Part A.Answer each question briefly(5 marks each). Al Predict the signs of aH and as and under what conditions the reaction is spontaneous Is the reaction Process AH°(Check One) AS"(Check One!) spontaneous below some Positive Negative Positive Negative T,above HOg→HO0 Below NaCls→Nag+CTg Above 2C2H2g+502g→4C02g+2 Below H2O() AtallT (The solution is observed to get hot) 16 CO2()+18 H2O(g)-2 CsHi8g)+ Never 2502e A2.Why is sodium(Na)higher than gold (Au)on the activity series? Sodium is more easily oxidized. A3.Write the oxidation numbers of the indicated atoms in each substance Oxidation Species Atom Number Ga +3 GaAs As 3 U +3 UH3 -1 Hgo) 0 Br20) Br 0 SO; +4
Part A. Answer each question briefly (5 marks each). A1. Predict the signs of ΔHo and ΔSo , and under what conditions the reaction is spontaneous: ΔHo (Check One!) ΔSo Process (Check One!) Positive Negative Positive Negative Is the reaction spontaneous below some T, above H2O(g) → H2O(l) 3 3 Below NaCl(s) → Na+ (g) + Cl- (g) 3 3 Above 2 C2H2(g) + 5 O2(g) → 4 CO2(g) + 2 H2O(g) 3 3 Below CaBr2(s) ⎯H O ⎯⎯2 → Ca+2 (aq) + 2 Br- (aq) (The solution is observed to get hot) 3 3 At all T 16 CO2(g) + 18 H2O(g) → 2 C8H18(g) + 25 O2(g) 3 3 Never A2. Why is sodium (Na) higher than gold (Au) on the activity series? Sodium is more easily oxidized. A3. Write the oxidation numbers of the indicated atoms in each substance: Species Atom Oxidation Number Ga +3 GaAs As -3 U +3 UH3 H -1 Hg(l) Hg 0 Br2(l) Br 0 SO3 –2 S +4
0 -2 Al +3 -1 A4.Which direction(left or right)will the equilibrium reaction CO+CO)shift if we do the following: Direction that the Change equilibrium will shift (check one!) Left Right Lower the temperature Add some O2() Remove some COxg) Increase the pressure Dilute the system by adding some Heg Note the word"Dilute"in the last question.This means that the partial pressures of all gases are decreased.Thus Q>K,since the denominator in Q decreases faster than the numerator.Thus the left shift. A5.Why are condensed phases such as solids and pure liquids not included in an expression for an equilibrium constant? The concentrations of these species do not change as they are used up or created in a reaction both the quantity and the volume change,but concentration is the quotient of the two,which does not change A6.Define all variables in the very important equation AG=-RT In(Keq).Be specific. AGis the free energy change of the reaction under standard conditions(29 K,all gases at atm pressure,all solution species at 1 mol/L) Tis the absolute temperature
O -2 Al +3 AlCl3 Cl -1 A4. Which direction (left or right) will the equilibrium reaction CO(g) + ½ O2(g) ∏ CO2(g) shift if we do the following: Direction that the equilibrium will shift (check one!) Change Left Right Lower the temperature 3 Add some O2(g) 3 Remove some CO2(g) 3 Increase the pressure 3 Dilute the system by adding some He(g) 3 Note the word “Dilute” in the last question. This means that the partial pressures of all gases are decreased. Thus Q>Keq, since the denominator in Q decreases faster than the numerator. Thus the left shift. A5. Why are condensed phases such