CHEM 217 Problems Chapter 5, Fall 2011

~t ~ For the reaction N2(g) + 3H2(g) = 2NH3(g), K = 1.60 x 10-4 at 400 °C. Calculate (a) ~rGo and (b) ~rG when the pressures of N2 and H2 are maintained at 10 and 30 bar, respectively, and NH3 is removed at a partial pressure of 3 bar. (c) Is the reaction spontaneous under the latter conditions? SOLUTION 5.1 (a) ~rGO = -RTln K = -(8.314 J K-l mol-l)(673 K) In 1.60 x 10-4 = 48.9 kJ mol-l (b) I Yes (c) 5.2 At 1 :3 mixture of nitrogen and hydrogen was passed over a catalyst at 450 °C. It was found that 2.04% by volllme of ammonia was fonned when the total pressure was maintained at 10.13 bar. Calculate the value ofKfor~ H2(g) + ~N2(g) = NH3(g) at this temperature. SOLUTION At equilibrium PH2 + PN2 + PNH3 =10.13 bar PNH3 = (10.13 bar)(0.0204) = 0.207 bar PH2 + PN2 = 10.13 bar -0.207 bar = 9.923 bar PH2 = 3PN2 because this initial ratio is not changed by reaction Chapter 5/Chemical Equilibrium 66 5.3 At 55 °C and 1 bar the average molar mass of partially dissociated N2O4 is 61.2 9 mol-l. Calculate (a) ~ and (b) Kfor the reaction N2O4(g) = 2NO2(g). (c) Calculate ~ at 55 °C if the total pressure is reduced to 0.1 bar. SOLUTION 92.0- 61.2 = 0.503 61.2 4(0.503)2 = 1.36 = 1 -0.5032 1= -MI -M2 - ~ -M2 - 4C}(P/PO) K= (a) l-C} ~ (c) K 1.36 =~ -- f;2 1 -f;2 -4(PlPO) C; = 0.879 (Note that C; is the dimensionless extent of reaction.) A 11iter reaction vessel containing 0.233 mol of N2 and 0.341 mol of PCl5 is heated to 250 °C. The total pressure at equilibrium is 29.33 bar. Assuming that all the gases are ideal, calculate K for the only reaction that occurs. PCI5(g) = PCI3(g) + Cl2(g) SOLUTION 5.4 PCl3 Cl2 PC1S + = 0 0 initial 0.341 9 9 total = 0.341 + 9 0.341 -~ eq. 67 Chapter 5/Chemical Equilibrium An evacuated tube containing 5.96 x 10-3 mol L -I of solid iodine is heated to 973 K. The experimentally determined pressure is 0.496 bar. Assuming ideal gas behavior, calculate K for 12(g) = 21(g). 5.5 SOLUTION n P=VRT = 0.0288 4gz(PlPO) K= 4(0.0288)2(0.496) = 1.64 x 10-3 = 1 -0.02872 1-1:;2 5.6 Nitrogen trioxide dissociates according to the reaction N203(g) = NO2(g) + NO(g) When one mole of N203(g) is held at 25 °C and 1 bar total pressure until equilibrium is reached, the extent of reaction is 0.30. What is LlrGo for this reaction at 25 °C? SOLUTION NO2(g) + NO(g) N203(g) = 0 0 init. 1 total = 1 + ~ 9 ~ 1-~ eq. mole fr . c; 1+C; p = 29.33 bar -(0.233 mol)(0.08314 ~ ~ar K-lmol-l )(52 (0.341 + ~(0.08314)(523) = 19.20 bar = 1 ~ = 0.1005 K = (0.1005)2(19.2) = 1 83 0.3412- 0.10052 . 68 Chapter 5/Chemical Equilibrium *5.7 For the reaction 2lfl(g) = lI2(g) + 12(g) at 698.6 K, K = 1.83 x 10-2. (a) lIow many grams of hydrogen iodide will be formed when 10 9 of iodine and 0.2 9 of hydrogen are heated to this temperature in a 3 L vessel ? (b) What will be the partial pressures of lI2, Iz, and lII? SOLUTION (a) Pressures due to reactants prior to reaction: (10 g)(0.08314 Lbar K-l mol-l)(698.6 K) = 0.762 bar PI2 = (254 9 mol-l)(3 L) (0.2)(0.08314 L bar K-l mol-l)(698.6 K) = 1.936 bar PH2= 2x3 -(0.762 -x)(1.936 -x) = 1.83 x 10-2 K- (2x)2 x = 0.730 bar (b) PH2 = 1.936- 0.730 = 1.206 bar Pl2 = 0.762 -0.730 = 0.032 bar PHI = 1.460 bar 5.8 Express K for the reaction CO(g) + 3H2(g) = ClI4(g) + H2O(g) in tenns of the equilibrium extent of reaction ~ when one mole of CO is mixed with one mole of hydrogen. SOLUTION co 3H2 CH4 H2O + + = initial 1 1 0 0 l-C; 1-3~ ~ total = 2 -2~ equilibrium ; (-L)2(~)2 '2 -2C;" , PO" K= ( ~ )(~)3( ~ )4 2-2C; 2-2C; po 69 Chapter 5/Chemical Equilibrium What are the percentage dissociations of H2(g), O2(g), and h(g) at 2000 K and a total pressure of 1 bar? SOLUTION 5.9 H2(g) = 2H(g) /lrGo = 2(106,760 J mol-l) = -RT In K = -(8.3145 J K-l mol-l)(2000 K) In K 4l!