Part A.Answer all eight questions(5 marks each). 1.Why is Ka2<Ka for a diprotic acid such as H2SO? In the second hydrolysis,a proton is being removed from a doubly charged SO4 ion,which requires more energy than the first hydrolysis,where a proton is being removed from only a singly charged HSOion. 2.A reaction has AH<0and AS0.Is this reaction spontaneous at high temperature,low temperature,both or neither? △G=△H°_T△S.Since△S<0,this reaction will be spontaneous at low temperature.At higher temperature, the-T△S°term will become sufficiently positive that△G°will become positive. 3.A solution is made by dissolving I mole of KClis)in I mole H2O.What is the mole fraction of H2O in this solution? The KCls dissociates into two ions.Thus,the water will be only 1/3 of the particles in solution,and thus its mole fraction will be 0.333. 5.A galvanic cell has the shorthand notation Pb(s)Pb)1 MI Pb)2 M Pbs).Is the cell potential positive or negative? The oxidation reaction is Pb(s)>Pb(ag)(1 M)+2 e.The reduction is Pb()(2 M)+2 e Pb(s).The overall reaction is therefore Pb(2 M)Pb(1 M).For this reaction,Q==0.5.According to the Nernst cquation,EER().Since Q1 In(Q)and so E>E,But since 0 nF 7.Melting ice is an endothermic process.Why is it spontaneous? Because the process has a positive entropy change. 8.For the reaction H2O=HO initially at equilibrium,which direction will the equilibrium shift (left or right)if the pressure is increased? 1
Part A. Answer all eight questions (5 marks each). 1. Why is Ka2 < Ka1 for a diprotic acid such as H2SO4(aq)? In the second hydrolysis, a proton is being removed from a doubly charged SO4 -2 ion, which requires more energy than the first hydrolysis, where a proton is being removed from only a singly charged HSO4 - ion. 2. A reaction has ΔH0 <0 and ΔS0 <0. Is this reaction spontaneous at high temperature, low temperature, both or neither? ΔGo = ΔHo –TΔSo . Since ΔS < 0, this reaction will be spontaneous at low temperature. At higher temperature, the –TΔSo term will become sufficiently positive that ΔGo will become positive. 3. A solution is made by dissolving 1 mole of KCl(s) in 1 mole H2O(l). What is the mole fraction of H2O(l) in this solution? The KCl(s) dissociates into two ions. Thus, the water will be only 1/3 of the particles in solution, and thus its mole fraction will be 0.333. 5. A galvanic cell has the shorthand notation Pb(s) | Pb+2 (aq), 1 M || Pb+2 (aq), 2 M | Pb(s). Is the cell potential positive or negative? The oxidation reaction is Pb(s) → Pb+2 (aq) (1 M) + 2 e¯. The reduction is Pb+2 (aq) (2 M) + 2 e¯ → Pb(s). The overall reaction is therefore Pb+2 (aq) (2 M) → Pb+2 (aq) (1 M). For this reaction, Q = ½ = 0.5. According to the Nernst equation, o RT E E ln(Q) nF = − . Since Q<1, ln(Q) < 0, and so E > Eo , But since Eo = 0, E > 0 7. Melting ice is an endothermic process. Why is it spontaneous? Because the process has a positive entropy change. 8. For the reaction H2O(l) = H2O(g) initially at equilibrium, which direction will the equilibrium shift (left or right) if the pressure is increased? 1