as solids and pure liquids not included in an expression for an equilibrium constant? The concentrations of these species do not change as they are used up or created in a reaction – both the quantity and the volume change, but concentration is the quotient of the two, which does not change. A6. Define all variables in the very important equation ΔGo = -RT ln(Keq). Be specific. ΔGo is the free energy change of the reaction under standard conditions (298 K, all gases at1 atm pressure, all solution species at 1 mol/L) T is the absolute temperature
Kis the equilibrium constant for the reaction at temperature T Part B.Answer each question.Best two answers will count.(20 marks each) B1.Balance the following reaction in a basic solution:S2O+ClO2(SO+Clg) S,062e→2S042am S2062em+2H200→2S042a S2O6+2 H2O)>2 SO4)+4 H(o) S2062m+2H200→2S042ag+4fam+2e Then the reduction reaction: 2CI0:eo→Cl2g 2C102am→C12g+4H20 2ClO2 (ag)+8 H'(ag)>Cl2(g)+4 H2O0) 2ClOz(+8 H'ag)+6eCl2(g)+4 H2O) Multiply the oxidation reaction by 3 to make the electrons cancel and add the two half reactions: 3S2062a+6H20→6S042m+12Hfam+6c 2 ClO2 (ag)+8 H'(ag)+6 e>Cl2(g)+4 H2O) 3S2062eg+2H0u+2C10ze侧→6S042am+4Hm+C12g To convert to basic solution,add 4OH to both sides 3 S206(ag)+2 H2O+2 CIO2(g)+4 OH(og)6 SO4(ag)+4 H(ag)+4OH(ag)+Cl2(g) and cancel the resultant waters:
Keq is the equilibrium constant for the reaction at temperature T Part B. Answer each question. Best two answers will count. (20 marks each) B1. Balance the following reaction in a basic solution: S2O6 -2 (aq) + ClO2 − (aq) → SO4 −2 (aq) + Cl2(g) S2O6 -2 (aq) → 2 SO4 -2 (aq) S2O6 -2 (aq) + 2 H2O(l) → 2 SO4 -2 (aq) S2O6 -2 (aq) + 2 H2O(l) → 2 SO4 -2 (aq) + 4 H+ (aq) S2O6 -2 (aq) + 2 H2O(l) → 2 SO4 -2 (aq) + 4 H+ (aq) + 2 eThen the reduction reaction: 2ClO2 - (aq) → Cl2(g) 2ClO2 - (aq) → Cl2(g) + 4 H2O(l) 2ClO2 - (aq) + 8 H+ (aq) → Cl2(g) + 4 H2O(l) 2ClO2 - (aq) + 8 H+ (aq) + 6 e- → Cl2(g) + 4 H2O(l) Multiply the oxidation reaction by 3 to make the electrons cancel and add the two half reactions: 3 S2O6 -2 (aq) + 6 H2O(l) → 6 SO4 -2 (aq) + 12 H+ (aq) + 6 e- 2 ClO2 - (aq) + 8 H+ (aq) + 6 e- → Cl2(g) + 4 H2O(l) ___________________________________________ 3 S2O6 -2 (aq) + 2 H2O(l) + 2ClO2 - (aq) → 6 SO4 -2 (aq) + 4 H+ (aq) + Cl2(g) To convert to basic solution, add 4 OH- (aq) to both sides: 3 S2O6 -2 (aq) + 2 H2O(l) + 2 ClO2 - (aq) + 4 OH- (aq) → 6 SO4 -2 (aq) + 4 H+ (aq) + 4 OH- (aq)+ Cl2(g) and cancel the resultant waters:
3 S206ag)+2 CIOz (ag+4 OH(g>6 SO4ag)+Claig+2 H2O) B2.Hydrogen gas can be produced from methane according to the reaction CH+H2OY CO+3 H2)This reaction has Kp=0.02 at a particular temperature. If a reaction vessel is charged with CH and HO at this temperature such that initially P1atm,find the quiibrim partial pressures of all four gases. CHgH0 CO(g H细 Initial pressure,atm 1.0 1.0 0 0 Change in pressure,atm 一X -X 十X +3x Final pressure,atm 1.0-x1.0-x 3 Thus at equilibrium, PcoPit=K PCH,PH2O x(3x)3 10-x1.0-3=K, 27x4 10-x=K。 √K,(1.0-x)=V27x 27x2+K,x-K,=0
