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第三章 化学平衡 (Chemical Equilibrium)

第三章 化学平衡 (Chemical Equilibrium). Outline 1. 化学平衡、 平衡常数、 平衡常数关系式 2. K P vs. K C ( 气相反应 ) 3. 转化率 (  ) 4. Q vs. K 5. 平衡常数与自由能变化的关系 Van’t Hoff 等温式 :  G = G + RT ln Q G rxn  =  RT ln K 6. 影响化学平衡的移动的因素 ( 吕  查德里原理 ). H 2 O(g) + CO(g) H 2 (g) + CO 2 (g).

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第三章 化学平衡 (Chemical Equilibrium)

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  1. 第三章 化学平衡 (Chemical Equilibrium)

  2. Outline 1. 化学平衡、平衡常数、平衡常数关系式 2. KP vs. KC (气相反应) 3. 转化率 () 4. Q vs. K 5. 平衡常数与自由能变化的关系 Van’t Hoff 等温式: G = G + RT lnQ Grxn =  RT ln K 6. 影响化学平衡的移动的因素 (吕查德里原理)

  3. H2O(g) + CO(g) H2(g) + CO2(g) H2O(g) + CO(g) H2(g) + CO2(g) H2(g) + CO2(g) H2O(g) + CO(g) 可逆反应 (Reversible Reaction)  化学平衡 (Chemical Equilibrium)： 在可逆反应中，正逆反应速率相等，反应物和生成物的浓度不再随时间而改变。  化学平衡是一种动态平衡。

  4. H2O(g) + CO(g) H2(g) + CO2(g)

  5. H2O(g) + CO(g) H2(g) + CO2(g)

  6. 平衡的形态 反应 平衡常数 平衡 Ka (酸电离常数) HA + H2O H3O+ + A 酸(碱)电离 溶解 MA(s) Mn+(aq) + An(aq) Ksp (溶度积) 配合物的形成 Kf (形成常数) Mn++ aLb MLa(nab)+ Keq (反应平衡常数) 氧化还原 Ared + Box Aox + Bred AH2O Aorg 相平衡 KD (分布系数)

  7. H2(g) + I2(g) 2 HI(g) (T = 698.5 K) Kc=[HI]2/[H2][I2]

  8. a A + b B c C + d D The Reaction Quotient (反应商) and Equilibrium Constant (平衡常数) Under Any Reaction Conditions [C]c[D]d Reaction quotient = Q = [A]a[B]b When Reaction is at Equilibrium Reaction quotient = Equilibrium constant = K = [C]c[D]d [A]a[B]b

  9. N2 (g) + 3H2 (g) 2NH3 (g) [NH3] K'C = 1/2 N2 (g) + 3/2 H2 (g) NH3 (g) [N2]1/2[H2]3/2 Writing Equilibrium Constant Expressions (平衡常数关系式) (1) 平衡常数的数值与化学方程式书写的方式有关。 [NH3]2 KC = [N2][H2]3 KC = (K'C)2

  10. (2) 如果反应体系中有固体或纯液体参与反应, 它们的浓度不写在平衡常数关系式中。 Fe3O4(s) + 4H2(g) 3Fe(s) + 4H2O(g) [H2O]4 KC = [H2]4

  11. Cr2O72 (aq) + H2O (l) 2CrO42 (aq) + 2H+ (aq) (3) 在稀溶液中进行的反应,如反应有水参加,由于在稀的水溶液中,水的浓度近似于一个常数 (1000/18.016 = 55.5 M), 可将其合并到平衡常数关系式中,水的浓度不写进去. [CrO42]2[H+]2 KC = [Cr2O72 ]

  12. a A + b B c C + d D (4) 气相反应平衡常数 (KP ) 反应物和生成物为气体 PCcPDd KP = PAaPBb PA, PB, PC, PD各为 A, B, C, D 的平衡分压

  13. [C]c[D]d KC = [A]a[B]b a A + b B c C + d D [A], [B], [C], [D] 各为 A, B, C, D 的平衡浓度

  14. Zn(s) + 2 H+(aq) H2(g) + Zn2+(aq) (5) 复相反应的平衡常数：杂平衡常数 (Kx ) KX= [Zn2+]·PH2/[H+]2

  15. Equilibrium Constant Expression for Gases KP vs. KC (气相反应) a A + b B c C + d D (PV = nRT  P = RT = cRT) {[C]RT}c {[D]RT}d PCc PDd = KP = {[A]RT}a {[B]RT}b PAa PBb n [C]c [D]d (RT) [(c+d)(a+b)] = [A]a [B]b  KP = KC(RT)n (n = total moles of gaseous products  total moles of gaseous reactants) n V