/- ~ K = 2.65 x 10-6 = ~ = (~) 1/2 K = 0.000814 or 0.0814% 9 = 0.0335% ( ) 1/2 g= K+4 4 02(g) = 20(g) ~rGo = 2(121,552 J mol-l) K = 4.48 x 10-7 h(g) = 2I(g) ~rGo = 2(- 29,410 J mol-l) K = 34.37 ~ = 94.6% In order to produce more hydrogen from "synthesis gas" (CO + H2) the water gas shift reaction is used. CO(g) + H20(g) = CO2(g) + H2(g) Calculate K at 1000 K and the equilibrium extent of reaction starting with an equimolar mixture of CO and H2O. SOLUTION ~GO = -395,886- 1- 200,275) -(- 192,590) = -3021 J mol-l = -(8.314 J K-l mol-l)(1000 K) In K PH2PCO2 f} K=1.44= p p = CO H20 ( 1 -~2 ~ = 0.545 (Note that this reaction is exothermic so that there will be a larger extent of reaction at lower temperatures. In practice this reaction is usually carried out at about 700 K.) Calculate the extent of reaction 9 of 1 mol of H2O(g) to form H2(g) and O2(g) at 2000 K and 1 bar. (Since the extent of reaction is small, the calculation may be simplified by assuming that PH20 = 1 bar.) SOLUTION 5.11 ( 2 !c; 1- 9 ~ eq. L\rGO = 135,528 J mol-l = -(8.1315 J K-1 moI-1)(2000 K) In K K = 2.887 x 10-4 ( ~ ) ( ~ ) 112 Po Po PO2 PH2 K -(~) , Po -2PO ( ~ ) ( ~ ) 112 = 1- ( ~ ) 312 Po 2PO .v2 Po ~ = (~ ) = (.v2 K)213 = 0.0055 5.12 At 500 K CH30H, ClI4 and other hydrocarbons can be formed from CO and H2. Until recently the main source of the CO mixture for the synthesis of CH30H was methane. CH4(g) + H20(g) = CO(g) + 3H2(g) When coal is used as the source, the "synthesis gas" has a different composition. C(graphite) + H20(g) = CO(g) + H2(g) Suppose we have a catalyst that catalyzes only the foImation of CH3OH. (a) What pressure is required to convert 25% of the CO to CH3OH at 500 K if the "synthesis gas" comes from ClI4? (b) If the synthesis gas comes from coal? SOLUTION (a) CO + 2H2 = CH30H 70 Chapter 5/Chemical Equilibrium 1 fl2()(g) = fl2(g) + ~2(g) init. 1 0 0 71 Chapter 5/Chemical Equilibrium L\rGO = -134.27- (- 155.41) = 22.14 kJ mol- K= 6.188 x 10-3 p = [ -~(4 -2/:;)2-- ] 112 K(l -/:;)(3 -2/:;)2 [ (0.25)(3.5)2 ] 112 = 6.188 ~ 10-3(0.75)(2.5)2 = 10.3 bar CO + 2H2 = CH30H (b) 0 1 1 Initial Total = 2 -2~ 1-2~ ~ 1- 9 Equil. p = [ (0.25)(1.5)2 ] 1/2 K(O.75)(0.5)2 = 22.0 bar Many equilibrium constants in the literature were calculated with a standard state pressure of 1 atm (1.01325 bar). Show that the corresponding equilibrium constant with a standard pressure of 1 bar can be calculated using K(bar) = K(atm)(1.01325)~Vi where the Vi are the stoichiometric numbers of the gaseous reactants. SOLUTION K(atm) = O [Pi/(1 atm)) vi I - ( ~ ) LVi -Ibar ~- n ( ~ ) Vi K(atm) -; 1 bar = 1.01325Lvi Older tables of chemical thennodynamic properties are based on a standard state pressure of 1 atm. Show that the corresponding L\fG~ with a standard state pressure of 1 bar can be calculated using o o L\tGj (bar) = L\tGj (atm) -(0.109 x 10-3 kJ K-l mol-l) TLvi Chapter 5/Chemical Equilibrium 72 where the Vi are the stoichiometric numbers of the gaseous reactants and products in the formation reaction. SOLUTION Show that the equilibrium mole fractions of n-butane and iso-butane are given by 5.15 Yn= e-6.tGnoIRT + e-6.tGisooIRT e-6.tGisooIRT Yiso = SOLUTION 4C(graphite) + SH2(g) = C4HlO(g,n) K - p /p S --J1.fGnoIRT n -n H2 -e 4C(graphite) + SH2(g) = C4HlO(g,iso) e-~tGnO/RT ~ *5.16 Calculate the molar Gibbs energy of butane isomers for extents of reaction ofO.2, 0.4, 0.6, and 0.8 for the reaction 73 Chapter 5/Chemical Equilibrium n-butane = iso-butane at 1000 K and I bar. At 1000 K, L\tG° (n-butane) = 270 kJ mol-l L\tG° (iso-butane) = 276.6 kJ mol-l Make a plot and show that the minimum co1Tesponds with the equilibrium extent of reaction. SOLUTION n-butane = iso-butane moles at t = 0 0 1 9 moles at equilibrium; nT = 1 mol 1-~ G = nn'Un + nj'Uj = nn'UnO + nnRT In(xn) + nj'UjO + njRT In(xj) o o = (1 -~) ~fGn + (1 -f;)RT In(l -g) + g~tGi + gRT In(f;) = 270 + (276.6- 270) 9 + RT[(l -f;)ln(l -f;) + gln(f;)] G(~) = 270 + 6.6~ + 8.314[(1 -~)ln(l -~) + ~ln(~)] kJ mol-l G(0.2) = 267.2 G(0.6) = 268.4 G(0.4) = 267.0 G(0.8) = 271.1 in kJ/mol o o 1 ~Go = ~tG. -~tG = 6.6 kJ mol- = -RT In K 1 n 9 [ -~rGO ] K=-=exp RT =0.4521 (T= 103K) l-g 9 = 0.311 G(0.311) = 266.9 kJ/mol corresponds to the minimum of the following graph of G(f;) versus extent of reaction GJ(kJJmol) 2761 2741 2'72 270' 268 . extent of rx 0.6 0.8 1 0.2 0.4 ~ Chapter 5/Chemical Equilibrium 74 In the synthesis of methanol by CO(g) + 2H2(g) = CH3OH(g) at 500 K, calculate the total pressure required for a 90% conversion to methanol if CO and H2 are initially in a 1:2 ratio. Given: K= 6.09 x 10-3. SOLUTION 5.17 CO(g) + 2H2(g) = CH30H(g) initial moles 1 2 0 equil. moles 0.1 0.2 0.9 total 1.2 K = (PCH30H/ PO) (PCdPO)(PH2/PO)2= 6.09 x 10-3 0.9 p -- 1.2 po -QlE- (~ E- )2 1.2po 1.2 po p -I (0.9)(1.2)2 :iJ() = -\' (0.1)(0:04)(6.09 x 10-3) = 231 p = 231 bar = total pressure for 90% conversion to CH30H ).l~ At 1273 K and at a total pressure of 30.4 bar the equilibrium in the reaction CO2(g) + C(s) = 2CO(g) is such that 17 mole % of the gas is CO2. (a) What percentage would be CO2 if the total pressure were 20.3 bar? (b) What would be the effect on the equilibrium of adding N2 to the reaction mixture in a close" vessel until the partial pressure of N2 is 10 bar? (c) At what pressure of the reactants wi1125% of the gas be CO2 ? SOLUTION (a) PCO2 = (30.4 bar)(0.17) = 5.2 bar PCO = (30.4 bar)(0.83) = 25.2 bar K=~= 122 Let x = mole fraction CO2 K -[20.3(1 -x)]2 - 122 -20.3x - x = 0.127 Percentage CO2 at equilibrium= 12.7% (b) No effect for ideal gases because the partial pressures of the reactants are not affected. . 75 Chapter 5/Chemical Equilibrium K = [0.75(P/PO)]2 0.25(P/PO) P = 54 bar (c) When alkanes are heated up, they lose hydrogen and alkenes are produced. For example, K = 0.36 at 1000 K C:2lI6(g) = C:2lI4(g) + lI2(g) If this is the only reaction that occurs when ethane is heated to 1000 K, at what total pressure will ethane be (a) 10% dissociated and (b) 90% dissociated to ethylene and hydrogen? SOLUTION C2H6 = C2H4 + H2 0 1 0 Init. ~ total = 1 + ~ 1-~ ~ Equil, l;2(PlPO) (I + ~)(I -~ [}-(P/po) 1-[}- K= = (a) (b) At 2000 °C water is 2% dissociated into oxygen and hydrogen at a total pressure of 1 bar. 1 (a) Calculate K for H2O(g) = H2(g) + 2 O2(g) (b) Will the extent of reaction increase or decrease if the pressure is reduced? (c) Will the extent of reaction increase or decrease if argon gas is added, holding the total pressure equal to 1 bar? (d) Will the extent of reaction change if the pressure is raised by addition of argon at constant volume to the closed system containing partially dissociated water vapor? (e) Will the extent of reaction increase or decrease if oxygen is added while holding the total pressure constant at 1 bar? Chapter 5/Chemical Equilibrium 76 SOLUTION 1 + 2 02(g) O H2(g) H20(g) (a) == 0 1 initial ~ 1:)2 9 1+2 p PO2= ~/2(PlPO)1/2 -{2(1 + f)2)1/2(1 -~ 1 + Cj2 po 0.023/211/2 = 2.03 x 10-3 (b) (c) {2 (1.01)1/2 (0.98) If the total pressure is reduced, the extent of reaction will increase because the reaction will produce more molecules to fill the volume. If argon is added at constant pressure, the extent of reaction will increase because the partial pressure due to the reactants will decrease. If argon is added at constant volume, the extent of reaction will not be changed because the partial pressure due to the reactants will not change. If oxygen is added at constant total pressure, the extent of reaction of H20 will decrease because the reaction will be pushed to the left. (d) (e) 5.21 At 250 °C PCIS is 80% dissociated at a pressure of 1.013 bar, and so K= 1.80. What is the extent of reaction at equilibrium after sufficient nitrogen has been added at constant pressure to produce a nitrogen partial pressure of 0.9 bar? The total pressure is maintained at 1 bar. SOLUTION OA -1 -!:;2 9 = 0.973 or 97.3% dissociated. 