Part A.Answer all eight questions(5 marks each). 1.Use VSEPR theory to predict the shape of the NF3 ion. 5+(3x7)+1 =27 electrons.Making single bonds,completing the octets around the F atoms uses 24 electrons The other three are placed as a lone pair plus a single electron on the N atom.The molecule is therefore of the form AX3E2.The shape is therefore T-shaped 2.For the reaction C+CO=2 CO which direction(left or right)will the equilibrium shift if the pressure is decreased?Why? The reaction will shift to the right in order to increase the pressure,according to le Chatelier's principle 3.Which of HCIO and HCIO2 is the stronger acid?Why? HCIO2 is the stronger acid.The extra Oatom draws electrons towards itself,withdrawing electrons from the O- H bond,weakening it and making it a stronger acid. 4.Name three ways to prevent or slow the corrosion of an object made of iron metal. 1.Paint it to prevent contact with oxygen and water. 2.Use passive cathodic protection. 3.Use active cathodic protection. 5.Define the term"state function"and give three examples of one A state function is one whose value does not depend on the path taken from reactants to products,nor on the rate.Three examples are H,S and G. Part B.Answer all four questions(20 marks each). 1. Calculate the lattice energy (in kJ mol)of CaO from the following data: (i)The bond energy of the O2 molecule is +495.0 kJ mol
Part A. Answer all eight questions (5 marks each). 1. Use VSEPR theory to predict the shape of the NF3 ¯ ion. 5 + (3x7) + 1 = 27 electrons. Making single bonds, completing the octets around the F atoms uses 24 electrons. The other three are placed as a lone pair plus a single electron on the N atom. The molecule is therefore of the form AX3E2. The shape is therefore T-shaped 2. For the reaction C(s) + CO2(g) = 2 CO(g), which direction (left or right) will the equilibrium shift if the pressure is decreased? Why? The reaction will shift to the right in order to increase the pressure, according to le Chatelier’s principle. 3. Which of HClO and HClO2 is the stronger acid? Why? HClO2 is the stronger acid. The extra O atom draws electrons towards itself, withdrawing electrons from the OH bond, weakening it and making it a stronger acid. 4. Name three ways to prevent or slow the corrosion of an object made of iron metal. 1. Paint it to prevent contact with oxygen and water. 2. Use passive cathodic protection. 3. Use active cathodic protection. 5. Define the term “state function” and give three examples of one. A state function is one whose value does not depend on the path taken from reactants to products, nor on the rate. Three examples are H, S and G. Part B. Answer all four questions (20 marks each). 1. Calculate the lattice energy (in kJ mol-1) of CaO(s) from the following data: (i) The bond energy of the O2(g) molecule is +495.0 kJ mol-1