  16. KP = 4.60/1.013 = 4.54 atm  KP = 4.60  100 = 460 kPa bar [例] C(s) + CO2(g) 2CO(g), KP = 4.60 (1040 K) ，将此平衡常数用KP 、KP 表示，并计算此温度的KC 、KC 及 KC 。 atm bar atm kPa kPa [解] C(s) + CO2(g) 2CO(g) PCO2 KP =  单位：压力 PCO2 又 1 atm = 1.013  102 kPa = 1.013 bar 1 bar = 105 Pa

  17. KP = KC(RT)n, n = 1 bar bar Kc = KP (RT)1 = 4.60  (0.0831  1040)1 = 0.0532 atm atm Kc = KP (RT)1 = 4.54  (0.0821  1040)1 = 0.0533 kPa kPa Kc = KP (RT)1 = 460  (8.31 1040)1 = 0.0532 P R bar 0.0831 atm 0.0821 kPa 8.31

  18. (6) 当几个反应相加得到总反应时, 总反应的平衡常数等于各相加反应的平衡常数之乘积。(多重平衡规则) [NO2]2 2NO + O2 2NO2 (1) K1 = [NO]2[O2] [N2O4] 2NO2 N2O4 (2) K2 = [NO2]2 [N2O4] K3 = 2NO + O2 N2O4 (3) [NO]2[O2] (1) + (2) = (3) [N2O4] [N2O4] [NO2]2 = K3  = K1 K2 = [NO]2[O2] [NO2]2 [NO]2[O2]

  19. (7) 平衡常数与温度有关.

  20. 平衡常数的分类 (1) 经验平衡常数 (实验平衡常数)： Kc , Kp , Kx (2) 相对平衡常数：Kr(或标准平衡常数Kө) r: relative

  21. (1) 经验平衡常数 (实验平衡常数)： Kc , Kp , Kx 例1：Cr2O72 (aq) + H2O (l) 2 CrO42 (aq) + 2 H+ (aq) Kc = ([CrO42]2 [H+]2) / [Cr2O72] = 2.0 × 105 (mol·dm3)3有量纲!

  22. 例2： N2O4 (g) 2 NO2 (g) Kp = P2(NO2)/P(N2O4) = 1116 kPa (373 K) 有量纲! KP = KC(RT)n  KC= KP/ (RT)n = 1163 / (8.31 373) = 0.37 mol·dm3

  23. (2) 相对平衡常数：Kr(或标准平衡常数Kө） 为了澄清经验(实验)平衡常数多值性、量纲≠1引起的混乱，引入“相对平衡常数”的概念。 定义：“标准压力”为 pө = 1 ×105 Pa (or 1 atm) “标准(物质的量)浓度”为 cө = 1 mol·dm 3

  24. 混合气体中气体 i 的相对分压为:(r : relative, 相对的) Pir = Pi / Pө 溶液中溶质i 的相对（物质的量）浓度为： cir = ci / cө  Pir 、cir 量纲为 1！

  25. 对于溶液反应：a A(aq) + b B(aq) d D(aq) + e E(aq) 相对浓度平衡常数为： Kcr = ( [cDr ]d [cEr ]e) / ( [cAr ]a [cBr ]b) =Kc×( cө)-△n (△n = d + e  a  b) • 对于气相反应：a A(g) + b B(g) d D(g) + e E(g) 相对压力平衡常数为: Kpr = ([PDr ]d [PEr ]e) / ( [pAr ]a [pBr ]b) =Kp× (pө) △n = Kc (RT)△n (pө) △n (△n = d + e  a  b)

  26. 对于多相反应 a A(g) + b B(aq) d D(s) + e E(g) 相对杂（混合）平衡常数为： K杂r = ([PEr ]e) / ( [PAr ]a [cBr ]b)  Kcr、Kpr、KXr 统一为 Kr： 优点:① 量纲为1; ② 单值。

  27. 关于“平衡常数”的要求： (1) 两套平衡常数都要掌握。 (2) 计算“平衡转化率”或某反应物(或产物)浓度(或分压)时，两套平衡常数均可以用，且用 Kc、Kp、Kx (经验平衡常数)更方便，故IUPAC未予废除。 (3) 只有Kr (或Kө) 才具有热力学含义。 例如，△Gø = RT lnK ， 只能用Kr (或Kө)。

  28. NO (g) + O3 (g) NO2 (g) + O2 (g) [NO2][O2] K = = 6 x 1034 at 25C [NO][O3] Meaning of the Equilibrium Constant, K K >> 1, at equilibrium, [NO2][O2] >> [NO][O3]  Reaction is product-favored (Equilibrium concentrations of products are greater than equilibrium concentrations of reactants.)