5.22 The following exothermic reaction is at equilibrium at 500 K and 10 bar. CO(g) + 2H2(g) = CH30H(g) Assuming the gases are ideal, what will happen to the amount of methanol at equilibrium when (a) the temperature is raised, (b) the pressure is increased, (c) an 0.02j/:l 11/:l :=2.0 ~ (1.01 )1/2 (0.98) 77 Chapter 5/Chemical Equilibrium inert gas is pumped in at constant volume, (d) an inert gas is pumped in at constant pressure, and (e) hydrogen gas is added at constant pressure? SOLUTION (a) (b) (c) (d) (e) n(CH30H) decreases because heat is involved. n(CH30H) increases because there are fewer molecules of gas in the product. No effect. n(CH30H) decreases because the volume increases. n(CH30H) increases because there is more of a reactant. Note: the effect of the addition of CO is more complicated. The following reaction is nonspontaneous at room temperature and endothermic. 3C(graphite) + 2H2O(g) = CH4(g) + 2CO(g) As the temperature is raised, the equilibrium constant will become equal to unity at some point. Estimate this temperature using data from Appendix C.3. SOLUTION At 1000 K /).rGo = 19.492 + 2(- 200.275) -2(- 192.590) = 4.122 kJ mol-l = -(8.3145 x 10-3 kJ K-l moI-l)(1000 K) In K The measured density of an equilibrium mixture of N2O4 and NO2 at 15 °C and 1.013 bar is 3.62 9 L -I, and the density at 75 °C and 1.013 bar is 1.84 9 L -I. What is the enthalpy change of the reaction N2O4(g) = 2NO2(g)? SOLUTION ~ Chapter 5/Chemical Equilibrium 78 4!:}P 1-!:} 4(0.0753)2(1.013) = 0.0231 1 -0.07532 K= = At 75 °C M -(0.08314)(348)(1.84) - 52 55 1-1 -1.013 -.9 mo ~=~=0.751 K=5.24 A H -RTIT2 1 ~-(8.314)(288)(348) 1 ~r -(T2 -Tl) n Kl -60 n 0.0231 5.24 = 75 kJ mol-l Ca]cu]ate Kc for the reaction in prob]em 5.19 at 1000 K and describe what it is equa] to. 5.25 SOLUTION Kc = Kp( ~ ) ~Vi coRT 1 bar = 0.36 1 1 1 (1 L mol- )(0.08314 L bar K- mol- )(238 K) = 0.0145 (~)(~) - [C2H6J cO where [ ] indicates concentrations in moles per liter and CO = 1 mol L -I ~ *5.26 The equilibrium constant for the reaction ~2(g) + 3II2(g) = 2~II3(g) is 35.0 at 400 K when partial pressures are expressed in bars. Assume the gases are ideal. (a) What is the equilibrium volume when 0.25 mol ~2 is mixed with 0.75 mol II2 at a temperature of 400 K and a pressure of 1 bar? (b) What is the equilibrium composition and equilibrium pressure if this mixture is held at a constant volume of 33.26 L at 400 K? SOLUTION 2NH3 3H2 (a) N2 + = 0.75 0 initial 0.25 total = 1 -2~ 0.75- 3g ~ 0.25 -9 equil. 6 1 bar (1 L mol-l)(O.O8314 L bar K-l mol 79 Chapter 5/Chemical Equilibrium The method of successive approximations or the use of an equation solver yields 9 = 0.1652. P(Nv/PO = (0.25- g)/(1 -2g) = 0.1266 P(Hv/PO = 3(0.1266) = 0.3798 P(NH3)/PO = 2l:j(1 -2g) = 0.493 Since the total pressure is 1 bar, these numbers also give the equilibrium mole fractions. V = (1 -2g)RT/P = (0.6696 mol)(0.8314 L bar K-l mol-l)(400 K)/(l bar) = 22.27 L When MathematicaTM is used to calculate the extent of reaction, four solutions are obtained; two of them are not satisfactory because they are imaginary and one is 0.334 mol, which is impossible because the extent of reaction has to be less than 0.25 mol. The fourth solution is correct. In order to calculate the equilibrium composition at constant volume, it is convenient to use Kc. Kc = (PO/cDR1)~Vi Kp = [ 1 bar . ] -2350 (1 