# giải chi tiết bài tập cân bằng hóa học (cao học)

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7.1 At 2257 K and 1.00 atm total pressure, water is 1.77 per cent dissociated at equilibrium by way of the reaction 2H2O(g) = 2H¬2,(g) + O2,(g). Calculate (a) K, (b) ∆G , and (c) ∆G at this temperature.K= 2.85 x 106; (b) ∆G = +240 kJ.mol1 (c) ∆G = 0 (the system is at quilibrium).7.2 Dinitrogen tetroxide is 18.46 per cent dissociated at 25°C and 1.00 bar in the quilibrium N2O4(g) = 2NO2(g). Calculate (a) K at 25°C, (b) ∆G , (c) K at 100°C given that ∆H = +57.2 kJ.mol1 over the temperature range. Chemical equilibrium Exercises 7.1 At 2257 K and 1.00 atm total pressure, water is 1.77 per cent dissociated at equilibrium by way of the reaction 2H2O(g) = 2H2,(g) + O2,(g) Calculate (a) K, (b) ∆G or , and (c) ∆G r at this temperature K= 2.85 x 10-6; (b) ∆G or = +240 kJ.mol-1 (c) ∆G r = (the system is at quilibrium) 7.2 Dinitrogen tetroxide is 18.46 per cent dissociated at 25°C and 1.00 bar in the quilibrium N2O4(g) = 2NO2(g) Calculate (a) K at 25°C, (b) ∆G or , (c) K at 100°C given that ∆H or = +57.2 kJ.mol-1 over the temperature range (a) K = 0.1411; (b) ∆G or = +4.855 kJ.mol-1 (c) K (l00°C) = 14.556 7.3 From information in the Data section, calculate the standard Gibbs energy and the equilibrium constant at (a) 298 K and (b) 400K for the reaction PbO(s) + CO(g) = Pb(s) + CO2(g) Assume that the reaction enthalpy is independent of temperature (a) ∆G or =-68.26 kJ.mol-1, K= 9.2 x l011; (b) K (400K) = 1.3 x 109, ∆G or (400K) = -69.7 kJ.mol-1 7.4 In the gas-phase reaction 2A + B = 3C + 2D, it was found that, when 1.00 mol A, 2.00 mol B, and 1.00 mol D were mixed and allowed to come to equilibrium at 25°C, the resulting mixture contained 0.90 mol C at a total pressure of 1.00 bar Calculate (a) the mole fractions of each species at equilibrium, (b) Kx, (c) K, and (d) ∆G or (a) Mole fractions: A: 0.087, B: 0.370, C: 0.196, D: 0.348, Total: 1.000; (b) Kx = 0.33; (c) Kp = 0.33; (d) ∆G or =+2.8 x 103 Jmol-1 7.5 The standard reaction enthalpy of Zn(s) + H 2O(g) → ZnO(s) + H2(g) is approximately constant at +224 kJ.mol-1 from 920K up to 1280K The standard reaction Gibbs energy is +33 kJ.mol -1 at 1280K Estimate the temperature at which the equilibrium constant becomes greater than T = 1500 K 7.6 The equilibrium constant of the reaction 2C 3H6(g) = C2H4(g) + C4H8(g) is found to fit the expression lnK =A + B/T + C/T between 300 K and 600 K, with A = -1.04, B = -1088 K, and C= 1.51 x 10 K2 Calculate the standard reaction enthalpy and standard reaction entropy at 400K ∆H or = +2.77 kJ mol-1, ∆S or = -16.5 J.K-1.mol-1, 7.7 The standard reaction Gibbs energy of the isomerization of borneol (C 10H17OH) to isoborneol in the gas phase at 503 K is +9.4 kJ.mol -1 Calculate the reaction Gibbs energy in a mixture consisting of 0.15 mol of borneol and 0.30 mol of isoborneol when the total pressure is 600 Torr ∆Go =+12.3 kJ.mol-1 7.8 Calculate the percentage change in Kx for the reaction H2CO(g) = CO(g) + H2(g) when the total pressure is increased from 1.0 bar to 2.0 bar at constant temperature 50 per cent 7.9 The equilibrium constant for the gas-phase isomerization of borneol (C 10H17OH) to isoborneol at 503 K is 0.106 A mixture consisting of 7.50 g of borneol and 14.0 g of isoborneol in a container of volume 5.0 dm is heated to 503 K and allowed to come to equilibrium Calculate the mole fractions of the two substances at equilibrium nB = xB=0.904, x1 =0.096 n 7.10 What is the standard enthalpy of a reaction for which the equilibrium constant is (a) doubled, (b) halved when the temperature is increased by 10 K at 298 K? (a) ∆H or = +53 kJ.mol-1; (b) ∆H or = -53 kJ.mol-1 7.11 The standard Gibbs energy of formation of NH 3(g) is -16.5 kJ.mol-1 at 298 K What is the reaction Gibbs energy when the partial pressures of the N 2, H2, and NH3 (treated as perfect gases) are 3.0 bar, 1.0 bar, and 4.0 bar, respectively? What is the spontaneous direction of the reaction in this case? ∆G r = -14.38 kJ.mol-1, spontaneous direction of reaction is towards products 7.12 Estimate the temperature at which CaCO3(calcite) decomposes T= 1110K 7.13 For CaF2(s) = Ca2+(aq) + 2F-(aq), K = 3.9x10-11 at 25°C and the standard Gibbs energy of formation of CaF2(s) is -1167 kJ.mol-1, Calculate the standard Gibbs energy of formation of CaF2(aq) ∆G of = -1108 kJ.mol-1 Equilibrium electrochemistry Exercises 7.14 Write the cell reaction and electrode half-reactions and calculate the standard emf of each of the following cells: (a) Zn│ZnSO4(aq)││AgNO3(aq) │Ag (b) Cd│CdCl2(aq) ││ HNO3(aq) │H2(g) │Pt (c) Pt│K3[Fe(CN)6](aq),K4[Fe(CN)6] (aq) ││CrCl3(aq) │Cr 7.15 Devise cells in which the following are the reactions and calculate the standard emf in each case: (a) Zn(s) + CuSO4(aq) → ZnSO4(aq) + Cu(s) (b) 2AgCl(s) + H2(g) → 2HCl(aq) + 2Ag(s) (c) 2H2(g) + O2(g) → 2H2O(l) 7.16 Use the Debye-Huckel limiting law and the Nernst equation to