  29. 3/2 O2 (g) O3 (g) [O3] K = = 2.5 x 1029 at 25C [O2]3/2 K << 1, at equilibrium, [O3] << [O2]3/2  Reaction is reactant-favored. (Equilibrium concentrations of reactants are greater than equilibrium concentrations of products.)

  30. [Isobutane] Kc = = 2.50 at 298 K [Butane] Meaning of the Reaction Quotient, Q Butane Isobutane CH3CH2CH2CH3 CH3CH(CH3)CH3 正丁烷 异丁烷

  31. Comparing Q and K Relative Magnitude Direction of Reaction (反应方向) Q < K Reactant (反应物)  Product (生成物) Q = K Reaction at equilibrium Q > K Product  Reactant

  32. Determining an Equilibrium Constant Example. An aqueous solution of ethanol and acetic acid, each with a concentration of 0.810 M, is heated to 100C. At equilibrium, the acetic acid concentration is 0.748 M. Calculate K at 100C for the reaction C2H5OH(aq) + CH3CO2H(aq) CH3COOC2H5(aq) + H2O(l) ethanol acetic acid ethyl acetate Solution C2H5OH(aq) + CH3CO2H(aq) CH3COOC2H5(aq) + H2O(l) I 0.810 0.810 0 C 0.062 0.062 +0.062 E 0.748 0.748 0.062 K = = 0.062/(0.748)(0.748) = 0.11 (I = Initial; C = Change; E = Equilibrium) [CH3CO2C2H5] [C2H5OH][CH3CO2H]

  33. Example. Suppose a tank initially contains H2S with a pressure of 10.00 atm at 800 K. When the reaction 2 H2S(g) 2 H2(g) + S2(g) reach equilibrium, the partial pressure of S2 vapor is 0.020 atm. Calculate Kp. Solution 2 H2S(g) 2 H2(g) + S2(g) I 10.00 0 0 C 2(0.020) +2(0.020) +0.020 E 9.96 0.040 0.020 Kp = = (0.040)2(0.020)/(9.96)2 = 3.2 x 107 (I = Initial; C = Change; E = Equilibrium) P2H2PS2 P2H2S

  34. Using Equilibrium Constants in Calculations Example. The equilibrium constant K = 55.64 for H2(g) + I2(g) 2 HI(g) has been determined at 425C. If 1.00 mol each of H2 and I2 are placed in a 0.500-L flask at 425C, what are the concentrations of H2, I2, and HI when equilibrium has been achieved? Solution H2(g) + I2(g) 2 HI(g) I 1.00/0.500 1.00/0.500 0 = 2.00 M = 2.00 M C x x +2x E 2.00x 2.00x 2x K = 55.64 = (2x)2/(2.00x)(2.00x) = (2x)2/(2.00x)2  K1/2 = 7.459 = 2x / (2.00x)  x = 1.58  [H2] = [I2] = 2.00  x = 0.42 M; [HI] = 2X = 3.16 M

  35. Calculating an Equilibrium Concentration Using an Equilibrium Constant Example. The reaction: N2(g) + O2(g) 2 NO(g) contributes to air pollution whenever a fuel is burned in air at one high temperature, as in a gaoline engine. At 1500 K, K = 1.0x105. Suppose a sample of air has [N2] = 0.8 mol/L and [O2] = 0.20 mol/L before any reaction occurs. Calculate the equilibrium concentrations of reactants and products after the mixture has been heated to 1500 K. Solution N2(g) + O2(g) 2 NO(g) I 0.80 0.20 0 C x x +2x E 0.80x 0.20x 2x K = 1.0 105 = (2x)2/(0.80x)(0.20x)  (2x)2/(0.80)(0.20)  x = 6.3 104 (condition: 100 K < 0.80 & 0.20)

  36.  [N2] = 0.80  6.3 104  0.80 M [O2] = 0.20  6.3  104  0.20 M [NO] = 2x = 1.26x103M Note: if you don’t use the approximation (1.0 105)(0.80x)(0.20x) = 4x2 (41.0105)x2 + (1.0  105)x + (0.16105) = 0 ax2 bx c  x = 6.3  104(or 6.3  104)

  37. K A B + C I A0 0 0 C x x x E A0x x x Using approximation K = [B][C]/[A] = x2/(A0x) x2/(A0) (When K is very small, the value of x will be much less than A0)  If 100 K < A0, The approximate expression will give acceptable values of equilibrium concentrations.

