Part A.Answer all eight questions(5 marks each). 1.Why does the pH of a buffered solution not change much if small amounts of acid or base are added to it? The large quantity of the base(or acid)re-equibrates with the weak acid (or base). restoring the pH to close to its original value. 2.Define the term"rate limiting step." The rate limiting step is the slowest step in the mechanism of a multistep reaction. 3.A solution is made by dissolving 2 moles of KCls)in 1 mole H2O.What is the mole fraction of K' in this solution? 2 moles of KCI gives 2 moles of K'ions and 2 moles of CI ions.The mole fraction of K*is therefore 21(2+2+1)=0.40. 4.Chemical reactions may release or absorb energy.Why do we not observe a change in mass, according to△E=△mc2? There is a change in mass,but it is too small to be observed directly This reaction has a positive AS,which tends to increase the entropy of the Universe.It also has a negative AH,which causes the surroundings to heat up,increasing the entropy of the surroundings. and increasing the entropy of the Universe as well. 7.Name two reasons,related to the lead acid battery.why it is difficult to start your car on a cold day. According to the Nernst equation,lowering the temperature can lower the battery voltage.Also,the electrolye becomes very viscous when cold,decreasing the flow of ions through it,decreasing the current available from the battery. 8.Why is the Henry's law constant,KH,for SO2 so much larger than that for CO2? the
Part A. Answer all eight questions (5 marks each). 1. Why does the pH of a buffered solution not change much if small amounts of acid or base are added to it? The large quantity of the conjugate base (or acid) re-equilibrates with the weak acid (or base), restoring the pH to close to its original value. 2. Define the term “rate limiting step.” The rate limiting step is the slowest step in the mechanism of a multistep reaction. 3. A solution is made by dissolving 2 moles of KCl(s) in 1 mole H2O(l). What is the mole fraction of K+ (aq) in this solution? 2 moles of KCl gives 2 moles of K+ ions and 2 moles of Cl- ions. The mole fraction of K+ is therefore 2/(2+2+1) = 0.40. 4. Chemical reactions may release or absorb energy. Why do we not observe a change in mass, according to ΔE = Δmc2 ? There is a change in mass, but it is too small to be observed directly. 6. Name two reasons that the reaction C3H8(g) + 5 O2(g) → 3 CO2(g) + 4 H2O(g) results in an increase of the entropy of the Universe. This reaction has a positive ΔS, which tends to increase the entropy of the Universe. It also has a negative ΔH, which causes the surroundings to heat up, increasing the entropy of the surroundings, and increasing the entropy of the Universe as well. 7. Name two reasons, related to the lead acid battery, why it is difficult to start your car on a cold day. According to the Nernst equation, lowering the temperature can lower the battery voltage. Also, the electrolye becomes very viscous when cold, decreasing the flow of ions through it, decreasing the current available from the battery. 8. Why is the Henry’s law constant, KH, for SO2 so much larger than that for CO2? SO2 is bent (draw the Lewis structure to prove this). It is therefore polar. CO2 is linear and therefore non-polar. Thus, there will be strong forces of attraction between SO2 and water, hence a high solubility and a high KH value