(ii)The first and second electron affinities of the O atom are-141.8 and +844.4 kJ mol,respectively (iii)The enthalpy of sublimation of calcium metal is+192.5 kJ mol (iv)The first and second ionization potentials of calcium are+589.5 and+1145.2 kJmol,respectively (v)The enthalpy of formation of is-635.5 kJ mol Cas→Cag +192.5kJmo Cag→Cag+e +589.5 kJ mol Cag→Cag+e +1145 2 kI mol h02g→0g 2(+495.0)=+247.5 kJ mol1 Og+e→0g -141.8 kJ mol-l 0e+e→02g +844.4 kJ mol! Ca”g+02g→Ca0s Cas+⅓02g→Ca0s-635.5 kJmol Thus,-U=-635.5-(192.5+589.5+1145.2+247.5-141.8+844.4)=-3512.8kJmo Thus,lattice energy=U=+3512.8 kJ mol Part C.Answer any three of the six questions.If you answer more than three,the best three will be used to calculate your mark(20 marks each) 1.(a)Balance the following REDOX reaction in acidic solution: 2
2 (ii) The first and second electron affinities of the O atom are -141.8 and +844.4 kJ mol-1, respectively. (iii) The enthalpy of sublimation of calcium metal is +192.5 kJ mol-1 (iv) The first and second ionization potentials of calcium are +589.5 and +1145.2 kJ mol-1, respectively. (v) The enthalpy of formation of CaO(s) is -635.5 kJ mol-1 Ca(s) → Ca(g) +192.5 kJ mol-1 Ca(g) → Ca+ (g) + e- +589.5 kJ mol-1 Ca+ (g) → Ca+2 (g) + e- +1145.2 kJ mol-1 ½ O2(g) → O(g) ½ (+495.0) = +247.5 kJ mol-1 O(g) + e- → O- (g) -141.8 kJ mol-1 O- (g) + e- → O-2 (g) +844.4 kJ mol-1 Ca+2 (g) + O-2 (g) → CaO(s) -U _____________________________________________________ Ca(s) + ½ O2(g) → CaO(s) -635.5 kJ mol-1 Thus, -U = -635.5 – (192.5 + 589.5 + 1145.2 + 247.5 – 141.8 + 844.4) = -3512.8 kJ mol-1 Thus, lattice energy = U = +3512.8 kJ mol-1 Part C. Answer any three of the six questions. If you answer more than three, the best three will be used to calculate your mark (20 marks each). 1. (a) Balance the following REDOX reaction in acidic solution:
C2O(g)+NO3 ()CO2(g)+NO(g) C2042am→C02g C2042am→2C02g C2042am)→2C02g+2e NO3q→NOg NO3a→NOg+2H2Ou NO3e+4Hem→NOg+2H,0o NO3(a+4 H'+3-NO)+2 H2Oo) 3C2042m→6C02g+6e 2 NO3g+8 H)+6e-2 NO(g)+4 H2O) 3 C204()+2 NO3(g)+8H'()6CO2(8)+2NO(g)+4 H2O() (b)The half reaction Sn+2eSn has a standard reduction potential of +0.15 V.Assuming a temperature of 25C.and that [Sn=1.00 M,calculate [Sn when the reduction potential of this half reaction is0.00 V. Q=ISn"l [Sn“a】 -6-ho 1.00 thus,0.00V=0.15V-8.314 JK-mol'(298K)n 2(96487Cmo1-) 1.00 -2(96487Cmol-） (8.314 JK-mol-(298KI1.68 R9-law [Sn“m】=8.43×10M 3
3 C2O4 -2 (aq) + NO3 − (aq) → CO2(g) + NO(g) C2O4 -2 (aq) → CO2(g) C2O4 -2 (aq) → 2 CO2(g) C2O4 -2 (aq) → 2 CO2(g) + 2 e- NO3 - (aq) → NO(g) NO3 - (aq) → NO(g) + 2 H2O(l) NO3 - (aq) + 4 H+ (aq) → NO(g) + 2 H2O(l) NO3 - (aq) + 4 H+ (aq) + 3 e- → NO(g) + 2 H2O(l) 3 C2O4 -2 (aq) → 6 CO2(g) + 6 e- 2 NO3 - (aq) + 8 H+ (aq) + 6 e- → 2 NO(g) + 4 H2O(l) _____________________________________________________ 3 C2O4 -2 (aq) + 2 NO3 - (aq) + 8 H+ (aq) → 6 CO2(g) + 2 NO(g) + 4 H2O(l) (b) The half reaction Sn+4 (aq) + 2 e− → Sn+2 (aq) has a standard reduction potential of + 0.15 V. Assuming a temperature of 25o C, and that [Sn+2 (aq)] = 1.00 M, calculate [Sn+4 (aq)] when the reduction potential of this half reaction is 0.00 V. 2 (aq) 4 (aq) 0 1 1 1 4 (aq) 1 1 4 1 1 (aq) 4 (aq) [Sn ] Q [Sn ] RT E E ln(Q) nF 8.314 J K mol (298K) 1.00 thus, 0.00V 0.15V ln 2(96487 C mol ) [Sn ] 1.00 2(96487 C mol ) ln 0.15J C 11.68 [Sn ] 8.314J K mol (298K) 1.00 118600 [Sn ] [Sn + + − − − + − − + − − + = = − = − ⎛ ⎞ − = − ⎜ ⎟ = ⎝ ⎠ = 4 6 (aq) ] 8.43 10 M + − = ×