mol L -1)(0.08314 L bar K-l mol-l)(400 K) . = 3.871 x 104 Kc = [C(NH3)/cD]2/[c(Nv/cD][c(H2)/cDr = (2~2 33.262/(0.25 -~(0.75 -3~3 = 3.871 x 104 The method of successive approximation or the use of an equation solver yields ~ = 0.151. n(N2) = 0.25 -~ = 0.0990 n(H2) = 0.75- 3~ = 0.2970 n(NH3) = 2~ = 0.302 n(total) = 0.698 p -(0.698 mol)RT - 0 698 b -33.26 L -.ar (b) 5.27 Show that to a first approximation the equation of state of a gas that dimerizes to a small extent is given by PV - RT = 1 -KdV SOLUTION 2A(g) = A2(g) Kc = [A2]/[A]2 = nA2 V/nA 2 80 Chapter 5/Chemical Equilibrium (a) assuming an ideal gas mixture. 2 nO = Total number of moles of A = nA + 2nA2 = nA + 2KcnA/V This can be solved for nA using the quadratic formula for ax2 + bx + c = 0: -b .t (b2 -4ac)1/2 x= 2a This yields nA = no(1 -2Kcn0/V) (b) when the approximation (I + x)1/2 = I + x/2 -x2/8 + ...is used. The gas law indicates that PV 2 Rt = nA + nA2 = nA + Kcn A/V = nA(1 + KcnNV) (c) Substituting equation b in equation c yields PV Rt= nO(1 -2Kcn0/V)(1 + KcnA/V) (d) Since nA is just a little smaller than no, it can be replaced by no in a term that is small compared to unity. Thus equation d can be written as PV -- Rt= no(1 -2Kc/V)(1 + Kc/V) (e) When this is multiplied out ignoring higher order terms and the no is moved to the left-hand side of the equation, we obtain PV ,- "' ~ = 1 -KdV 5.28 Water vapor is passed over coal (assumed to be pure graphite in this problem) at 1000 K. Assuming that the only reaction occurring is the water gas reaction C(graphite) + H2O(g) = CO(g) + H2(g) K = 2.52 calculate the equilibrium pressures of H2O, CO, and H2 at a total pressure of 1 bar. [Actually the water gas shift reaction CO(g) + H2O(g) = CO2(g) + H2(g) occurs in addition, but it is considerably more complicated to take this additional reaction into account.] SOLUTION (PColPO)(PH2/PO) -~ -~ = 2.52 K= (PH2o1PO) -y -1- 2x 2x+y= 1 X2 = 2.52- 5.04x X2 + 5.04x -2.52 = 0 x = -5.04 .:t "V' 5.~42 + 4(2.52) = 0.458 -~-~ -pa-pa "",,""' 81 Chapter 5/Chemical Equilibrium -2t=O.O84=~ po 1 PH20 = 0.084 bar Pco = 0.458 bar PH2 = 0.458 bar What is the standard change in entropy for the dissociation of molecular oxygen at 298.15 K and 1000 K? Use Appendix C.3. SOLUTION = 2 O(g) LlrS(298.15 K) = 2(161.058) -205.147 = 116.969 J K-l mol-l LlrS(1000 K) = 2(186.790) -243.578 = 130.002 J K-l mol-l Using molar entropies from Appendix C.2, calculate I1.rS° for the following reactions at 25 °C. 1 (a) H2(g) + -02(g) = H20(!) (b) H2(g) + tl2(g) = 2HCl(g) 1 (c) Methane (g) + 2 02(g) = methanol(!) SOLUTION (a) (b) ~rSO = 69.91- 130.68- ~ (205.13) = -163.34 J K-l mol-l ~rSo = 2(186.908) -130.681- 223.066 = 20.066 J K-l mol-l ~rSo = 126.8- 186.264- ~(205.138) = -162.0 J K-l mol-l (c) What is L\rSO (298 K) for H20(1) = H+(ao) + OH-(ao) Why is this change negative and not positive? SOLUTION IlrSO = -10.75- 69.92 = -80.67 J K-l mol-l The ions polarize neighboring water molecules and attract them. For this reason the product state is more ordered than the reactant state. Mercuric oxide dissociates according to the reaction 2HgO(s) = 2Hg(g) + O2(g). At 420 °C the dissociation pressure is 5.16 x 104 Pa, and at 450 °C it is 10.8 x 104 Pa. Calculate (a) the equilibrium constants, and (b) the enthalpy of dissociation per mole of HgO. SOLUTION 5.29 WI 02(g) 5.30 Chapter 5/Chemical Equilibrium 82 (a) (b) L\Ho =!I&ln~ r T2- Tl Kl (8.314 J K-l mol-l)(693 K)(723 K) 1 0.1794 = -30K n 0.0196 = 308 kJ mol-l for the reaction as written = 154 kJ mol-l of HgO(s) The decomposition of silver oxide is represented by 2Ag20(s) = 4Ag(s) + 02(g) Using data from Appendix C.2 and assuming ~rCp = 0 calculate the temperature at which the equilibrium pressure of 02 is 0.2 bar. This temperature