Part A.Answer all questions(5 marks each) A1.Why is light of only certain wavelengths observed in an atomic spectrum? The light is a result of transitions of electrons from one quantized energy to another.Since the energy levels are quantized,so are the differences between them,and so are the photon energies. A2.Why is a molecule with a triple bond in it likely to be very reactive? A triple bond consists of one sigma-bond plus two pi-bonds.The pi-electrons are delocalized (far from the nuclei)and so are not tightly bound.Since they are easily removed,this makes the molecule very reactive. A3.What type of hybrid orbitals is the As atom using to bond with the Cl atoms in AsCl3? 5+(3x7)=26 electrons.Thus it is AXE,i.e.the As atoms has four electron groups around it.Thus,it is using sphybrid orbitals A4.Can Cas be used as a reducing agent for Ni?Why or why not? Yes.Ca is above Ni in the activity series.Thus,Ca is more easily oxidized that Ni.Thus,Ca gets oxidized and Nigets reduced. A5.Explain why a reaction having AH<tends to make the entropy of the universe increase. The heat from the system moves into the surroundings.This increases the entropy of the surroundings.This in turn increases the entropy of the Universe,since the surroundings are part of the Universe. A6.Suppose the reaction Zn+Sn=Zn+Sn(s)is at equilibrium.Will diluting the solution with water have an effect on the position of the equilibrium?Why or why not? Dilution will have no effect since Q=[Zn][Sn]and dilution will not change the value ofQ A7.Dissolving CuCls in HO to make Cu and Cl ions has AS<0.Why?
1 Part A. Answer all questions (5 marks each). A1. Why is light of only certain wavelengths observed in an atomic spectrum? The light is a result of transitions of electrons from one quantized energy to another. Since the energy levels are quantized, so are the differences between them, and so are the photon energies. A2. Why is a molecule with a triple bond in it likely to be very reactive? A triple bond consists of one sigma-bond plus two pi-bonds. The pi-electrons are delocalized (far from the nuclei) and so are not tightly bound. Since they are easily removed, this makes the molecule very reactive. A3. What type of hybrid orbitals is the As atom using to bond with the Cl atoms in AsCl3? 5 + (3 x 7) = 26 electrons. Thus it is AX3E, i.e. the As atoms has four electron groups around it. Thus, it is using sp3 hybrid orbitals. A4. Can Ca(s) be used as a reducing agent for Ni+2 (aq)? Why or why not? Yes. Ca is above Ni in the activity series. Thus, Ca is more easily oxidized that Ni. Thus, Ca gets oxidized and Ni+2 gets reduced. A5. Explain why a reaction having ΔH < 0 tends to make the entropy of the universe increase. The heat from the system moves into the surroundings. This increases the entropy of the surroundings. This in turn increases the entropy of the Universe, since the surroundings are part of the Universe. A6. Suppose the reaction Zn(s) + Sn+2 (aq) = Zn+2 (aq) + Sn(s) is at equilibrium. Will diluting the solution with water have an effect on the position of the equilibrium? Why or why not? Dilution will have no effect since Q = [Zn+2] / [Sn=2], and dilution will not change the value of Q. A7. Dissolving CuCl2(s) in H2O(l) to make Cu+2 (aq) and Cl− (aq) ions has ΔSo < 0. Why?