(a)The normal boiling point of water is 100.0C.Its vapor pressure at 90C is 525.76 mm Hg.Calculate the enthalpy of vaporization of water(kJ mol). )贤 thus,AH.=RIn (P儿TT, =8341 Ko'(725g0+2771m+2 1 1 =41.5×103Jmol- =41.5kJ mol-1 (b)10.0g of dissolved in 1 kg of water.Calculate the vapor pressure of this solution at 90C 2(23.0)+321+64gmor=0.0704mol→021 molions 10.0g 1kg water= 18.0gmoFT=55.5mol 1000g 55.5 X0-55+02m=0,96 Pu4,0=p°4，oX4,o=525.76mmHg(0.996)=523.77mmHg 3. (a)Kp for the equilibrium FeOs)+COFe(s)+CO2 at 1000C is 0.403.If CO()at a pressure of 1.000 atm and excess FeO are mixed in a container at 1000C,what are the partial pressures of CO and CO)when equilibrium is attained? Initial,atm 1.00 0 Change,atm x +x Equilibrium,atm 1.00-x 4
4 2. (a) The normal boiling point of water is 100.0o C. Its vapor pressure at 90o C is 525.76 mm Hg. Calculate the enthalpy of vaporization of water (kJ mol-1). 2 vap 1 1 2 1 2 vap 1 12 1 1 3 1 1 p H 1 1 ln p R TT p 1 1 thus, H R ln p TT 760mmHg 1 1 8.314 J K mol ln 525.76 90 273K 100 273K 41.5 10 J mol 41.5kJ mol − − − − − ⎛⎞ ⎡ ⎤ Δ ⎜ ⎟ = − ⎢ ⎥ ⎝⎠ ⎣ ⎦ ⎛ ⎞⎡ ⎤ Δ= − ⎜ ⎟⎢ ⎥ ⎝ ⎠⎣ ⎦ ⎛ ⎞⎛ = − ⎜ ⎟⎜ ⎝ ⎠⎝ + + = × = ⎞ ⎟ ⎠ (b) 10.0 g of Na2SO4(s) is dissolved in 1 kg of water. Calculate the vapor pressure of this solution at 900 C. 2 2 22 1 1 H O 0 HO HO HO 10.0g 0.0704mol 0.211molions 2(23.0) 32.1 64g mol 1000g 1kg water 55.5mol 18.0g mol 55.5 X 0.996 55.5 0.211 p p X 525.76mmHg(0.996) 523.77 mmHg − − = → + + = = = = + == = 3. (a) Kp for the equilibrium FeO(s) + CO(g) ∩ Fe(s) + CO2(g) at 1000o C is 0.403. If CO(g), at a pressure of 1.000 atm and excess FeO(s) are mixed in a container at 1000o C, what are the partial pressures of CO(g) and CO2(g) when equilibrium is attained? CO(g) CO2(g) Initial, atm 1.00 0 Change, atm -x +x Equilibrium, atm 1.00-x x
K-==00 x=0.403-0.403x 1.403x=0.403 x=0.287 Thus,pco=1.00-x=0.713 atm Pco2=X=0.287 (b)Calculate the value of G(kJmol)for the reaction at 1000C.Is this reaction spontaneous at 1000℃？ △G°=-RTIn(Kp) =-8.314 JK mol(1273K)ln(0.403) =+9619Jmo =+9.62kmol Since AG>0,the reaction is not spontaneous at 1000C. 4 (a)Calculate the pH of a 2.00 M solution of trimethylamine,N(CH3)3.Kp for this base is 7.4 x 105. N(CH3)ag)+H2ON(CH3)3H'+OHQ K.-[N(CH,),74x10 [N(CH,)3] N(CH3)3(g)N(CH3)3H'()OH) Initial,M 2.00 0 0 Change,M -X +x +x Equilibrium,M 2.00-x X Here,Kb <<0.001,thus x <<2.00. 5
5 CO2 p CO p x K 0.403 p 1x x 0.403 0.403x 1.403x 0.403 x 0.287 === − = − = = Thus, pCO = 1.00 – x = 0.713 atm pCO2 = x = 0.287 (b) Calculate the value of ΔGo (kJ mol-1) for the reaction at 1000o C. Is this reaction spontaneous at 1000o C? ΔG0 = -RTln(Kp) = -8.314 J K-1 mol-1 (1273 K)ln(0.403) = +9619 J mol-1 = +9.62 kJ mol-1 Since ΔG0 > 0, the reaction is not spontaneous at 1000o C. 4. (a) Calculate the pH of a 2.00 M solution of trimethylamine, N(CH3)3. Kb for this base is 7.4 x 10-5. N(CH3)3(aq) + H2O(l) ∩ N(CH3)3H+ (aq) + OH- (aq) 3 3 (aq) (aq) 5 b 3 3 [N(CH ) H ][OH ] K 7 [N(CH ) ] + − .4 10− = = × N(CH3)3(aq) N(CH3)3H+ (aq) OH- (aq) Initial, M 2.00 0 0 Change, M -x +x +x Equilibrium, M 2.00 - x x x Here, Kb <<0.001, thus x <<2.00