3 S2O6 -2 (aq) + 2 ClO2 - (aq) + 4 OH- (aq) → 6 SO4 -2 (aq) + Cl2(g) + 2 H2O(l) B2. Hydrogen gas can be produced from methane according to the reaction CH4(g) + H2O(g) Ý CO(g) + 3 H2(g). This reaction has Kp = 0.02 at a particular temperature. If a reaction vessel is charged with CH4(g) and H2O(g) at this temperature such that initially 1 atm, find the equilibrium partial pressures of all four gases. CH H O p p = 4 2 = CH4(g) H2O(g) CO(g) H2(g) Initial pressure, atm 1.0 1.0 0 0 Change in pressure, atm -x -x +x +3x Final pressure, atm 1.0 - x 1.0 - x x 3x Thus at equilibrium, CO H p CH H O p p p p p p p p K p p x( x) K ( . x)( . x) x K ( . x) x K . x K ( . x) x x Kx K = = − − = − = − − = + −= 2 4 2 3 3 4 2 2 2 2 3 10 10 27 1 0 27 1 0 1 0 27 27 0
here, a=-√27 b=K c=-K, x=-b±6-4ac.K±K,-427-K 2a 2(√27列 =-0.141±6.02-45196j-0140 2(5.196) -0.141±1718-01520r-0.179 10.392 Thus, Pal,=1.0-x=0.848atm P40=1.0-x=0.848atm Pco =x=0.152 atm PH,=3x=0.456atm x(3x)3 27x4 27(0.152) Ceka0-M10-Na0-x100152-020-K B3.Nitric oxide,NO is produced in vehicle engines according to N+Y2NO Using the data in the table below. H's kJ S,JKmol mol-i Nze) 0 192 02细 0 205 NO +90.0 211
p p p p here, a b K c K b b ac K K ( )( K x a ( ) . . ( . )( . ) (. ) . . . or . . = = = − −±− − −± − = = −± − − = − ± = − 2 27 4 27 4 2 2 27 0 141 0 02 4 5 196 0 141 2 5 196 0 141 1 718 0 152 0 179 10 392 p ) Thus, CH H O CO H p p . x . atm p . x . atm p x . atm p x . atm x( x) x ( . ) Check : . K ( . x)( . x) ( . x) ( . . ) = −= = −= = = = = == = −− − − 4 2 2 34 4 2 2 1 0 0 848 1 0 0 848 0 152 3 0 456 3 27 27 0 152 0 020 1 0 1 0 1 0 1 0 0 152 = B3. Nitric oxide, NO(g), is produced in vehicle engines according to N2(g) + O2(g) Ý 2 NO(g). Using the data in the table below, Ho f, kJ mol-1 So , J K-1 mol- 1 N2(g) 0 192 O2(g) 0 205 NO(g) +90.0 211
(a)Calculate(in kJ/mol)for the reaction. H=2HNOe）-HN2e）-H(O2e =180.0-0-0 =180.0 kJ/mol (b)Calculate S(in JKmol)for the reaction. S=2(NO()-S(N2()-S(O2(g) =2(211)-192-205 =+25 JK!mol (c)Calculate G(in kJ/mol)for the reaction at 500C G=H°.TS° =180,000J/mol-(500+273)K(25JKmo =+160675J/mol =+160.7kJ/mol (d)Calculate the value of Kp at 500C. G°=H°-298(S) =180,000J/mol-298K(25 JK!mol) =+172550Jmol -172550Jmo1 8.314JK-mo-1(500+273)K =2.19×10-12 (e)If po2=0.01 atm and pN2=0.7 atm,calculate pNo (in atm)at equilibrium at 500C. K品 Thus,PNO(KPN P =(2.19x102(0.7x0.01)12=1.24x10atm
(a) Calculate ° (in kJ/mol) for the reaction. H° = 2 Hf°(NO(g)) - Hf°(N2(g)) - Hf°(O2(g)) = 180.0 – 0 – 0 = 180.0 kJ/mol (b) Calculate S° (in J K-1 mol-1) for the reaction. S° = 2 S°(NO(g)) - S°(N2(g)) - S°(O2(g)) = 2(211) – 192 – 205 = +25 J K-1 mol-1 (c) Calculate G (in kJ/mol) for the reaction at 500o C . G = H° - T S° = 180,000 J/mol – (500+273) K (25 J K-1 mol-1) = +160675 J/mol = +160.7 kJ/mol (d) Calculate the value of Kp at 500°C. Go = H° - 298( S°) = 180,000 J/mol – 298 K (25 J K-1 mol-1) = +172550 J/mol Kp = o G J mol exp exp . RT . J K mol ( )K − − − − ⎡⎤⎡ −Δ − ⎤ = = ⎢⎥⎢ ⎥ ⎣⎦⎣ + ⎦ 1 12 1 1 172550 2 19 10 8 314 500 273 × (e) If pO2 = 0.01 atm and pN2 = 0.7 atm, calculate pNO (in atm) at equilibrium at 500°C. ON 22 NO 2 p pp p K = Thus, pNO = (Kp pN2 pO2) 1/2 = (2.19 x 10-12 (0.7 x 0.01))1/2 = 1.24 x 10-7 atm
