Part A.60 marks.Answer each question(5 marks each). Does the position of the equilibrium in the reaction 2 CO+O2 COz shift left or right when we: Increase the pressure Righ址 Increase the temperature Left Add some O2() Right Remove some CO2() Righ址t Remove some COg) Left Explain how the effective nuclear charge changes going DOWN a group in the periodic table,and explain why. The effective nuclear charge decreases going down a group because inner electrons shield valence electrons. State whether each of the following processes has a positive or negative entropy change (a)CgH18+12.502g→8C02g+9H20g Positive (b)C02g→C02g Positive (c)H2Og→H2O Negative (d)Na" +Br NaBr( Negative (e)H0at5C→H0at10C Positive A reaction has negative values of both H and S.Will this reaction be spontaneous at high temperature,low temperature,both or neither?How do you know? We know that a reaction is spontaneous if AG<0,and we know that AG=AH-TAS.A negative value of AH tends to make AG negative.However,a negative value of AS tends to make AG positive,unless the temperature is low.The reaction is therefore spontaneous at a low temperature. 5 Which of HNO or HNO is a stronger oxidant?Why? HNO. insronger oidant.The N atom in HNas an oxidation state of whereas itonly 6 Why is the H-C-H bond angle in CH3Cl greater than the ideal tetrahedral angle of 109.5? The Cl atom is very electronegative.This pulls electrons towards itself,lowering the density of electrons in the C-Cl bond.The repulsion between these electrons and those in the C-H bonds is therefore smaller and the H atoms move apart,increasing the H-C-H bond angle
Part A. 60 marks. Answer each question (5 marks each). 1. Does the position of the equilibrium in the reaction 2 CO(g) + O2(g) ∏ 2 CO2(g) shift left or right when we: Increase the pressure Right Increase the temperature Left Add some O2(g) Right Remove some CO2(g) Right Remove some CO(g) Left 2. Explain how the effective nuclear charge changes going DOWN a group in the periodic table, and explain why. The effective nuclear charge decreases going down a group because inner electrons shield valence electrons. 3. State whether each of the following processes has a positive or negative entropy change: (a) C8H18(l) + 12.5 O2(g) → 8 CO2(g) + 9 H2O(g) Positive (b) CO2(s) → CO2(g) Positive (c) H2O(g) → H2O(l) Negative (d) Na+ (g) + Br¯(g) → NaBr(s) Negative (e) H2O(l) at 5°C → H2O(l) at 10°C Positive 4. A reaction has negative values of both H and S. Will this reaction be spontaneous at high temperature, low temperature, both or neither? How do you know? We know that a reaction is spontaneous if ΔG<0, and we know that ΔG = ΔH – TΔS. A negative value of ΔH tends to make ΔG negative. However, a negative value of ΔS tends to make ΔG positive, unless the temperature is low. The reaction is therefore spontaneous at a low temperature. 5. Which of HNO3(aq) or HNO2(aq) is a stronger oxidant? Why? HNO3(aq) is a stronger oxidant. The N atom in HNO3(aq) has an oxidation state of +5, whereas it is only +3 in HNO2(aq). 6. Why is the H-C-H bond angle in CH3Cl greater than the ideal tetrahedral angle of 109.5o ? The Cl atom is very electronegative. This pulls electrons towards itself, lowering the density of electrons in the C-Cl bond. The repulsion between these electrons and those in the C-H bonds is therefore smaller and the H atoms move apart, increasing the H-C-H bond angle