Left(so as to decrease the pressure as per le Chatelier's principle). Part B.Answer any five of the eight questions(CI-C8).If you answer more than five,the best five will be used to calculate your mark (20 marks each). B1.(a)Aluminum metal is produced by the reduction of aluminum ions in a molten solution according to Al+3eAl What mass of aluminum metal can be produced using a current of 100,000 A over a period of8 hours? t105C1s(8hx3600sh-=29,849mole n-F 96487C(mol e-)-1 but each mole of Al requires 3 moles of the electrons to be reduced to the metal.Thus,the number of moles of Al produced in this time is 29.849/3=9950 moles Als Mass Al(s)27.0 g mol"x 9950 mol 2.69 x 105g=269 kg B1.(b)For the galvanic reaction+CZnCa 50C.calculate the cell potential (in Volts)if [Cd"]=0.068 M and [Zn]=1.00 M. E°=-0.40V+0.76V=+0.36V Q=IZn"1.00M ICd10.069M=14.7 T=50℃+273=323K E=Er-Rno=038v.8aaenCg23 2(96487Cmo') =0.32V 2
Left (so as to decrease the pressure as per le Chatelier’s principle). Part B. Answer any five of the eight questions (C1 – C8). If you answer more than five, the best five will be used to calculate your mark (20 marks each). B1. (a) Aluminum metal is produced by the reduction of aluminum ions in a molten solution according to Al+3 (l) + 3 e⎯ → Al(l). What mass of aluminum metal can be produced using a current of 100,000 A over a period of 8 hours? 5 1 1 it 10 C/ s(8h 3600sh ) n 2 F 96487C(mol e ) − 9,849 mol e− − − × = = = but each mole of Al+3 requires 3 moles of the electrons to be reduced to the metal. Thus, the number of moles of Al(s) produced in this time is 29,849 / 3 = 9950 moles Al(s). Mass Al(s) = 27.0 g mol-1 x 9950 mol = 2.69 x 105 g = 269 kg B1. (b) For the galvanic reaction Zn(s) + Cd+2 (aq) → Zn+2 (aq) + Cd(s) at 50o C, calculate the cell potential (in Volts) if [Cd+2 (aq)] = 0.068 M and [Zn+2 (aq)] = 1.00 M. Eo = -0.40 V + 0.76 V = +0.36 V 2 (aq) 2 (aq) [Zn ] 1.00M Q 1 [Cd ] 0.068M + + = == 4.7 T = 50o C + 273 = 323 K 1 1 o 1 RT 8.314JK mol (323K) E E ln(Q) 0.36V nF 2(96487C mol ) 0.32 V − − − =− = − = 2
B2.(a)Aspirin,a common pain reliever,is a monoprotic acid with the molecular formula HCHO.K for this weak acid is 3.27x 104 Calculate the pH of a solution made by dissolving 0.325 g of Aspirin in 200.0 mL of water. 0.325g =0.00181mol 180gmol-1 0.00181mol =0.0090M 0.200L Initial 0.0090 0 0 Change +X Equilibrium 0.0090.X Thus at equilibrium, x(x) =3.27×10 0.0090-× Note that although the acid is weak.we can not assume that x<<00090.since there is little acid dissolved in this solution.Thus we must use the quadratic formula to solve this one x2=3.27x10(0.0090)-3.27x104x x2+3.27x104x-2.94x106=0 ×=-b±B2-490-3.27x10-±327×10y-40-2.94×10可 2a 2(① -0.000327±0.00344 =0.001560r-0.00188 thus,×=0.00156 Thus,[H3O'(]=0.00156 M pH=-log1o[H30a=-log10(0.00156)=2.81
B2. (a) Aspirin, a common pain reliever, is a monoprotic acid with the molecular formula HC9H7O4. Ka for this weak acid is 3.27 x 10-4. Calculate the pH of a solution made by dissolving 0.325 g of Aspirin in 200.0 mL of water. 1 0.325g 0.00181mol 180gmol 0.00181mol 0.0090M 0.200L − = = [HC9H7O4(aq) ] [H3O+ (aq) ] [C9H7O4¯(aq) ] Initial 0.0090 0 0 Change -x +x +x Equilibrium 0.0090 - x x x Thus at equilibrium, x(x) 4 3.27 10 0.0090 x − = × − Note that although the acid is weak, we can not assume that x<<0.0090, since there is little acid dissolved in this solution. Thus we must use the quadratic formula to solve this one. x 2 = 3.27 x 10-4(0.0090) – 3.27 x 10-4 x x 2 +3.27 x 10-4 x – 2.94 x 10-6 = 0 2 4 4 2 b b 4ac 3.27 10 (3.27 10 ) 4(1)( 2.94 10 ) x 2a 2(1) 0.000327 0.00344 2 0.00156or 0.00188 thus,x 0.00156 − − −± − −×± × −−× = = − ± = = − = −6 Thus, [H3O+ (aq)] = 0.00156 M pH = -log10[H3O+ (aq)] = -log10(0.00156) = 2.81 3