Part B.Answer all three questions B1,B2 and B3(20 marks each). B1.For the reaction 2+S2O+2SO()the following data were collected: [I ]mol L'[S2O82 ]mol L'Initial rate,mol L min 0.125 0.150 0.0045 0.375 0.150 0.0135 0.125 0.050 0.0015 (a)Determine the rate law for this reaction. eoeaeC8ms8scoe6mens,uesaehatf们stpled,heaepes The Comparing the first and third experiments,we see that if [S2Oa]is tripled,the rate also triples. Therefore the reaction is also first order in [ The rate law is therefore rate =k[][S2O8] (b)Calculate the rate constant(with the correct units)for this reaction. Using the first experiment(we could use any of the three),we have: rate 0.0045 mol L-'min- .0.125m.150mLmor'min (c)The decomposition reaction of NOBr is second order in [NOBr].with a rate constant of 25 L mol1 min1.If the initial concentration of NOBr is 0.025 mol L-1,find (i)the time at which the concentration of NOBr will be 0.010 mol L 1 NOBr]"NOBri,+ thus,t=[NOBr][NOBr), k 1 1 0.010 mol L-1 0.025 mol L-1 =2.4min 25 L mol-min (ii)the concentration of NOBr after 125 min of reaction
Part B. Answer all three questions B1, B2 and B3 (20 marks each). B1. For the reaction 2I− (aq) + S2O8 2− (aq) → I2(aq) + 2SO4 2− (aq), the following data were collected: [I− ], mol L- 1 [S2O8 2− ], mol L- 1 Initial rate, mol L-1 min-1 0.125 0.150 0.0045 0.375 0.150 0.0135 0.125 0.050 0.0015 (a) Determine the rate law for this reaction. Comparing the first and second experiments, we see that if [I- ] is tripled, the rate triples. Therefore the reaction is first order in [I- ]. Comparing the first and third experiments, we see that if [S2O8 -2] is tripled, the rate also triples. Therefore the reaction is also first order in [I- ]. The rate law is therefore rate = k[I- ][S2O8 -2] (b) Calculate the rate constant (with the correct units) for this reaction. Using the first experiment (we could use any of the three), we have: 1 1 1 1 2 11 2 8 rate 0.0045 mol L min k 0 [I ][S O ] 0.125 mol L (0.150 mol L ) − − .24 L mol min − − − − − − = = = (c) The decomposition reaction of NOBr is second order in [NOBr], with a rate constant of 25 L mol-1 min-1. If the initial concentration of NOBr is 0.025 mol L-1, find (i) the time at which the concentration of NOBr will be 0.010 mol L-1 0 0 1 1 1 1 1 1 kt [NOBr] [NOBr] 1 1 [NOBr] [NOBr] thus, t k 1 1 0.010 mol L 0.025 mol L 2.4 min 25 L mol min − − − − = + − = − = = (ii) the concentration of NOBr after 125 min of reaction
1 [NOBr]-[NOBr),+kt 0.025 molT+25Lmol min(125 min) 1 =3165L mol N081=3165Lmo 1 =0.000316molL C7).If you answer more than four,the C1(a).Achemical e ine if it is o make nol(C2HsOH)by reacing ethvlene with waternto C+HOCHaOHn.Use thefolong dat o determine if this reaction is spontaneous under standard conditions: △H°kJmo) s°(JKmo) C2H4(g) 52.0 2190 H200 -286.0 70.0 C2H5OH0-278.0 161.0 22602微8o260S+Ho》 =-44.0 kJ mor AS=SCzH:OHm)-(S(CzHa) +S(H20)》 =161.0 (219.0 +(70.0)JK =-128 J AG°=AH.TAS9 =-44,000 J mol'-(298K(-128 J Kmor) =-5856Jm0 Since AG<0,this reaction is spontaneous under standard conditions. C1(b).Determine the equilibrium constant Ke for the same reaction at 25C. -AG K.=exp RT) 5.856Jmol- -exp.314JK-mol-(298K) =exp(2.36) =10.6 C2(a).Calculate the pH of 0.5 M sodium acetate,CHaCOONa.Ka of acetic acid,CHaCOOH, is1.8x105