is of interest because Ag20 will decompose to yield Ag at temperatures above this value if it is in contact with air. SOLUTION The dissociation of ammonium carbamate takes place according to the reaction (NH2)CO(ONH4)(S) = 2NH3(g) + CO2(g) When an excess of ammonium carbamate is placed in a previously evacuated vessel, the partial pressure generated by NH3 is twice the partial pressure of the CO2, and the partial pressure of (NH2)CO(ONH4) is negligible in comparison. Show that K=(~)2(~)=~(~)3 where p is the total pressure. Chapter 5/Chemical Equilibrium 83 SOLUTION P = PNH3 + PCO2 = 3PCO2 since PNH3 =2PCO2 P 2 PCO2 = 3 PNH3 = 3P K= (~)2(~) = (~~)2(~~) =~(~)3 At 1000 K methane at 1 bar is in the presence of hydrogen. In the presence of a sufficiently high partial pressure of hydrogen, methane does not decompose to form graphite and hydrogen. What is this partial pressure? SOLUTION ClI4(g) = C(graphite) + 2H2(g) L\Go = -RT In K = -19.46 kJ mol-l K= 10.39 = (~)2 ~ Po PH = PO[(10.39)(1)]1/2 = 3.2 bar 2 For the reaction Fe203(s) + 3CO(g) = 2Fe(s) + 3CO2(g) the following values of K are known. 1000 t/OC 250 K 100 0.0721 At 1120 °C for the reaction 2C02(g) = 2CO(g) + 02(g), K = 1.4 x 10-12. What equilibrium partial pressure of 02 would have to be supplied to a vessel at 1120 °C containing solid Fe203 just to prevent the formation of Fe? SOLUTION ( K2 ) L\rHO(T2- TI) In Kl = RTIT2 L\ H o -(8.314)(523)(1273) In (0.0721/100) r -(750) = -53.4 k J mol-l 1 (K1393) (53400)(120) n Kim = 8.314 (1393)(1273) 5.36 5.35 Chapter 5/Chemical Equilibrium 83 SOLUTION P = PNH3 + PCO2 = 3PCO2 since PNH3 =2PCO2 P 2 PCO2 = 3 PNH3 = 3P K= (~)2(~) = (~~)2(~~) =~(~)3 At 1000 K methane at 1 bar is in the presence of hydrogen. In the presence of a sufficiently high partial pressure of hydrogen, methane does not decompose to form graphite and hydrogen. What is this partial pressure? SOLUTION ClI4(g) = C(graphite) + 2H2(g) L\Go = -RT In K = -19.46 kJ mol-l K= 10.39 = (~)2 ~ Po PH = PO[(10.39)(1)]1/2 = 3.2 bar 2 For the reaction Fe203(s) + 3CO(g) = 2Fe(s) + 3CO2(g) the following values of K are known. 1000 t/OC 250 K 100 0.0721 At 1120 °C for the reaction 2C02(g) = 2CO(g) + 02(g), K = 1.4 x 10-12. What equilibrium partial pressure of 02 would have to be supplied to a vessel at 1120 °C containing solid Fe203 just to prevent the formation of Fe? SOLUTION ( K2 ) L\rHO(T2- TI) In Kl = RTIT2 L\ H o -(8.314)(523)(1273) In (0.0721/100) r -(750) = -53.4 k J mol-l 1 (K1393 ) (53400)(120) n Kim = 8.314 (1393)(1273) Chapter 5/Chemical Equilibrium 85 Calculate the partial pressure of CO2(g) over CaCO3(calcite) -CaO(s) at 500 °C using the equation in Example 5.11 and data in Appendix C.3. SOLUTION CaCO3(calcite) = CaO(s) + CO2(g) L\rHo = -635.09 + (- 393.51) -(- 1206.92) = 178.32 kJ mol-l L\r~ = 42.80 + 37.11- 81.88 = -1.97 J K-l mol-l L\rSo= 39.75 + 213.74- 92.9 = 160.6 J K-l mol-l Substituting in the equation in Example 5.11, 178320 160.6 1.97 In K773 = -(8.314)(773.15) + 8:314 + 8:314 (1 - 298.15--1 ~ ) 773.15 n 298.15 Pco K773 = --1 = 20 x 10-5 po The NBS Tables contain the following data at 298 K: ~fHo/kJ mol-l -771.36 -1085.83 -1684.31 -241.818 /1tG°1kJ mol-l -661.8 -918.11 -1399.96 -228.572 CUSO4(S) CuSO4oH20(S) CuSO4o3H20(S) H20(g) (a) What is the equilibrium partial pressure of H20 over a mixture of CUSO4(S) and CuSO4'H20(S) at 25 °C? (b) What is the equilibrium partial pressure of H20 over a mixture of CuSO4'H20(S) and CuSO4'3H20(S) at 25 °C? (c) What are the answers to (a) and (b) if the temperature is 100 °C and LlC~ is assumed to be zero? SOLUTION (a) CuSO4.H20(S) = CUSO4(S) + H20(g) L\Go = -228.572- 661.8 + 918.11 = 27.7 kJ mol-l ( -L\GO ) PH o K= exp RT = 1.4 x 10-5 = -j!;- PH2o= 1.4 x 10-5 bar Chapter 5/Chemical