B2.(b)Calculate the pH of a basic buffer made by dissolving 0.125 moles of NHCls and 0.25 moles of NH)in 500 mL of water.Ko for NH)is 1.8 x 10 pK=14-pKb=14-(log1o(K)=14-(log101.8x10=9.25 [acid][NH"(ag)]=0.125 mol /0.5 L=0.25 M [base]=[NH3()=0.25 mol /0.5 L=0.50 M pH=k-loans明-925-log.25-955 B3.Calculate the standard enthalpy of formation of strontium carbonate,SrCOs given the following data: Srg+%02e→Sr0g △H°=-592 kJ mol SrC03s→SrOs+C02gl △H°=+234 kJmol C(onpe)+02g→C0e AH=-394 kJ mol The reaction whose enthalpy we want is S+C+1.This reaction is obtained by adding the above reactions as follows: Srs+hO2g→SrOs △H°=-592 kJ mol" C(gaphite))+O2g)→CO2g △H°=-394 kJmol SrOs+CO2g→SrCO3s △H°=-(+234kmo) Srg+C(graphite)+1.502g→SrC03g△H°=（←592-394-234）=-1220 kJmol
B2. (b) Calculate the pH of a basic buffer made by dissolving 0.125 moles of NH4Cl(s) and 0.25 moles of NH3(aq) in 500 mL of water. Kb for NH3(aq) is 1.8 x 10-5. pKa = 14 – pKb = 14 – (-log10(Ka) = 14 – (-log10(1.8 x 10-5) = 9.25 [acid] = [NH4 + (aq)] = 0.125 mol / 0.5 L = 0.25 M [base] = [NH3(aq)] = 0.25 mol / 0.5 L = 0.50 M a 10 10 [acid] .25 pH pK log 9.25 log 9.55 [base] .5 =− = − = B3. Calculate the standard enthalpy of formation of strontium carbonate, SrCO3(s) given the following data: Sr(s) + ½ O2(g) → SrO(s) ΔHo = −592 kJ mol-1 SrCO3(s) → SrO(s) + CO2(g) ΔHo = +234 kJ mol-1 C(graphite) + O2(g) → CO2(g) ΔHo = −394 kJ mol-1 The reaction whose enthalpy we want is Sr(s) + C(graphite) + 1.5 O2(g) → SrCO3(s). This reaction is obtained by adding the above reactions as follows: Sr(s) + ½ O2(g) → SrO(s) ΔHo = −592 kJ mol-1 C(graphite) + O2(g) → CO2(g) ΔHo = −394 kJ mol-1 SrO(s) + CO2(g) → SrCO3(s) ΔHo = − (+234 kJ mol-1) ________________________________________________ Sr(s) + C(graphite) + 1.5 O2(g) → SrCO3(s) ΔHo = (– 592 – 394 – 234) = – 1220 kJ mol-1 4
B4.Use the data below to calculate the equilibrium constant at 1000 K for the reaction 2Hlg)→H2e+l2e △H°kJmo) S(J mol) 26.5 206.6 H2xg) 0.0 130.6 12g0 62.4 260.7 △H°=△H'Hg)+△HI2g）-2△H'(Hg) =0+62.4-2(26.5)=9.4 kJ mol △S°=△Sr'Hg)+△SrL2g）-2△Sr'Hlg) =130.6+260.7-2(206.6)=-21.9JKmo △G°=△H°.TAS9 =9400-1000(-21.9=+31,300Jmor K-d0.em0023 (b)If p=0.002 atm at equilibrium,calculate Pe and pr2. PnP=0.023 P p,P,=0.023p=0.023(0.002atm)2=9.27×10-8atm Thus,p4,=P2=(9.27×108atm2)v2=3.04×10atm B5.(a)What hybridization is each of the four indicated atoms using in the molecule of Aspirin shown below? 5
B4. Use the data below to calculate the equilibrium constant at 1000 K for the reaction 2 HI(g) → H2(g) + I2(g) ΔHf o (kJ mol-1) So (J mol-1) HI(g) 26.5 206.6 H2(g) 0.0 130.6 I2(g) 62.4 260.7 ΔHo = ΔHf 0 (H2(g)) + ΔHf 0 (I2(g)) – 2 ΔHf 0 (HI(g)) = 0 + 62.4 – 2(26.5) = 9.4 kJ mol-1 ΔSo = ΔSf 0 (H2(g)) + ΔSf 0 (I2(g)) – 2 ΔSf 0 (HI(g)) = 130.6 + 260.7 – 2(206.6) = –21.9 J K-1 mol-1 ΔGo = ΔHo - TΔSo = 9400 – 1000(-21.9) = +31,300 J mol-1 0 1 1 1 G 31300 Jmol RT 8.314 JK mol (1000K) Ke e 0.023 − − − ⎡⎤ ⎡ −Δ − ⎤ ⎢⎥ ⎢ ⎥ ⎢⎥ ⎢ ⎣⎦ ⎣ ⎥⎦ == = (b) If pHI = 0.002 atm at equilibrium, calculate pH2 and pI2. 2 2 2 2 H I 2 HI 2 2 H I HI p p 0.023 p p p 0.023p 0.023(0.002atm) 9.27 10 atm − = = = =× 8 2 Thus, 2 2 8 2 1/ 2 4 p p (9.27 10 atm ) 3.04 10 atm H I − − == × = × B5. (a) What hybridization is each of the four indicated atoms using in the molecule of Aspirin shown below? 5