Although the ions in CuCl have low entropy,making the aqueous ions reduces the entropy of the associated water molecules even more. A8.CuCla(s spontaneously dissolves in water even though AS<0.Why? Because△H°<0 A9.Why are the ionization potentials of phosphorus(P)and sulphur(S)approximately equal? P has three unpaired p-electrons.Although S has four,and the fourth should be more easily removed since it is paired with one other,thus decreasing the ionization potential,the S atom also has one more proton,making the effective nuclear charge greater,increasing the ionization potential.The net effect is no change. Part B.Answer all questions on this page(20 marks total). BI.Ammonium nitrate can be used as an explosive.It decomposes according to NH4NO3g→N2g+2HOg+⅓O2e Assuming the gases behave ideally,calculate the total volume of the products at 300C and 1 atm when 12.0 g of ammonium nitrate decomposes. 12g 80g/mol =0.15 mol NH NO x 3.5 mol gas mol NHaNO3 =0.525 mol gas V-RT0.525 mol(0.082Latm mol-K300+273)K247L p 1atm B2.A sample of an unknown gas diffuses in 11.1 minutes.An equal volume of H2dg)in the same apparatus under 2
Although the ions in CuCl2(s) have low entropy, making the aqueous ions reduces the entropy of the associated water molecules even more. A8. CuCl2(s) spontaneously dissolves in water even though ΔSo < 0. Why? Because ΔHo < 0 A9. Why are the ionization potentials of phosphorus (P) and sulphur (S) approximately equal? P has three unpaired p-electrons. Although S has four, and the fourth should be more easily removed since it is paired with one other, thus decreasing the ionization potential, the S atom also has one more proton, making the effective nuclear charge greater, increasing the ionization potential. The net effect is no change. Part B. Answer all questions on this page (20 marks total). B1. Ammonium nitrate can be used as an explosive. It decomposes according to: NH4NO3(s) → N2(g) + 2 H2O(g) + ½ O2(g). Assuming the gases behave ideally, calculate the total volume of the products at 300o C and 1 atm when 12.0 g of ammonium nitrate decomposes. g . mol NH NO g / mol ⎛ ⎞ ⎜ ⎟ = ⎝ ⎠ 4 3 12 0 15 80 x 3.5 mol gas / mol NH4NO3 = 0.525 mol gas nRT . mol( . L atm mol K )( )K V . p atm − − + = = = 1 1 0 525 0 082 300 273 24 7 1 L B2. A sample of an unknown gas diffuses in 11.1 minutes. An equal volume of H2(g) in the same apparatus under 2
the same conditions effuses in 2.42 minutes.Calculate the molecular weight of the unknown gas. time raten MW MW timen MW,气time， time =2.02gmol1 11.1)2 =42.5 g mol- timeu 2.42 B3.Calculate the pressure exerted by 1.00 mol of CH ina 500 mL vessel at 25.0Cusing (a)The ideal gas law nRT 1.00 mol(.082Latm K-'molX25+273)K.9 atm P= 0.500L (b)The van der Waals equation.For CH4)a=2.25 L2atm mol2 and b=0.0428 L mol p=- 1.00 mol(0.082 Latm K-'mol-25+273)KLatm1.00 mol 0.500L-1.00mol(0.0428Lmol) mo20.500L =53.45atm-9.00atm =44.5atm Part C.Answer any five of questions C1-C7(20 marks each).If you answer more than five,the best five will be used to calculate your total mark. C1.Nitrogen dioxide decomposes according to 2 NO2)=2 NO)+Ox()At a certain temperature,Kp is extremely small(4.48x).If0.75 atm of NO is placed in a vessel at this temperature,find the equilibrium partial pressures of all three gases p(NOz).atm p(NO),atm p(O2),atm 3
