑱 𝑲 𝒎𝒐𝒍 ൈሺ𝟕𝟎𝟎.𝟎ା𝟐𝟕𝟑.𝟐ሻ𝑲 െ 𝟒𝟐.𝟎 𝑱/ሺ𝑲 𝒎𝒐𝒍ሻ 𝟖.𝟑𝟏 𝑱/ሺ𝑲 𝒎𝒐𝒍ሻ = 5.094 – 5.054 = 0.040 K = 𝒆 𝟎.𝟎𝟒𝟎 = 1.0 Reaction equilibrium constant, K C , at 700.0 o C __ 1.0 __

11 Some Physical Constants and conversion factors Gas constant, R = 0.08206 L×atm/(mol×K) = 8.314 J /(mol×K) Important formulas The reduced mass of particles 1 and 2 : 𝝁 𝟏𝟐 ൌ 𝑴 𝟏 𝑴 𝟐 𝑴 𝟏 ା𝑴 𝟐 , kg/mol The average relative speed of particles 1 and 2 : 𝒗 𝒓𝒆𝒍 തതതതത ൌ ට 𝟖𝑹𝑻 𝝅𝝁 𝟏𝟐 , m/s The kinetic collision rate constant of particles 1 and 2 : 𝒌 ൌ 𝑷𝑵 𝑨 𝝅ሺ𝒓 𝟏 ൅ 𝒓 𝟐 ሻ 𝟐 𝒗 𝒓𝒆𝒍 തതതതത 𝐞𝐱𝐩 ቀ ି𝑬 𝒂 𝑹𝑻 ቁ ൈ 𝟏𝟎 𝟑 ሺ𝑳/𝒎 𝟑 ሻ , 𝐿 𝑚𝑜𝑙 ିଵ 𝑠 ିଵ Equilibrium constant K C for reaction a A + b B  c C + d D : 𝑲 𝑪 ൌ ሾ𝑪ሿ @𝒆𝒒𝒖𝒊𝒍𝒊𝒃𝒓𝒊𝒖𝒎 𝒄 ሾ𝑫ሿ @𝒆𝒒𝒖𝒊𝒍𝒊𝒃𝒓𝒊𝒖𝒎 𝒅 ሾ𝑨ሿ @𝒆𝒒𝒖𝒊𝒍𝒊𝒃𝒓𝒊𝒖𝒎 𝒂 ሾ𝑩ሿ @𝒆𝒒𝒖𝒊𝒍𝒊𝒃𝒓𝒊𝒖𝒎 𝒃 ሺ𝑴ሻ 𝒂ା𝒃ି𝒄ି𝒅 Relationship between K C and K P values for reaction a A( g ) + b B( g )  c C( g ) + d D( g ) : 𝑲 𝑪 ൌ ሾ𝑪ሿ 𝒄 ሾ𝑫ሿ 𝒅 ሾ𝑨ሿ 𝒂 ሾ𝑩ሿ 𝒃 ൌ 𝑲 𝑷 ሺ𝑹𝑻ሻ 𝒂ା𝒃ି𝒄ି𝒅 Reaction quotient Q C f or a reversible reaction a A + b B  c C + d D : 𝑸 𝑪 ൌ ሾ𝑪ሿ 𝒄 ሾ𝑫ሿ 𝒅 ሾ𝑨ሿ 𝒂 ሾ𝑩ሿ 𝒃 ሺ𝑴ሻ 𝒂ା𝒃ି𝒄ି𝒅 pK a and pK b scales: pK a = - logK a ; pK b = - logK b ; pK a (HA) + pK b (A - ) = 14.00 in water at 25 o C pH and pOH scales: pH = - log[H + ] ; pOH = - log[OH - ] ; pH + pOH = 14.00 in water at 25 o C The Henderson-Hasselbach equation: 𝒑𝑯 ൌ 𝒑𝑲 𝒂 ሺ𝒘𝒆𝒂𝒌 𝒂𝒄𝒊𝒅ሻ ൅ 𝒍𝒐𝒈 ቀ ሾ𝒄𝒐𝒏𝒋𝒖𝒈𝒂𝒕𝒆 𝒃𝒂𝒔𝒆ሿ 𝟎 ሾ𝒘𝒆𝒂𝒌 𝒂𝒄𝒊𝒅ሿ 𝟎 ቁ Molar solubility of an ionic compound M a X b : ට 𝑲 𝒔𝒑 𝒂 𝒂 𝒃 𝒃 𝒂శ𝒃 Entropy change for a phase transition: ∆𝑺 𝒔𝒚𝒔 ൌ ∆𝑯 𝒔𝒚𝒔 𝑻 𝒔𝒚𝒔 ; ∆𝑺 𝒔𝒖𝒓 ൌ െ ∆𝑯 𝒔𝒚𝒔 𝑻 𝒔𝒖𝒓 The standard Gibbs energy change of a reaction: ∆𝑮 𝒓𝒙𝒏 𝒐 ൌ ∑ 𝒏 𝒊 ∆𝑮 𝒇 𝒐 ሺ𝒑𝒓𝒐𝒅𝒖𝒄𝒕𝒔ሻ െ ∑ 𝒏 𝒋 𝑮 𝒇 𝒐 ሺ𝒓𝒆𝒂𝒄𝒕𝒂𝒏𝒕𝒔ሻ A non-standard Gibbs energy change of a reaction: ∆𝑮 𝒓𝒙𝒏 ൌ ∆𝑮 𝒓𝒙𝒏 𝒐 ൅ 𝑹𝑻𝒍𝒏𝑸 A Gibbs energy change for the system: ∆𝑮 𝒔𝒚𝒔 ൌ ∆𝑯 𝒔𝒚𝒔 െ 𝑻∆𝑺 𝒔𝒚𝒔 A reaction standard Gibbs energy and equilibrium constant:  G o = -RTln ( K ) The van’t Hoff’s equation: 𝒍𝒏ሺ𝑲ሻ ൌ െ ∆𝑯 𝒐 𝑹𝑻 ൅ ∆𝑺 𝒐 𝑹 Standard entropy change of a reaction: ∆𝑺 𝒓𝒙𝒏 𝒐 ൌ ∑ 𝒏 𝒊 𝑺°ሺ𝒑𝒓𝒐𝒅𝒖𝒄𝒕𝒔ሻ െ ∑ 𝒏 𝒋 𝑺°ሺ𝒓𝒆𝒂𝒄𝒕𝒂𝒏𝒕𝒔ሻ