estimate the potential of the cell Ag│AgBr(s)│KBr(aq, 0.050 mol kg-1) ││Cd(NO3)2(aq, 0.010 mol kg-1) │Cd at 25°C E=-0.62V 7.17 Calculate the equilibrium constants of the following reactions at 25°C from standard potential data: (a) Sn(s) + Sn4+(aq) → 2Sn2+(aq) (b) Sn(s) + AgCl(s) → SnCl2(aq) + 2Ag(s) (a)K = 6.5x 109; (b) K = 1.5 x 1012 7.18 The emf of the cell Ag│AgI(s) │AgI(aq) │Ag is +0.9509V at 25°C Calculate (a) the solubility product of AgI and (b) its solubility (a) 9.2 x 10-9 M, (b) 8.5 x 10-17 M 7.1 Tại 2257K, 1.00 atm Ta có phản ứng: 2H2O(k) = 2H2(k) + O2(k) H2O phân hủy 1.77% Tại trạng thái cân tính: a/ Tính K b/ ∆G0r c/ ∆Gr T = 2257K Bài làm: a/ Gỉa sử số mol ban đầu H2O 1mol 2H2O(k) = 2H2(k) + O2(k) [ ]0 [] 0 1-α α 0,5α KP = (P2H2 PO2) / (P2H2O) = KN.P∆n = (α2.0,5 α) / [(1- α)2.(1+0,5α)] Thay α = 0.0177 vào biểu thức tính KP ta được: KP = (0,01772.0,5.0,0177) / [(1-0,0177)2.(1+0,5.0,0177)] = 2,85.10-6 b/ ∆G0r = - R.T.lnKP = - 8,314.2257.ln(2,85.10-6) = 239591 J/mol = 240 Kj/mol c/ Vì hệ cân nhiệt độ 2257K nên ∆Gr (2257) = 7.2 1bar ≈ 1atm Gỉa sử số mol N2O4 ban đầu 1mol N2O4(k) = 2NO2(k) [ ]0 [] 1-α 2α Kp = KN .P∆n , ∆n = 1, P = 1atm KP = KN = (2α)2 / (1-α) Thay α = 0,1846 vào biểu thức ta KP = (2.0,1846)2 / (1-0,1846) = 0,1672 Nếu tính xác 1bar = 0,9869atm Khi KP = KN.P = [(2.0,1846)2 / (1-0,1846)].0,9869 = 0,1650 ∆G0r = - R.T.lnKP = -8,314.298.ln(0,1650) = 4431,77 J/mol = 4,431 kJ/mol c/ Ap dụng công thức: ln(KP2(373) / KP1(298)) = ∆H.(T2-T1) / R.T1.T2  KP2(373) = KP1(298) Exp[∆H.(T2-T1) / R.T1.T2]  KP2(373) = 0,1650 exp[57,2.103(373-298) / 8,314.373.298] = 17,122 Lời giải: N2O4(k) = 2NO2(k) CB: 1-α 2α Nồng độ phần mol chất: N(N2O4) = 1−α 1+α Σni = 1-α + 2α = + α N(NO2) = 2α 1+α K = KN.P∆ν = KN.p (∆ν = 2-1=1 p = 0,99 atm) 2 4.0,1846 4α 0,99 = 0,14 p K = KN.p = = - 0,18462 1-α ∆Gro = -RTlnK = -8,31.298.ln0,14 = +4,869 kJ.mol-1 K2 ln K = ∆H(T2 - T1 ) 57200(373 - 298) = RT1T2 8,31.298.373 K2 = 104 → K2 = 14,556 K1 7.3 Tính ∆G, KP 298K: PbO(s) + CO(g) = Pb(s) + CO2(g) ∆G0r = ∆G0f(CO2) - ∆G0f(CO) - ∆G0f(PbO) = -314,36 - (-137,17) - (-188,93) = -68,26 kJ/mol KP = exp (-∆G/R.T) = exp[-(-68260) / 8,314.298] = 9,2.1011 b/ Tính ∆G, KP 400K: Tính ∆G0r = ∆H0f(CO2) - ∆H0f(CO) - ∆H0f(PbO) = -393,51 -(-110.53) -(-218,99) = -63,99 kJ/mol Tính KP2(400) ln(KP2 / KP1) = ∆H.(T2-T1) / R.T1.T2  KP2 = KP1.exp[ ∆H.(T2-T1) / R.T1.T2 ]  KP2 = 9,2.1011.epx[ -63990.