7. In the van der Waals equation,why is the 'a'term for H2O larger than that for CO2? This is due to the fact that water is a polar molecule,causing intermolecular attractions that are stronger than those in CO2. 8. Can H'ions oxidize Cus?Why or why not? No,because hydrogen is above copper in the activity series. 9. Why do deep-sea divers breathe a mixture of oxygen and helium? Nitrogen is quite soluble in blood.As the diver scend ntrogen comes out of the blood as bubbles causing a very painfu l and sometimes fatal condition called the bends.Helium is not vey soluble in blood,and so this problem is avoided. 10. Consider the spontaneous reaction carried out at atmospheric pressure:C3H6)+4.5 O2g)3 COz)+3 H2O).State whether the following variables increase or decrease as the reaction progresses from left to right: Volume of the system Increase Temperature of the surroundings Increase Entropy of the system Increase Entropy of the surroundings Increase Entropy of the Universe Increase 11. In a C molecule.the highest energy electrons are in a bonding molecular orbital.Is the bond length of C2'greater or less than the bond length of C2?How do you know? n must be removed from this bondin C.Thus the bond lensth of C.must be greater than that of C. 12 The electronic configuration of the Mo atom is predicted by AUFBAU to be [Kr]5s24d.What do you think is the actual configuration and why? It is likely to be [Kr]5s'4d,since this configuration has all unpaired electrons,i.e.half-filled subshells. Part B.Answer both questions Bl and B2.(20 marks each) B1.[6 marks](a)At a particular temperature and pressure,it takes 4.55 minutes for a 1.5 L sample of He to agh a porous s membrane How long will it take for 1.5 L of F to effuse under the same
2 7. In the van der Waals equation, why is the ‘a’ term for H2O larger than that for CO2? This is due to the fact that water is a polar molecule, causing intermolecular attractions that are stronger than those in CO2. 8. Can H+ (aq) ions oxidize Cu(s)? Why or why not? No, because hydrogen is above copper in the activity series. 9. Why do deep-sea divers breathe a mixture of oxygen and helium? Nitrogen is quite soluble in blood. As the diver ascends, the nitrogen comes out of the blood as bubbles causing a very painful and sometimes fatal condition called the bends. Helium is not vey soluble in blood, and so this problem is avoided. 10. Consider the spontaneous reaction carried out at atmospheric pressure: C3H6(g) + 4.5 O2(g) ↓ 3 CO2(g) + 3 H2O(g). State whether the following variables increase or decrease as the reaction progresses from left to right: Volume of the system Increase Temperature of the surroundings Increase Entropy of the system Increase Entropy of the surroundings Increase Entropy of the Universe Increase 11. In a C2 molecule, the highest energy electrons are in a π2p bonding molecular orbital. Is the bond length of C2 + greater or less than the bond length of C2? How do you know? Since the electron must be removed from this bonding molecular orbital, the bond order of C2 + will be less than that of C2. Thus the bond length of C2 + must be greater than that of C2. 12. The electronic configuration of the Mo atom is predicted by AUFBAU to be [Kr] 5s2 4d4 . What do you think is the actual configuration and why? It is likely to be [Kr] 5s1 4d5 , since this configuration has all unpaired electrons, i.e. half-filled subshells. Part B. Answer both questions B1 and B2. (20 marks each) B1. [6 marks] (a) At a particular temperature and pressure, it takes 4.55 minutes for a 1.5 L sample of He(g) to effuse through a porous membrane. How long will it take for 1.5 L of F2(g) to effuse under the same conditions?