0 1 1 1 1 1 1 1 kt [NOBr] [NOBr] 1 25 L mol min (125 min) 0.025 mol L 3165 L mol 1 [NOBr] 0.000316 mol L 3165 L mol − − − − − = + = + = = = Part C. Answer any four of the seven questions (C1 – C7). If you answer more than four, the best four will be used to calculate your mark (20 marks each). C1 (a). A chemical engineer wants to determine if it is possible to make ethanol (C2H5OH) by reacting ethylene with water according to C2H4(g) + H2O(l) Ý C2H5OH(l). Use the following data to determine if this reaction is spontaneous under standard conditions: ΔHf o (kJ mol-1) So (J K-1 mol-1) C2H4(g) 52.0 219.0 H2O(l) -286.0 70.0 C2H5OH(l) -278.0 161.0 ΔH0 = ΔHf 0 (C2H5OH(l)) – (ΔHf 0 (C2H4(g)) + ΔHf 0 (H2O(l))) = – 278.0 – (52.0 + (-286.0)) kJ mol-1 = – 44.0 kJ mol-1 ΔS0 = S0 (C2H5OH(l)) – (S0 (C2H4(g)) + S0 (H2O(l))) = 161.0 – (219.0 + (70.0)) J K-1 mol-1 = -128 J K-1 mol-1 ΔG0 = ΔH0 - TΔS0 = – 44,000 J mol-1 – (298 K)(–128 J K-1 mol-1) = – 5856 J mol-1 Since ΔG0 < 0, this reaction is spontaneous under standard conditions. C1 (b). Determine the equilibrium constant Kc for the same reaction at 25o C. 0 c 1 1 1 G K exp RT 5,856 J mol exp 8.314 J K mol (298K) exp(2.36) 10.6 − − − ⎛ ⎞ −Δ = ⎜ ⎟ ⎝ ⎠ ⎛ ⎞ = ⎜ ⎟ ⎝ ⎠ = = C2 (a). Calculate the pH of 0.5 M sodium acetate, CH3COONa. Ka of acetic acid, CH3COOH, is 1.8 x 10-5
Sodium acetate will dissociate in water: CH3COONa(s)>CH3COO(ag)+Na'(ag) And the acetate ion will hydrolzye according to: CH3COO(ag)+H20>CH3COOH(aq)+OH(aQ) Since the acetate ion is the conjugate base of acetic acid,then the equilibrium constant for this hydrolysis reaction is Ko: K1.0x10-4 =5.56x1010 1.8×10 Thus: CH3COO(ag) CH3COOH(ag) OH'(eg) Initial,M 0.5 0 +x 0.5-x At equilibrium: [CH,COOH=K [CH,COO ) x）=5.56×10-0 0.5-x =05556x100-X0.5. 78× pOH =-log10[OH (ag)]=-log10(x)=-log10(1.67 x 105)=4.78 pH=14-p0H=9.22 C2(b).Calculate the pH of 2.0x107M HCl).This acid hydrolyzes fully in water.K=1.0x104. The HCI hydrolyzes fully: HClag)+H2O>H3O(aQ+Cr(aQ) Since this is such a low concentration of HaO'(ag from the HCI,we must take account of the autohydrolysis of water 2 H2O0H3O'(ag)+OH(ag Kw=1
Sodium acetate will dissociate in water: CH3COONa(s) → CH3COO- (aq) + Na+ (aq) And the acetate ion will hydrolzye according to: CH3COO- (aq) + H2O(l) → CH3COOH(aq) + OH- (aq) Since the acetate ion is the conjugate base of acetic acid, then the equilibrium constant for this hydrolysis reaction is Kb: 3 14 w 10 b 5 a(CH COOH) K 1.0 10 K 5 K 1.8 10 − − − × = = =× × .56 10 Thus: CH3COO- (aq) CH3COOH(aq) OH- (aq) Initial, M 0.5 0 0 Change, M -x +x +x Equilibrium, M 0.5-x x x At equilibrium: 3 (aq) (aq) b 3 (aq) 10 [CH COOH ][OH ] K [CH COO ] x(x) 5.56 10 0.5 x − − − = = × − Here, Kb is very small, so 0.5 – x ≈ 0.5. Thus, x2 = 0.5(5.56 x 10-10) x2 = 2.78 x 10-10 x = 1.67 x 10-5 pOH = -log10[OH- (aq)] = -log10(x) = -log10(1.67 x 10-5) = 4.78 pH = 14-pOH = 9.22 C2 (b). Calculate the pH of 2.0 x 10-7 M HCl(aq). This acid hydrolyzes fully in water. Kw = 1.0 x 10-14. The HCl hydrolyzes fully: HCl(aq) + H2O(l) → H3O+ (aq) + Cl- (aq) Since this is such a low concentration of H3O+ (aq) from the HCl, we must take account of the autohydrolysis of water: 2 H2O(l) Ý H3O+ (aq) + OH- (aq) Kw = 1.0 x 10-14
Thus. [HgOg OH)]=1.0x10- Or H]=1.0x10 [H,Otaa】 But the total negative charge must equal the total positive charge: [H,Oia】=[oHa】+[Ctaa】 H,0J=1.0x10 以O+2.0x10, [H,0a2-2.0×10-H,01】-1.0×104=0 H,0]=20x10'±2.0x10y-40-1.0x105 2(0 2.0×10-7±2.83×10-7 =2.41x10-70r-4.15×10-8 pH=-log1oH,0aa】=-logo(2.41×10-7)=6.62 C3(a).What volume of fluorine gas,measured at STP,is produced by electrolysis of molten KF using a current of 10.0 A for 2.00 hours? KF0→K+F 2F→F2+2e it =nF n=其-10.0Cs'2.00×3600s F- 96,487Cmo -0.746mole 1mol F2=0.373 molFa0) 0.746mole×2mole 0.373molF2g 22.4L=8.36 mol at STP mol C3(b).What mass of potassium metal is produced in this same reaction? K*+e→Ke 0.746 mol e>0.746 mol Kis)