(400-298) / 8,314.400.298] = 1,3.109 ∆G0r = - R.T.lnKP2 = -8,314.400.ln(1,3.109) = -69,79 kJ/mol Solution: From information in the Data section: PbO(s) + = CO(g) = Pb(s) + CO2(g) (∆G of )298 (kJ.mol-1) -187,89 -137,17 -394,36 (∆H of )298 (kJ.mol-1) -217,32 -110,53 -393,51 S298 (J.K-1.mol-1) 68,7 197,67 64,81 213,74 a) C1: ∆G or = (0 - 394,36) – (-187,89 - 137,17) = - 69,3 kJ.mol-1 69300 ∆G or (∆G )298 = -RTlnK298 → lnK298 = = 8,31.298 = 14.1011 RT o r C2: ∆H or = (-393,51) - (-217,32 - 110,53) = -65,66 kJ.mol-1 ∆S or = (64,81 + 213,74) - (68,7 + 197,67) = 12,18 J.K-1.mol-1 (∆G or )298 = -65,66 – 298 12,18.10-3 = -69,29 kJ.mol-1 69290 ∆G or = 8,31.298 = 14.1011 RT ∆H(T2 - T1 ) 57200(373 - 298) = RT T = → K400 = 1,3.109 8,31.298.373 (∆G or )298 = -RTlnK298 → lnK298 = b) K 400 ln K 298 (∆G or )400 = -RTlnK400 = -8,31.400.ln(1,3.109) = -69700 J = -69,7 kJ 7.4 2A + B = 3C + 2D Bđ Pư 0,6 0,3 0,9 1+ 0,6 CB 0,4 1,7 0.9 1,6 ∑n= 4,6 a ) Tính phân số mol A, B, C, D lúc cân NA=0,4/ 4,6= 0,087 NB=1,7/ 4,6= 0,37 NC=0.9/ 4,6= 0,196 ND= –( NA + NB+ NC)=0,347 b) Tính Kx Kx= NC3 ND2/ NA2 NB = 0,33 C) Kp= Kx.p∆n =0,33.12= 0,33 d) ∆G0r= - RTlnKp=- 8,314.298ln0,33=2,8.103 J/mol 7.5 Tính Kp 1280K Kp= exp (-∆G0r/ RT)= exp(-33.103/8,314.1280)= 0,045 Gọi Kp’> nhiệt độ T ln Kp’/ Kp=∆H(T’-T)/ R.T’.T theo đề Kp’>  ∆H(T’-T)/ R.T’.T > ln1/0,045=3,10  224.103(T’-1280)/8,314 T’.1280 >3,10  T’ >1501  Vậy ước tính T’ =1501thì có phản ứng Kp’> 7.6 dlnK=∆H/RT2 => ∆H=( dlnK).RT2 lnK=A + B/T +C/T2 => dlnK=-B/ T2 + (-2C/ T3) Vậy ∆H=- RT2 (B/ T2 + 2C/ T3) Thay B=-1080, C=1,51.105, R=8,314, T=400K Ta được: ∆H=2768,562 J/mol=2,768562KJ/mol Và K(400K)=exp(A + B/T +C/T2)= exp(-1,04 - 1088/400 +1,51.105/4002)=0,0598 ∆G0r= - RTlnK=- 8,314.400lnK=9,367 J/mol ∆G0=∆H0-T ∆S0 =>∆S0=∆H0-∆G0/T=2,768-9,367/400=-0,0165KJ/mol=16,5J/mol Lời giải: Từ biểu thức: lnK = ∫ ∆HdT +a RT Nếu đặt ∆H = R(b + cT-1), Ta có: So sánh với: c (b + cT −1 ) dT + a = a + b( - ) + ( - ) ∫ T2 T T 1 lnK = -1,04 + 1088( - ) - 1,51.105( - ) T T lnK = Suy ra: a = -1,04; b = 1088; c = -2 1,51.105 = -3,02.105 → ∆H400 = R(1088 - 3,02.105T-1) = 8,31.(1088 - 3,02.105.400-1) = +2.77 kJ mol-1 → lnK400 = -1,04 - 1088 1,51.105 + = -1,04 – 2,72 + 0,943 = -2.81625 400 4002 → ∆G400 = -RT.lnK400 = -8,31.400.lnK400 = 9361 J.mol-1 ∆G400 = ∆H400 – T.