timer rateus MWe MW. =4.55min 38.0gmol▣ V4.00g mol- =14.0min (b)[6 marks]A sample of Freon-12(CF2Cl)occupies 24.5 L at 298 K and 253.3 kPa.Find its volume at STP. PV-P:V nT nT in this example.nVor VV TT TP2 Note that STP means I atm(101,325 Pa)and C(273 K).Thus, V=253,30Pa245LX273K=561L 298K(101,325Pa) (c)[8 marks]Chlorine is produced electrochemically from seawater and collected in a sealed container.If a 15.0 L container holds 0.580 kg of Clag at 200C,calculate the pressure (in atm)using the van der Waals equation.For Cla(),a=6.49 atm L'mol and b=0.0562 L mol n= 580g 2(35.5g mo8.17mol 15.0L-8.17mol(0.0562Lmol) 15.0L -21.79-1.93atm =19.86atm B2.(a)[5 marks]Beginning with a sulphide ore such as NiSs).show the reactions by which acid rain is produced. +50二s00+s0 S02g)+0 (b)[5 marks]Beginning with NO show the reactions that produce ozone at ground level,i.e.in smog. NOg+⅓O2g→NO2e NO2g+hv→NOg+Oe 0g+02g)→03g (c)[5 marks]Show the reactions in the catalytic cycle by which Cl atoms destroy ozone in the stratosphere
2 2 2 2 2 F F He He F He 1 F F He 1 He time rate MW time rate MW MW 38.0 g mol thus, time time 4.55 min MW 4.00 g mol 14.0 min − − = = = = = (b) [6 marks] A sample of Freon-12 (CF2Cl2) occupies 24.5 L at 298 K and 253.3 kPa. Find its volume at STP. 11 2 2 11 2 2 pV pV nT nT = In this example, n1 = n2, thus 11 2 2 1 2 pV pV T T = , or 112 2 1 2 p VT V T p = Note that STP means 1 atm (101,325 Pa) and 0o C (273 K). Thus, 2 253,300 Pa(24.5 L)(273)K V 5 298K(101,325 Pa) = = 6.1L (c) [8 marks] Chlorine is produced electrochemically from seawater and collected in a sealed container. If a 15.0 L container holds 0.580 kg of Cl2(g) at 200o C, calculate the pressure (in atm) using the van der Waals equation. For Cl2(g), a = 6.49 atm L2 mol-2 and b = 0.0562 L mol-1 1 580 g n 8 2(35.5 g mol ) − = = .17 mol 2 2 1 1 2 2 1 nRT n p a V nb V 8.17 mol(0.082 L atm K mol )(200 273)K 8.17 mol 6.49 atm L mol 15.0 L 8.17 mol(0.0562 L mol ) 15.0 L 21.79 1.93 atm 19.86 atm − − − − ⎛ ⎞ = − ⎜ ⎟ − ⎝ ⎠ + ⎛ ⎞ = − ⎜ ⎟ − ⎝ ⎠ = − = B2. (a) [5 marks] Beginning with a sulphide ore such as NiS(s), show the reactions by which acid rain is produced. NiS(s) + 1.5 O2(g) → NiO(s) + SO2(g) SO2(g) + ½ O2(g) → SO3(g) SO3(g) + H2O(l) → H2SO4(aq) (b) [5 marks] Beginning with NO(g), show the reactions that produce ozone at ground level, i.e. in smog. NO(g) + ½ O2(g) → NO2(g) NO2(g) + hν → NO(g) + O(g) O(g) + O2(g) → O3(g) (c) [5 marks] Show the reactions in the catalytic cycle by which Cl atoms destroy ozone in the stratosphere. 3