Thus,2.00=7.4x10- solving,x=0.0122 Thus,[OHag】=0.0122 p0H=-log10lOH(a】=-log10(0.0122)=1.91 pH=14-p0H=14-1.91=12.1 (b)Calculate the pH of a solution which is 2.00 M trimethylamine,N(CH3)3 and 1.00 M trimethylamine hydrochloride,N(CH3)3HCI. [acid]=[N(CH3)3HCI]=1.00 M [base=[N(CH)3]=2.00M pK=14-pKb=14-(log1o7.4x105)=9.87 pH-pK.-log base] [acid] -9.87-1og0200 1.00 =10.2 Silver crystallizes in a cubic unit cell.The edge length of the unit cell is 408 pm.The density of silver metal is 10.6g cm What type of cubic unit cell does silver form? Volume of the unit cell=V=(408x 1010 cm)3=6.79x 1023 cm3 Mass of unit cell =m=pV=10.6 g cmx 6.67x 1023 cm3=7.20 x 1022g Mass of one Ag atom= 107.9g mol-1 602x10m0F=1.79x10-g Number of atoms per unit cell= 701g/unit cell=atoms/unit cell 1.79×10g/atom 6
6 Thus, 2 x 5 7.4 10 2.00 solving, x 0.0122 − = × = Thus, [OH- (aq)] = 0.0122 pOH = -log10[OH- (aq)] = -log10(0.0122) = 1.91 pH = 14 – pOH = 14 – 1.91 = 12.1 (b) Calculate the pH of a solution which is 2.00 M trimethylamine, N(CH3)3 and 1.00 M trimethylamine hydrochloride, N(CH3)3HCl. [acid] = [N(CH3)3HCl] = 1.00 M [base] = [N(CH3)3] = 2.00 M pKa = 14 – pKb = 14 – (-log10(7.4 x 10-5)) = 9.87 a 10 10 [acid] pH pK log [base] 1.00 9.87 log 2.00 10.2 = − = − = 5. Silver crystallizes in a cubic unit cell. The edge length of the unit cell is 408 pm. The density of silver metal is 10.6 g cm-3. What type of cubic unit cell does silver form? Volume of the unit cell = V = (408 x 10-10 cm)3 = 6.79 x 10-23 cm3 Mass of unit cell = m = ρV = 10.6 g cm-3 x 6.67 x 10-23 cm3 = 7.20 x 10-22 g Mass of one Ag atom = 1 22 23 1 107.9g mol 1.79 10 g 6.02 10 mol − − − = × × Number of atoms per unit cell = 22 22 7.20 10 g / unit cell 4 atoms / unit cell 1.79 10 g / atom − − × = ×
Only FCC(face centered cubic)has this many atoms per unit cell 6.(a)Name four physical properties of molecules that can be predicted using molecular orbital theory that can not be predicted using valence bond theory or hybrid orbital theory. 1.Bond length 2.Bond energy 3.Magnetic properties 4.Ionization potentials (c)Describe the bonding between the two carbon atoms in a molecule of acetylene,C2H2,in terms of the hybrid and other orbitals used by each atom,and by the types of bonds that result. The carbon atoms are bound by: 1.A sigma bond,formed by overlap of an sp hydrib orbital on one carbon with an sp hybrid orbital on the other carbon. 2.A pi bond,formed by the sideway overlap of two parallel p-orbitals,one on each carbon atom 3.Another pi bond,formed by the sideway overlap of the other two parallel p-orbitals,one on each carbon atom. 7
7 Only FCC (face centered cubic) has this many atoms per unit cell. 6. (a) Name four physical properties of molecules that can be predicted using molecular orbital theory that can not be predicted using valence bond theory or hybrid orbital theory. 1. Bond length 2. Bond energy 3. Magnetic properties 4. Ionization potentials (c) Describe the bonding between the two carbon atoms in a molecule of acetylene, C2H2, in terms of the hybrid and other orbitals used by each atom, and by the types of bonds that result. The carbon atoms are bound by: 1. A sigma bond, formed by overlap of an sp hydrib orbital on one carbon with an sp hybrid orbital on the other carbon. 2. A pi bond, formed by the sideway overlap of two parallel p-orbitals, one on each carbon atom 3. Another pi bond, formed by the sideway overlap of the other two parallel p-orbitals, one on each carbon atom