Thus, 14 3 (aq) (aq) 14 (aq) 3 (aq) [H O ][OH ] 1.0 10 or 1.0 10 [OH ] [H O ] + − − − + = × × = − But the total negative charge must equal the total positive charge: 3 (aq) (aq) (aq) 14 7 3 (aq) 3 (aq) 2 7 14 3 (aq) 3 (aq) 7 7 2 3 (aq) 7 7 7 [H O ] [OH ] [Cl ] 1.0 10 [H O ] 2.0 10 [H O ] [H O ] 2.0 10 [H O ] 1.0 10 0 2.0 10 (2.0 10 ) 4(1)( 1.0 10 ) [H O ] 2(1) 2.0 10 2.83 10 2 2.41 10 or 4.15 + −− − + − + + − + − − − + − − − = + × = +× −× −× = × ± × − −× = ×± × = =× − 8 7 10 3 (aq) 10 10 pH log [H O ] log (2.41 10 ) 6.62 − + − × = − =− × = −14 C3 (a). What volume of fluorine gas, measured at STP, is produced by electrolysis of molten KF using a current of 10.0 A for 2.00 hours? KF(l) → K+ + F- 2 F- → F2 + 2 eit = nF 1 1 2(g) 2(g) 2(g) 2(g) it 10.0 C s (2.00 3600)s n F 96,487 C mol 0.746 mol e 1mol F 0.746 mol e 0.373 mol F 2 mol e 22.4 L 0.373 mol F 8.36 mol F at STP mol − − − − − × = = = × = × = C3 (b). What mass of potassium metal is produced in this same reaction? K+ + e- → K(s) 0.746 mol e- → 0.746 mol K(s)
0.746 mol K(s)x 39.1 g/mol 29.2 g K(s)produced Whatyrdich ofh four indicated atomsngnhmolecule ofllns H 0 Atom 1:_sp2 Atom 2:sp Atom3:_sp3」 Atom 4:_sp2 C5(b).Use VSEPR theory to predict the shape of the KrF4 molecule omn the octets onemoes lave electrons,placed as two lone pairs on the Kr atom.It is therefore an AX4E2 structure,which is square planar. C5(c).Why do atomic radii decrease going from left to right across a period? The effective nuclear charge increases going left to right Or Electrons are be added to the same principal shell,hence have ergies.How ver,protons are als 00 d.which collectively exert a greater atiraction for these electrons,pulling them ing adde inwards,reducing the radius C6 (a).At 90C,the equilibrium constant is 0.0068 for the reaction H2()+S(s)=H2S().If 0.15 mol H2(g)and 1.0 mol S(s)are heated in a 1.0 L reaction chamber to 90C,what will be the partial pressures of H2(g)and H2S(g)at equilibrium? Note that the sulphur is a solid and is not considered in the equilibrium expression,and that since the other species are gases,the equilibrium constant is Kp,not K.But you are given molar quantities so you need to convert these to the initial partial pressures:
0.746 mol K(s) x 39.1 g/mol = 29.2 g K(s) produced. C5 (a). What hybridization is each of the four indicated atoms using in the molecule of Vanillin shown below? H O O CH3 1 OH 2 3 4 Atom 1:_sp2 ______ Atom 2:_sp3 ______ Atom 3:_sp3 ______ Atom 4:_sp2 ______ C5 (b). Use VSEPR theory to predict the shape of the KrF4 molecule. 8 + (7 x 4) = 36 electrons Making the four bonds and completing the octets on the F atoms uses 32 of these. This leaves 4 electrons, placed as two lone pairs on the Kr atom. It is therefore an AX4E2 structure, which is square planar. C5 (c). Why do atomic radii decrease going from left to right across a period? The effective nuclear charge increases going left to right Or Electrons are being added to the same principal shell, hence have similar energies. However, protons are also being added, which collectively exert a greater attraction for these electrons, pulling them inwards, reducing the radius. C6 (a). At 90o C, the equilibrium constant is 0.0068 for the reaction H2(g) + S(s) = H2S(g). If 0.15 mol H2(g) and 1.0 mol S(s) are heated in a 1.0 L reaction chamber to 90o C, what will be the partial pressures of H2(g) and H2S(g) at equilibrium? Note that the sulphur is a solid and is not considered in the equilibrium expression, and that since the other species are gases, the equilibrium constant is Kp, not Kc. But you are given molar quantities so you need to convert these to the initial partial pressures:
R.15mW0.02Latmk-m+K.atm 1.0L 0mol0.082LatmK-m0+273K0atm 1.0L TH2S(g Initial,atm Change,atm Equilibrium,atm 4.46-X P5@=K =0.0068 4.46-× ×=0.0068(4.46-x) ×=0.0304-0.0068x 1.0068x=0.0304 ×=0.0301 Therefore,PH2=4.46-x=4.43 atm and pH2s=x=0.0301 atm C6(b).Another 0.15 mol H2)are added to the equilibrium mixture from part(a).Calculate the new partial pressures Before adding the extra hydrogen, 4.43atm(1.0L) =RT(0.082LatmK-'mo)90+273)K 0.149mol Thus,after adding it,we have 0.149+0.15=0.299 mol H2,and its new partial pressure is: = nRT0.299mol(0.082LatmK-mol90+273K8.90 atm 1.0L Then the system re-equilibrates: Initial,atm 8.90 0.030 Change,atm -X Equilibrium,atm 8.90-× 0.0301+X
2 2 2 2 1 1 H H 1 1 H S H S n RT 0.15 mol(0.082LatmK mol )(90 273)K p 4.46 atm V 1.0 L n RT 0 mol(0.082LatmK mol )(90 273)K p 0 V 1.0 L − − − − + = = = + = = = atm H2(g) H2S(g) Initial, atm 4.46 0 Change, atm -x +x Equilibrium, atm 4.46 - x x 2 (g) 2(g) H S p H p K p x 0.0068 4.46 x x 0.0068(4.46 x) x 0.0304 0.0068x 1.0068x 0.0304 x 0.0301 = = − = − = − = = Therefore, pH2 = 4.46 – x = 4.43 atm and pH2S = x = 0.0301 atm C6 (b). Another 0.15 mol H2(g) are added to the equilibrium mixture from part (a). Calculate the new equilibrium partial pressures. Before adding the extra hydrogen, 2 2 H H 1 1 p V 4.43 atm(1.0 L) n 0 RT (0.082LatmK mol )(90 273)K − − = = = + .149 mol Thus, after adding it, we have 0.149 + 0.15 = 0.299 mol H2, and its new partial pressure is: 2 2 1 1 H H n RT 0.299 mol(0.082LatmK mol )(90 273)K p 8 V 1.0 L − − + = = = .90 atm Then the system re-equilibrates: H2(g) H2S(g) Initial, atm 8.90 0.0301 Change, atm -x +x Equilibrium, atm 8.90 - x 0.0301 + x
Ps@=K。 0.0301+×=0.0068 8.90-× 0.0301+×=0.0068(8.90-x) ×+0.0068x=0.06052-0.0301 1.0068x=0.03012 ×=0.0299 Thus,pH2=8.90-×=8.87 atm and p2s=0.0301+×=0.0600atm C6(c).More S(s)is added to the equilibrium mixture from part(b).What happens?Why? Nothing happens.The equilibrium position is not affected by the addition of a solid C7 (a)Lead(Pb)crystallizes in a face centred cubic unit cell with an edge length of 495 pm.Calculate the radius of a Pb atom(in pm). If the edge length=1=495 pm,we can calculate the radius knowing that the diagonal of the face is 4r. Thus,(4)2=2+2.Thus,162=22,or,r= 1_495pm=175.0pm 22-22 (b)Calculate the density of Pb(in g/cm3). The mass of the unit cell is that of four atoms mass=4 atoms(2072gmo)1.38x1g 6.02×1023mo1 volume-3-(495×10-10cm)3-1.21×10-22cm3 density= mass 1.38×10-1g=11.4gcm volume 1.21x10-22cm?
2 (g) 2(g) H S p H p K p 0.0301 x 0.0068 8.90 x 0.0301 x 0.0068(8.90 x) x 0.0068x 0.06052 0.0301 1.0068x 0.03012 x 0.0299 = + = − += − +=− = = Thus, pH2 = 8.90 – x = 8.87 atm and pH2S = 0.0301 + x = 0.0600 atm C6 (c). More S(s) is added to the equilibrium mixture from part (b). What happens? Why? Nothing happens. The equilibrium position is not affected by the addition of a solid. C7 (a) Lead (Pb) crystallizes in a face centred cubic unit cell with an edge length of 495 pm. Calculate the radius of a Pb atom (in pm). If the edge length = l = 495 pm, we can calculate the radius knowing that the diagonal of the face is 4r. Thus, (4r)2 = l2 + l2 . Thus, 16r2 = 2l2 , or, l 495 pm r 1 22 22 == = 75.0 pm (b) Calculate the density of Pb (in g/cm3 ). The mass of the unit cell is that of four atoms: 1 21 23 1 3 10 3 22 21 3 22 3 4 atoms(207.2 g mol ) mass 1.38 10 g 6.02 10 mol volume l (495 10 cm) 1.21 10 cm mass 1.38 10 g density 11.4g cm volume 1.21 10 cm − − − − − − 3 − − = = × == × = × × == = × ×