Clg+03g)→C10g+02g) CIOg+0g→Clg)+O2g) 器o+20e-20au+2c0e 2C10g→Cl202e C1202g+hv→2Clg+O2e (d)[5 marks]Show the four oxygen-only reactions in the stratosphere collectively known as the Chapman cycle 02g+hv→0g)+Oe 0g)+02g)→03g O3g)+hv→02g+Og 0g+03g→202g Part C.Attempt all five questions CI to C5.The best four will be used to calculate your mark.(20 marks each) C1.(a)[10marks]Ozone(O3)protects us from damaging radiation from the Sun because photons having a wavelength lower than 27 nm are absorbed causing a bond to break in the O3 molecule.Calculate the bond energy of this bond (kJ mol). E=hv= hc-663x104]3300x10ms-737×10 (perphoton）) 270×10°m ×6.02×1023mo=443,473Jmol=443 kJ mol1 (b)[8 marks]Calculate the frequency(s)of the photon emitted by a hydrogen atom when an electron falls from n=7 to m=3. 司 =01097m[- =9.95×10nm 1 1=g95x10-nm=105nm c_3.00×10ms1 v=元=105x10m =2.99x104s (c)[2 marks]Are the photons in part(b)in the UV,visible or IR portion of the electromagnetic spectrum? 1005 nm photons are in the IR portion of the spectrum. C2.(a)[5 marks]Two structures for Cl2O are possible,Cl-O-CI and O-CI-CI.Draw the Lewis structures for these and use formal charge to predict which structure is more likely
Cl(g) + O3(g) → ClO(g) + O2(g) ClO(g) + O(g) → Cl(g) + O2(g) or 2 Cl(g) + 2 O3(g) → 2 O2(g) + 2 ClO(g) 2 ClO(g) → Cl2O2(g) Cl2O2(g) + hν → 2 Cl(g) + O2(g) (d) [5 marks] Show the four oxygen-only reactions in the stratosphere collectively known as the Chapman cycle O2(g) + hν → O(g) + O(g) O(g) + O2(g) → O3(g) O3(g) + hν → O2(g) + O(g) O(g) + O3(g) → 2 O2(g) Part C. Attempt all five questions C1 to C5. The best four will be used to calculate your mark. (20 marks each) C1. (a) [10marks] Ozone (O3) protects us from damaging radiation from the Sun because photons having a wavelength lower than 270 nm are absorbed causing a bond to break in the O3 molecule. Calculate the bond energy of this bond (kJ mol-1). 8 1 34 19 9 23 1 1 1 hc 3.00 10 m s E h 6.63 10 J s 7.37 10 J (per photon) 270 10 m 6.02 10 mol 443, 473 J mol 443 kJ mol − − − − − −− × = ν= = × = × λ × ×× = = (b) [8 marks] Calculate the frequency (s-1) of the photon emitted by a hydrogen atom when an electron falls from n=7 to m=3. 2 2 1 2 2 4 1 4 1 8 1 14 1 9 1 11 R m n 1 1 0.01097 nm 3 7 9.95 10 nm 1 1005 nm 9.95 10 nm c 3.00 10 m s 2.99 10 s 1005 10 m − − − − − − − − ⎡ ⎤ = − ⎢ ⎥ λ ⎣ ⎦ ⎡ ⎤ = − ⎢ ⎥ ⎣ ⎦ = × λ = = × × ν= = = × λ × (c) [2 marks] Are the photons in part (b) in the UV, visible or IR portion of the electromagnetic spectrum? 1005 nm photons are in the IR portion of the spectrum. C2. (a) [5 marks] Two structures for Cl2O are possible, Cl-O-Cl and O-Cl-Cl. Draw the Lewis structures for these and use formal charge to predict which structure is more likely. 4
ci0-ci: ICLLOLCu Valence 67 Electrons assigned 7 67 Formal Charge 000 .0 OCI CI Valence 676 ns assigned 7 67 Formal Charge -11 Cl-O-CI is therefore the more likely structure because the formal charges are all zero. (b)[2 marks each]Name the type of hybrid orbitals used by the central atom in each of the following three structures: AsFs SiCL Br3 5+5x7)=40e 4+(4x7=32e (3x7+1=22e Thus,AXs(5 charge clouds on the As) Thus,AX(4 charge clouds on the Si) Thus,AX2E (5 charge clouds on the central Br) Thus,sp'd hybrids Thus,sp'hybrids Thus,sp'd hybrids (c)[3 marks each]Use VSEPR to predict the shapes of the following three species.Wrong name=zero marks. CNO AsF4 XeOF4 (the N atom is in the centre) 4+5+6+1=16 electrons 5+(4x7)=33e 8+6+(4x7)=42e Thus AXE Thus,AXsE Thus,see-saw Thus,square pyramid C3.[20 marks]Balance the following REDOX reaction in a basic solution: CN(ag)+MnO4 (ag)>CNO (ag)+MnO2(s) Oxidation: CN (ad) CNO CNm+H60u→ CNO() CN (ag)+H2O)CNO (+2 H(ag) CN(ag)+H2O()>CNO (ag)+2 H'(ag)+2e