Equilibrium 86 (b) CuSO4.3H20(S) = CuSO4.H20(S) + 2H20(g) LlGo = 2(- 228.572) -918.11 + 1399.46 = -24.71 kJ mol-l [ -24 710 ] PH20 = exp (2)(8.314)(298) = 6.83 x 10-3 bar (c) For the first reaction K1OO 72650(75) ln 1.4 x 10-5 = 8.314(298)(373) Mlo = -241.818- 771.36 + 1085.83 = 72.65 kJ mol-l K1OO ln 1.4 x 10-5 = 5.896 K1OO = 363.6(1.4 x 10-5) = 0.0051 bar For the second reaction MJO = 2(- 241.818) -1085.83 + 1684.31 = 114.84 kJ mol-l K1OO 114840(75) ln 4.7 x 10-5 = 8.314(298)(373) = 9.32 2 K 100 = 0.52 bar2 = PH20 PH20 = 0.72 bar 5.41 One micromole of CuO(s) and 0.1 ~mole of Cu(s) are placed in a 1 L container at 1000 K. Determine the identity and quantity of each phase present at equilibrium if tltG° of CuO is -66.66 kJ mol-l and that of CU2O is -77 .94 kJ mol-l at 1000 K. (From H. F. Franzen, I. Chem. Ed. 65, 146 (1988).) SOLUTION Cu(s) + CuO(s) = CU20(S) ~rGo = -77.94- (- 66.66) = -11.28 kJ mol-l Therefore, this reaction goes to completion to the right. The two solids are in equilibrium with 02(g). 1 2CuO(s) = CU20(S) + 2 02(g) ~rGo = -77.94- 2(-66.66) = 55.38 kJ mol-l = -RTln p~~ [ 2(55 380) ] -6 PO2 = exp -(8.314)(1000) = 1.64 x 10 The amount of 02(g) at equilibrium is PV (1.64 x 10-6)(1) 8 no2 = RT = (O.08314)(1000) = 1.97 x 10- mol Thus the amounts at equilibrium are essentially Chapter 5/Chemical Equilibrium 87 ncuo = 0.9 -4(0.02) = 0.82 J.!mol ncu2o = 0.1 + 2(0.02) = 0.14 J.!mol no2 = 0.02 J.!mol For the heterogeneous reaction CH4(g) = C(s) + 2H2(g) derive the expression for the extent of reaction in tenns of the equilibrium constant and the applied pressure, where the extent of reaction when graphite is in equilibrium with the gas mixture. Is this the same expression (equation 5.33) that was obtained for the reaction N204(g) = 2NO2(g)? SOLUTION CH4(g) = C(s) + 2H2(g) 1-~ ~ ~ ~p 1+~ This is the same as equation 5.33, but it only applies when graphite is in equilibrium with the gas. Calculate the equilibrium extent of the reaction N204(g) = 2NO2(g) at 298.15 K and a total pressure of 1 bar if the N204(g) is mixed with an equal volume of N2(g) before the reaction occurs. As shown by Example 5.3, K = 0.143. Do you expect the same equilibrium extent of reaction as in example? If not do you expect a larger or smaller equilibrium extent of reaction? SOLUTION If there is initially 1 mol of N2O4, the total amount of gas at equilibrium is 2 + ~. Thus the expression for the equilibrium constant is 4~2(p/po) K= (2+~)(1-~) where p is the total pressure. When the total pressure is 1 bar, the equilibrium extent of reaction obtained by solving this quadratic equation with the fonnula ~ = ::E;t(b2 -4ac)1/2 2a Chapter 5/Chemical Equilibrium 88 is 0.249. This equilibrium extent of reaction is smaller than that in Example 5.3 because the partial pressure of N204(g) plus NO2(g) is larger than 0.5 bar. The partial pressure of N2(g) was initially 0.5 bar, but it is less than this in the equilibrium mixture because of the expansion of the reaction mixture during the reaction at a constant pressure of 1 bar . (a) A system contains CO(g), CO2(g), H2(g), and H20(g). How many chemical reactions are required to describe chemical changes in this system? Give an example. (b) If solid carbon is present in the system in addition, how many independent chemical reactions are there? Give a suitable set. 5.44 SOLUTION (a) (b) CO CO2 H2 H20 C 1 1 0 0 O 1 2 0 2 H 0 0 2 2 To perform a Gaussian elimination, subtract the first row from the second, and divide the third row by 2. 1 1 0 0 0 1 0 1 0 0 1 1 Subtract the second row from the first. 