6 Periodic Table of the Elements MAIN-GROUP ELEMENTS MAIN-GROUP ELEMENTS 1 18 1 1 H 1.008 2 13 14 15 16 17 2 He 4.003 2 3 Li 6.941 4 Be 9.012 5 B 10.81 6 C 12.01 7 N 14.01 8 O 16.00 9 F 19.00 10 Ne 20.18 d-BLOCK ELEMENTS 3 11 Na 22.99 12 Mg 24.31 3 4 5 6 7 8 9 10 11 12 13 Al 26.98 14 Si 28.09 15 P 30.98 16 S 32.07 17 Cl 35.45 18 Ar 39.95 4 19 K 39.10 20 Ca 40.08 21 Sc 44.96 22 Ti 47.87 23 V 50.94 24 Cr 52.00 25 Mn 54.94 26 Fe 55.85 27 Co 58.93 28 Ni 58.69 29 Cu 63.55 30 Zn 65.41 31 Ga 69.72 32 Ge 72.61 33 As 74.92 34 Se 78.96 35 Br 79.90 36 Kr 83.80 5 37 Rb 85.47 38 Sr 87.62 39 Y 88.91 40 Zr 91.22 41 Nb 92.91 42 Mo 95.94 43 Tc (97.9) 44 Ru 101.1 45 Rh 102.9 46 Pd 106.4 47 Ag 107.9 48 Cd 112.4 49 In 114.8 50 Sn 118.7 51 Sb 121.8 52 Te 127.6 53 I 126.9 54 Xe 131.3 6 55 Cs 132.9 56 Ba 137.3 57-71 72 Hf 178.5 73 Ta 180.9 74 W 183.8 75 Re 186.2 76 Os 190.2 77 Ir 192.2 78 Pt 195.1 79 Au 197.0 80 Hg 200.6 81 Tl 204.4 82 Pb 207.2 83 Bi 209.0 84 Po ( 209.0 ) 85 At ( 210.0 ) 86 Rn ( 222.0 ) 7 87 Fr ( 223.0 ) 88 Ra ( 226.0 ) 89-103 104 Rf (261.1) 105 Db (262.1) 106 Sg (263.1) 107 Bh (262.1) 108 Hs (265) 109 Mt (266) 110 Ds (271) 111 Rg (272) 112 Cn (285) 113 Nh (284) 114 Fl (289) 115 Mc (288) 116 Lv (292) 117 Ts (294) 118 Og (294) LANTHANIDES AND ACTINIDES 6 Lanthanides 57 La 138.9 58 Ce 140.1 59 Pr 140.9 60 Nd 144.2 61 Pm ( 144.9 ) 62 Sm 150.4 63 Eu 152.0 64 Gd 157.3 65 Tb 158.9 66 Dy 162.5 67 Ho 164.9 68 Er 167.3 69 Tm 168.9 70 Yb 173.0 71 Lu 174.0 7 Actinides 89 Ac ( 227.0 ) 90 Th 232.0 91 Pa 231.0 92 U 238.0 93 Np ( 237.1 ) 94 Pu ( 244.1 ) 95 Am ( 243.1 ) 96 Cm ( 247.1 ) 97 Bk ( 247.1 ) 98 Cf ( 251.1 ) 99 Es ( 252.1 ) 100 Fm ( 257.1 ) 101 Md ( 258.1 ) 102 No (259.1) 103 Lr (260.1)

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Old Midterm Exam 2 - answer key # 9 units corrected

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