the same conditions effuses in 2.42 minutes. Calculate the molecular weight of the unknown gas. ?? H ?? H ?? H ?? ?? H H ?? ?? H H time rate MW time rate MW MW time MW time time . MW MW . g mol . g mol time . − − = = ⎛ ⎞ = ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎛ ⎞ ⎛ ⎞ == = ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ 2 2 2 2 2 2 2 2 2 2 1 1 11 1 2 02 42 5 2 42 B3. Calculate the pressure exerted by 1.00 mol of CH4(g) in a 500 mL vessel at 25.0o C using: (a) The ideal gas law nRT . mol( . L atm K mol )( )K p . V . L − − + = = = 1 1 1 00 0 082 25 273 48 9 0 500 atm (b) The van der Waals equation. For CH4(g), a = 2.25 L2 atm mol-2 and b = 0.0428 L mol-1 nRT n p a V nb V . mol( . L atm K mol )( )K L atm . mol . . L . mol( . L mol ) mol . L . atm . atm . atm − − − ⎛ ⎞ = − ⎜ ⎟ − ⎝ ⎠ + ⎛ ⎞ = − ⎜ ⎟ − ⎝ ⎠ = − = 2 2 1 1 2 1 2 1 00 0 082 25 273 1 00 2 25 0 500 1 00 0 0428 0 500 53 45 9 00 44 5 Part C. Answer any five of questions C1 – C7 (20 marks each). If you answer more than five, the best five will be used to calculate your total mark. C1. Nitrogen dioxide decomposes according to 2 NO2(g) = 2 NO(g) + O2(g). At a certain temperature, Kp is extremely small (4.48×10-13). If 0.75 atm of NO2(g) is placed in a vessel at this temperature, find the equilibrium partial pressures of all three gases. p(NO2), atm p(NO), atm p(O2), atm 3
Initial,atm 0.75 0 0 Change,atm -2x +2x +x Equilibrium,atm 0.75-2x 2x Atequilibrium. Pso Po=K PNo,2 (0.75-X7=448x103 (2x)2x Here,Kp is so small,we can make the approximation that x<<0.75,so the expression simplifies to 075=4.48x10-9 (2x)2x 4x3=(0.75)2(4.48×10-13) x3=(0.75)2(4.48×10-13)/4 x3=6.3×10-14 x=3.98×10-5 Thus,pNo2=0.75-2x=0.75 atm pw0=2x=7.96x103atm Po2=x=3.98 x 105 atm Check:(7.96x102(3.98x10)/(0.75-2(3.98x102=4.48x1013=K C2.Given the following information,calculate the ionization energy of lithium(in kJmol): Lattice energy of LiF(s) 1050 kJ mol! Electron affinity of fluorine -328 kJ mol △H°(Lifs) -617 kJ mol Enthalpy of sublimation of lithium metal 161 kJ mol
Initial, atm 0.75 0 0 Change, atm -2x +2x +x Equilibrium, atm 0.75 – 2x 2x x At equilibrium, NO O p NO p p K p ( x) x . ( . x) − = = × − 2 2 2 2 2 13 2 2 4 48 10 0 75 Here, Kp is so small, we can make the approximation that x << 0.75, so the expression simplifies to ( x) x . (. ) x (. )( . ) x ( . ) ( . )/ x . x . − − − − − = × = × = × = × = × 2 13 2 32 1 32 13 3 14 5 2 4 48 10 0 75 4 0 75 4 48 10 0 75 4 48 10 4 6 3 10 3 98 10 3 Thus, pNO2 = 0.75 – 2x = 0.75 atm pNO = 2x = 7.96 x 10-5 atm pO2 = x = 3.98 x 10-5 atm Check: (7.96 x 10-5) 2 (3.98 x 10-5) / (0.75 – 2(3.98 x 10-6))2 = 4.48 x 10-13 = Kp C2. Given the following information, calculate the ionization energy of lithium (in kJ mol-1): Lattice energy of LiF(s) 1050 kJ mol-1 Electron affinity of fluorine − 328 kJ mol-1 ΔHf o (LiF(s)) − 617 kJ mol-1 Enthalpy of sublimation of lithium metal 161 kJ mol-1 4
Bond dissociation energy of fluorine 159 kJ mol! Lig→Lie +161 kJ mol Lig→Litg+e I (Li) hFg→Fg 2(+159kmol) Fg+e→Fg -328 kJ mor Li计g+Fg)一LiFo -(1050)kJ mol Li6+⅓F2g→Lifo -617 kJ mol Thus,the ionization potential of lithium is found by difference: 11(Li)=-617-(+161+⅓(159)-328-1050)=520.5 kJ mol C4.Use the data in the table to answer the following questions about the reaction 2) NO2(g) N204g △H,kJ molT 33.2 9.16 S°，JK-mo 240.0 304.2 (a)Calculate AH(kJ mol) △Hr°=△HN204He）-2AHNO2e) =9.16-233.2) =-57.24 kJ mol
5 Bond dissociation energy of fluorine 159 kJ mol-1 Li(s) → Li(g) +161 kJ mol-1 Li(g) → Li+ (g) + e- I1 (Li) ½ F2(g) → F(g) ½ (+159 kJ mol-1) F(g) + e- → F- (g) -328 kJ mol-1 Li+ (g) + F- (g) → LiF(s) -(1050) kJ mol-1 _______________________________________ Li(s) + ½ F2(g) → LiF(s) -617 kJ mol-1 Thus, the ionization potential of lithium is found by difference: I1(Li) = -617 – (+161 + ½ (159) – 328 – 1050) = 520.5 kJ mol-1 C4. Use the data in the table to answer the following questions about the reaction 2 NO2(g) Ý N2O4(g). NO2(g) N2O4(g) ΔHf o , kJ mol-1 33.2 9.16 So , J K-1 mol-1 240.0 304.2 (a) Calculate ΔHo (kJ mol-1) ΔHo = ΔHf o (N2O4(g)) - 2 ΔHf o (NO2(g)) = 9.16 – 2(33.2) = -57.24 kJ mol-1