1 0 0 -1 0 1 O 1 0 O 1 1 The rank of the A matrix is 3, and so the number of independent reactions is R = N -rank A = 4 -3 = 1 where N is the number of species. The stoichiometric numbers for a suitable reaction is obtained by changing the sign of the numbers in the last column, and extending the vector with a 1; that is, 1, -1, -1,1. These are the stoichiometric numbers for the species across the top of A . CO -CO2 -H2 + H20 = 0 CO2 + H2 = CO + H20 CO CO2 H2 H20 C C 1 1 0 0 1 O 1 2 O 1 O H O O 2 2 O Subtract the first row from the second and divide the third row by 2. 1 1 O 0 1 0 1 0 1 -1 0 0 1 1 0 Subtract the second row from the first 1 0 0 -1 2 0 1 0 1 -1 0 0 1 1 0 Rank A = 3 and R = N- rank A = 5 -3 = 2. To obtain a suitable set of reactions, change the signs in the last two columns and put an identity Chapter 5/Chemical Equilibrium 89 matrix at the bottom. 1 -2 -1 1 -1 O 1 O O 1 These two independent reactions are CO -CO2 -H2 + H20 = O -2CO + CO2 + C = O or CO2 + H2 = CO + H20 2CO = CO2 + C *5.45 For a closed system containing C2H2, H2, C6ll6, and CIOH8, use a Gaussian elimination to obtain a set of independent chemical reactions. Perform the matrix multiplication to verify A V = 0. SOLUTION C2H2 H2 C6H6 CIOH8 C 2 0 6 10 H 2 2 6 8 3 3 5 4 1 1 O 1 1 O O 1 3 O 5 -1 -5 1 O 1 v= C2H2 -3 H2 O C6ll6 1 CIOH8 O 3C2H2 = C6ll6 5C2H2 = CIOH8 + H2 2 O 6 2 2 6 g] 10)[ -3 8 0 01 ~] [ 9 = 5.46 The reaction A + B = C is at equilibrium at a specified Tand P. Derive the fundamental equation for G in terms of components by eliminating .uc. ~ SOLUTION (I) (2) The fundamental equation for G is dG = -SdT + VdP + .uAdnA + .uBdnB + .ucdnc When the reaction is at equilibrium, .uA + .uB = .uc Eliminating .uc from equation 1 yields Chapter 5/Chemical Equilibrium 90 dG = -SdT+ VdP + JlA(dnA + dnC) + JlB(dnB + dnc) (3) This equation is written in terms of 2 components rather than 3 species because C = N- R = 3 -1 = 2. The two components can be referred to as the A,C pseudoisomer group with amount m' = n A + nc (4) and the B component with amount nB'=nB+nc (5) Thus at chemical equilibrium the fundamental equation can be written as dG = -SdT + VdP + JlAdnl' + JlBdnB' (6) Note that the chemical potentials of the components are the same as the chemical potentials of two of the species. *5.47 The article C. A. L. Figueiras, J. of Chem. Educ., 69,276 (1992) illustrates an interesting problem you can get into in trying to balance a chemical equation. Consider the following reaction without stoichiometric numbers: C1O3- + Cl- + H+ = C1O2 + C12 + H20 There is actually an infinite number of ways to balance this equation. The following steps in unraveling this puzzle can be carried out using a personal computer with a program like Mathematica TM, which can do matrix operations. Write the conservation matrix A and determine the number of components. How many independent reactions are there for this system of six species? What are the stoichiometric numbers for a set of independent reactions? These steps show that chemical change in this system is represented by two chemical reactions, not one. ~ SOLUTION The conservation matrix A for this system is CIO3- Cl- H 0 0 O 3 0 Cl 1 1 charge -3 -I H+ 1 0 O +1 H2Q 2 1 0 O CIO2 Cl2 0 0 2 0 1 2 0 0 Row reduction of this matrix yields CIO3- Cl- CIO3- 1 O Cl- O 1 H+ O O H20 O O H+ 0 0 1 0 H2O 0 0 0 1 CIO2 5/6 1/6 1 -1/2 Cl2 1/3 5/3 2 -1 This indicates that there are 4 components. Thus R = Ns -C = 6- 4 = 2. The last two columns with changed signs and augmented by a 2x2 unit matrix at the bottom give the stoichiometric numbers of two independent reactions. Another way to obtain a set of independent reactions is to calculate the null space of the A matlix. The null space V is Chapter 5/Chemical Equilibrium 91 CIO3- Cl- H+ H2O CIO2 C12 rx 1 rx 2 -1 -5 -5 -1 -6 -5 3 3 O 6 3 O A chemical reaction system contains three species: C2H4 (ethylene), C3B6 (propene), a