∆S400 → ∆S400 = 2770 - 9361 = -16,5 J.K-1.mol-1 400 7.7 Borneol↔isoborneol; 600tour=0,790atm ∆G=∆G0 +RT.lnQ Trong Q= PBorneol/Pisoborneol Pisoborneol=0,3/(0,157+0,3) ×0,790=0,527atm Pborneol=0,15/(0,15+0,3) ×0,790=0,263atm Vậy ∆G=9,4.103 +8,314.503ln0,527/0,263=12306J/mol=12,306KJ/mol Lời giải: borneol (C10H17OH)(gas) = isoborneol (C10H17OH)(gas) Vì phản ứng có Δn = 1-1 = nên: 0,3 Q = Qp = Qx = 0,15 = ∆G = ∆Go + RTlnQ = 9400 + 8,31.503.ln2 = 12297 J.mol-1 = 12,3 kJ.mol-1 7.8 Thầy giải giúp chúng em này, ko hiểu đề Lời giải: H2CO(g) = CO(g) + H2(g) Ta có quan hệ Kp = KN.PΔn Δn = – = Kp = KN.1 = KN’.2 → K 'x = = 0,5 hay 50 per cent Kx Bài 7.9 Borneol  isoborneol nborneol = 7,5/154=15/308 (mol) nisoborneol= 14/154= 1/11(mol) Borneol  isoborneol 15/308 1/11 α α 15/308-α 1/11+α = Áp suất hệ lúc cân = K = Kp =  =1,152 (atm) = = 1,06 + α = 1,06 - 1,06.α Tổng số mol hệ trước sau phản ứng 0,1396 mol Gọi NA phần số mol borneol 1-NA phần số mol isoborneol KP = = = = 0,106 NA=0,4854 Vậy phần số mol borneol 0,4854 Phần số mol isoborneol 1- 0,4854= 0,5146 (thưa thầy 7.9 có dư kiện không ) Lời giải: 7,5 14 = 0,0487 mol; niso = = 0,0974 mol 154 154 borneol (C10H17OH)(gas) = isoborneol (C10H17OH)(gas) 0,0487 + x 0,0974 – x Vì phản ứng có Δn = -1 = nên: 0,0974 - x K = K N = Kn = = 0,106 0,0487 + x → 0,106(0,0487 + x) = 0,0974 – x → 1,106.x = 0,0974 – 0,0052 = 0,092 → x = 0,0834 0,1321 → nbo = 0,1321 → Nbo = = 0,904 → Niso = 0.096 0,1461 Bài không cần đến kiện thể tích V = dm3 nbo = Bài 7.10 Áp dụng phương trình Ln = a,Theo giả thuyết K2=2K1 T2=3080K T1= 2980K  =528935,1 J.mol-1 =53 kJ.mol-1 = b, theo giả thuyết K2=0,5 K1  = -53 kJ.mol-1 Lời giải: K RT1T2ln K 400 ∆H(T2 - T1 ) K1 ln K = RT T → ∆H = 298 (T2 - T1 ) K2 8,31.298.308.ln2 Khi ln K = ln2 → ∆H = = 52868 J.mol-1 = +53 kJ.mol-1 10 K2 8,31.298.308.ln0,5 Khi ln K = ln0,5 → ∆H = = -52868 J.mol-1 = -53 kJ.mol-1 10 Bài 7.11 N2 + H2  NH3 (NH3) = = = -16.5 kJ/mol + RTln PNH3= bar PH2 = bar PN2 = bar = -16500 + 8,314.298.ln = -14426,3 J/mol = -14,43 kJ/mol phản ứng xảy theo chiều tạo sản phẩm Bài 7.12 CaCO3 CaO + CO2 = + = + ) - phản ứng tự diễn biến nên Lời giải: CaCO3(s) → CaO(s) + CO2(g) ∆H298(kJ.mol ) -1206,9 -635,09 -393,51 -1 -1 S298(J.K mol ) 92,9 39,75 213,74 ∆Hr = (-635,09 - 393,51) – (-1206,9) = 178,3 kJ.mol-1 ∆Sr = (39,75 + 213,74) – (92,9) = 160,59 J.K-1.mol-1 -1 178300 ∆GT = ∆HT - T∆ST < → T > 160,59 = 1110K Bài 7.13 CaF2(s) = Ca2+(aq) + 2F-(aq), = - RTlnK = - 8,314.298.ln = 61098 J Vậy Lời giải: ∆G1 = -RTlnK = - 8,31.298.ln3,9.10-11 = 59353 J.mol-1 = 59 kJ.mol-1 CaF2(s) = Ca2+(aq) + 2F-(aq) ∆G1 = 59 Ca(s) + F2(g) = CaF2(s) ∆G2 = -1167 (1) + (2) → Ca(s) + F2(g) = Ca2+(aq) + 2F-(aq) ∆G = ∆G1+∆G2 = -1108 kJ.mol-1
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