  38. 转化率 () 转化掉的部分  = 转化前的全部

  39. [例] 在密闭容器中, CO 和 H2O 在 850◦C 时建立下列平衡: CO(g) + H2O(g) CO2(g) + H2(g), Kc = 1.0 (1) 计算 CO 和 H2O 起始浓度分别为 2.0 M 和 3.0 M 时,CO 和 H2O的转化率及各组分的平衡浓度; (2) 计算 CO 和 H2O 起始浓度分别为 1.0 M 和 5.0 M 时,CO 和 H2O的转化率; (3) 计算 CO 和 H2O 起始浓度分别为 2.0 M 和 5.0 M 时,CO 和 H2O的转化率。

  40. [解] CO(g) + H2O(g) CO2(g) + H2(g) Kc = 1.0 (1) 起始浓度 2 3 0 0 平衡浓度 2x 3x x x x2  X = 1.2 Kc = = 1 (2x)(3x) [CO] = 0.8 M [H2O] = 1.8 M [CO2] = 1.2 M [H2] = 1.2 M 1.2 / 2  100% = 60% CO = 1.2 / 3 100% = 40% H2O =

  41. (2) 解得 x = 0.83 (M) 0.83 / 1  100% = 83% CO = 0.83 / 5  100% = 17% H2O = (3) 解得 x = 1.4 (M) 1.4 / 2  100% = 70% CO = 1.4 / 5  100% = 28% H2O =

  42. 平衡常数与自由能变化的关系 a A + b B c C + d D Van’t Hoff 等温式: (封闭体系、等温、不做非体积功) (推导：p55) G = G+ RT lnQ (1) G：在指定温度下, 反应物与生成物处于标准状态， 即固体和液体系纯净的固体和纯净的液体物质、气体分压为1bar，溶液的浓度为1时，反应物完全转变为生成物的自由能变化。 G ：气体的分压或溶液的浓度为任意数值时，反应的自由能变化。

  43. a A + b B c C + d D [C]c [D]d Q = [A]a [B]b 平衡时, G = 0  0 = G + RT lnK  Grxn =  RT ln K (2) (= 2.303 RT log K) R = 8.314 JK1mol 1

  44. Grxn =  RT ln K If Grxn < 0  product-favored reaction  K > 1 (Grxn 愈负，K愈大) If Grxn > 0  reactant-favored reaction  K < 1

  45. Free Energy, the Reaction Quotient, and the Equilibrium Constant  The free energy at equilibrium is lower than either the free energy of the pure reactants or the free energy of the pure products. Grxn =  Gf,生成物  Gf,反应物 =Hrxn TSrxn =  RT ln K

  46.  Grxn predicts the direction in which a reaction proceeds to reach equilibrium and may be calculated from Grxn = Grxn + RT ln Q = RTln (Q/K) When Grxn < 0, Q < K  reactants  products When Grxn > 0, Q > K  products  reactants When Grxn = 0, Q = K  at equilibrium

  47. [例] 分别计算合成氨反应: N2(g) + 3H2(g) 2NH3(g)在 298K 及 673K 时的平衡常数. [解] (1) 298 K G= 2  Gf = 2  (16.5) = 33.0 (kJ·mol1) G =  RT lnKp  Kp = 6.11  105 (R = 8.314  103kJK1mol1)

  48. (2) 673 K G◦673 = H◦298  TS◦298 H◦298 = 2  H◦f = 2  (46.1) = 92.2 (kJ·mol1) S◦298 = 2  S◦NH3  S◦N2  3  S◦H2 = 2  192.51 191.49 3  130.6 = 198.3 (J·mol 1·K 1)  G673 = 92.2  673  (198.3/1000) = 41.3 (kJ·mol1) G673 =  RT lnKp  Kp = 6.15  10 4

  49. 影响化学平衡的移动的因素 (1)浓度 (2)压力 (3)温度

  50. 吕查德里原理 (Le Châtelier’s principle) 改变平衡体系的外界条件时，平衡向着削弱这一改变方向移动。

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