5 Cl O Cl Valence 7 6 7 Electrons assigned 7 6 7 Formal Charge 0 0 0 O Cl Cl Valence 6 7 6 Electrons assigned 7 6 7 Formal Charge -1 1 -1 Cl-O-Cl is therefore the more likely structure because the formal charges are all zero. (b) [2 marks each] Name the type of hybrid orbitals used by the central atom in each of the following three structures: AsF5 SiCl4 Br3¯ 5+(5x7) = 40 e¯ 4+(4x7) = 32 e¯ (3x7) + 1 = 22 e¯ Thus, AX5 (5 charge clouds on the As) Thus, AX4 (4 charge clouds on the Si) Thus, AX2E3 (5 charge clouds on the central Br) Thus, sp3 d hybrids Thus, sp3 hybrids Thus, sp3 d hybrids (c) [3 marks each] Use VSEPR to predict the shapes of the following three species. Wrong name = zero marks. CNO- AsF4 XeOF4 (the N atom is in the centre) 4+5+6+1 = 16 electrons 5 + (4 x 7) = 33 e- 8 + 6 + (4 x 7) = 42 eC=N=O, with two lone pairs on each Thus, AX4E Thus, AX5E of the C and O atoms uses all 16 electrons Thus, see-saw Thus, square pyramid Thus, AX2 Thus, linear C3. [20 marks] Balance the following REDOX reaction in a basic solution: CN¯ (aq) + MnO4¯ (aq) → CNO¯ (aq) + MnO2(s) Oxidation: CN¯ (aq) → CNO¯ (aq) CN¯ (aq) + H2O(l) → CNO¯ (aq) CN¯ (aq) + H2O(l) → CNO¯ (aq) + 2 H+ (aq) CN¯ (aq) + H2O(l) → CNO¯ (aq) + 2 H+ (aq) + 2 e¯ .. .. .. .. .. .. O Cl Cl : : .. .. .. .. .. .. Cl O Cl : :
Reduction: MnO4(am→MnO2g MnO4aq→MnO2s+2HOm Mn0,四+4Heo+3e→Mn02g+2H0o Overall 3(CN+H:Od →CN0+2H(a+2e) +2(Mn04(a+4Hq+3e→Mn02s+2H0） 30侧t30w→3CN0四+6H四+6c ag)+8 H'(ag)+6e>2 MnOz(s)+4 H2O 3 CN(a)+2 MnO4()+2 H'>3 CNO ()+2 MnO2()+H2O) Convert to basic solution: 3 CN (ag)+2 MnO4(ag)+2 H'(ag)+2OH (ag)3 CNO ()+2 MnO2(s)+H2O)+2 OH ( 3CN (+2 MnO4()+2 H2O>3 CNO()+2 MnO2(s)+H2O+2OH() 3CN(ag)+2 MnO4()+H2O>3 CNO()+2 MnO2()+2 OH (Q) Check Left Right 3 N 3 3 Mn 0 0 0 Charge-5-5 C4.(a)[10 marks]The reaction Ba+SOBaSOs is carried out in a calorimeter.When 2.00 mol Ba" aremixedth00molLof water atthe temperature of the solution rises to 28.1C.Assuming that the calorimeter itself absorbs a negligible amount of heat,that the final solution has a density of 1.00g mL,and that the specific heat capacity of the solution is 4.18JCg calculate the enthalpy of reaction(AH)per mole of BasO)formed. 4.00L=4,000ml 4,000 mL x 1.00 g mL=4.000 g water q=mCp△T =4,000g(4.18J℃g)28.1-25.0)℃ =51.832J AH=-51,832 J/2 mol=-25,916 J/mol =-25.9 kJ/mol (negative because heat is lost by the system) (b)[10 marks]Calculate the standard enthalpy of formation of B2H given the following data 6
6 Reduction: MnO4¯ (aq) → MnO2(s) MnO4¯ (aq) → MnO2(s) + 2 H2O(l) MnO4¯ (aq) + 4 H+ (aq) + 3e¯ → MnO2(s) + 2 H2O(l) Overall: 3(CN¯ (aq) + H2O(l) → CNO¯ (aq) + 2 H+ (aq) + 2 e¯) + 2(MnO4¯ (aq) + 4 H+ (aq) + 3e¯ → MnO2(s) + 2 H2O(l)) = 3 CN¯ (aq) + 3 H2O(l) → 3 CNO¯ (aq) + 6 H+ (aq) + 6 e¯ 2 MnO4¯ (aq) + 8 H+ (aq) + 6e¯ → 2 MnO2(s) + 4 H2O(l) _________________________________________________________ 3 CN¯ (aq) + 2 MnO4¯ (aq) + 2 H+ (aq) → 3 CNO¯ (aq) + 2 MnO2(s) + H2O(l) Convert to basic solution: 3 CN¯ (aq) + 2 MnO4¯ (aq) + 2 H+ (aq) + 2 OH¯ (aq) → 3 CNO¯ (aq) + 2 MnO2(s) + H2O(l) + 2 OH¯ (aq) 3 CN¯ (aq) + 2 MnO4¯ (aq) + 2 H2O(l) → 3 CNO¯ (aq) + 2 MnO2(s) + H2O(l) + 2 OH¯ (aq) 3 CN¯ (aq) + 2 MnO4¯ (aq) + H2O(l) → 3 CNO¯ (aq) + 2 MnO2(s) + 2 OH¯ (aq) Check: Left Right C 3 3 N 3 3 Mn 2 2 O 9 9 H 2 2 Charge -5 -5 C4. (a) [10 marks] The reaction Ba+2 (aq) + SO4 -2 (aq) → BaSO4(s) is carried out in a calorimeter. When 2.00 mol Ba+2 (aq) are mixed with 2.00 mol SO4 -2 (aq) in 4.00 L of water at 25o C, the temperature of the solution rises to 28.1o C. Assuming that the calorimeter itself absorbs a negligible amount of heat, that the final solution has a density of 1.00 g mL-1, and that the specific heat capacity of the solution is 4.18 J o C-1 g -1, calculate the enthalpy of reaction (ΔH) per mole of BaSO4(s) formed. 4.00 L = 4,000 mL 4,000 mL x 1.00 g mL-1 = 4,000 g water q = mCpΔT = 4,000 g (4.18 J o C-1 g -1)(28.1 – 25.0)o C = 51,832 J ΔH = - 51,832 J / 2 mol = - 25,916 J/mol = - 25.9 kJ/mol (negative because heat is lost by the system) (b) [10 marks] Calculate the standard enthalpy of formation of B2H6(g) given the following data:
2Bg+3/202g→B20g △H=-1273 kJ mol +302() B,0+3H0g AH=-2035 kJ mol! He+⅓02g一H0 AH=-286 kJ mol H2O0→HOg AH=+44 kJ mol We want the reaction:2 B+3(g) B29+3H9g→B2H6g+30 AH=-(-2035 kJ mol)=+2035 kJ mol 2Bs+32024e→B2034 △H=-1273 kJ mol 3H2B+32024e→3H04 AH=3(-286 kJ mol)=-858 kJ mol 3H9e→3H9 AH=3(+44 kJ mol)=+132 kJ mol 2B+3Hg→BH6e △H=+36kJmo' C5.[20marks]For the reaction Fe ant Ke=1100 at 25C Calculat the ns of Fe SCN (ag)at equil (p and g) (ag) FeSCN() Initial,M 002 010 10 Change,M -X 一X +X Equilibrium,M 0.02-x 0.10-xx At equilibrium, [FeSCN2am】 e"✉lscN10 Thus. (0.02-xX0.1-3=100 x=1100(0.02-x)0.1-x) x=1100.002-02x-1x+x2) x=2.2-22x-110x+1100x 1100x2-133x+2.2=0 X=-b±y6-4c_133±V133-4010y22133t895 2a 21100) 2200 0.101or0.01977 x=0.101 would give negative concentrations,thus x=0.01977 Thus,Feo]=0.02-x=0.000227 [SCNol=0.10-x=0.08023 Fe(SCN]=x=0.01977
2 B(s) + 3/2 O2(g) → B2O3(s) ΔH = −1273 kJ mol-1 BB 2H6(g) + 3 O2(g) → B2O3(s) + 3 H2O(g) ΔH = −2035 kJ mol-1 H2(g) + ½ O2(g) → H2O(l) ΔH = −286 kJ mol-1 H2O(l) → H2O(g) ΔH = +44 kJ mol-1 We want the reaction: 2 B(s) + 3 H2(g) → B2H6(g) B2O3(s) + 3 H2O(g) → B2H6(g) + 3 O2(g) ΔH = −(−2035 kJ mol-1) = +2035 kJ mol-1 2 B(s) + 3/2 O2(g) → B2O3(s) ΔH = −1273 kJ mol-1 3 H2(g) + 3/2 O2(g) → 3 H2O(l) ΔH = 3(−286 kJ mol-1) = −858 kJ mol-1 3 H2O(l) → 3 H2O(g) ΔH = 3(+44 kJ mol-1) = +132 kJ mol-1 _______________________________________________________________________ 2 B(s) + 3 H2(g) → B2H6(g) ΔH = +36 kJ mol-1 C5. [20 marks] For the reaction Fe+3 (aq) + SCN- (aq) ∏ FeSCN+2 (aq), the equilibrium constant Kc = 1100 at 25o C. Calculate the concentrations of Fe+3 (aq), SCN- (aq) and FeSCN+2 (aq) at equilibrium if 0.0200 mol Fe+3 (aq) and 0.1000 mol SCN- (aq) are added to 1.000 L water at 25o C. Fe+3 (aq) SCN- (aq) FeSCN+2 (aq) Initial, M 0.02 0.10 0 Change, M -x -x +x Equilibrium, M 0.02 - x 0.10 - x x At equilibrium, 2 (aq) 3 (aq) (aq) [FeSCN ] 1100 [Fe ][SCN ] + + − = Thus, 2 2 2 2 2 x 1100 (0.02 x)(0.1 x) x 1100(0.02 x)(0.1 x) x 1100(.002 .02x .1x x ) x 2.2 22x 110x 1100x 1100x 133x 2.2 0 b b 4ac 133 133 4(1100)(2.2) 133 89.5 x 0.101 or 0.01977 2a 2(1100) 2200 = − − = −− = − −+ =− − + − += −± − ± − ± = = == x = 0.101 would give negative concentrations, thus x = 0.01977 Thus, [Fe+3 (aq)] = 0.02 – x = 0.000227 [SCN- (aq)] = 0.10 – x = 0.08023 [Fe(SCN)- ] = x = 0.01977 7