Chapter3Chemical EquilibriumTwoissues:.Spontaneity of reactions at nonstandard state·Chemicalequlibrium
Two issues: •Spontaneity of reactions at nonstandard state •Chemical equlibrium Chapter 3 Chemical Equilibrium
一、非标态下，化学反应方向的判据化学反应等温式△,G㎡=△,Gm+RTlnQQ：反应商aA+bB=gG+hH△G.0逆向自发进行
一、非标态下，化学反应方向的判据——化学反应等温式 △rGm = △rGm º+ RT㏑Q Q：反应商 △rGm 0 逆向自发进行 aA + bB = gG + hH
例1: C,H,OH(L) → C,H,OH(g) (298K), △,G°= 6.2(KJ/mol,当P, =1/2 P%,P, =1/10 P,P=1/20 P°时,反应能否自发进行？解：△,Gm（1)=△,Gm+RTln0=6.2+8.314×298n(1/2)= 6.2-1.7=4.5(KJ/mol)>0△Gm(2)=6.2+8.314×298×10-31n（1/10)=6.2-5.7=0.495(KJ/mol)>0△Gm(3)=6.2+8.314×298×10-31n（1/20）=6.2-7.4=一1.2（KJ/mol)<0蒸发可自发进行
例1：C2H5OH(l.) → C2H5OH(g.) (298K)，△rGm º= 6.2 (KJ/mol)，当P1 = 1/2 Pº，P2 = 1/10 Pº，P3 = 1/20 Pº时， 反应能否自发进行？ 解： △rGm (1)= △rGm º+ RT㏑() = 6.2 + 8.314×298㏑(1/2) = 6.2－1.7 = 4.5 (KJ/mol) > 0 △rGm(2) = 6.2 + 8.314×298×10－3㏑(1/10) = 6.2 －5.7 = 0.495 (KJ/mol) > 0 △rGm(3) = 6.2 + 8.314×298×10－3㏑(1/20) = 6.2－7.4= －1.2 (KJ/mol) < 0 蒸发可自发进行
例2：求298K时，Fe生锈的逆反应发生的条件解：2Fe (s.) +3/2 0, (g) →Fe,03(s.)0-741.0△,G(KJ/mol)0.:.△,G.°=-741.0KJ/mol1若要逆向进行，需△G0= -741.0 —3.72 ln(Po2 / P°) > 0ln(Po2 / P) < -199.4Po2 / P<2.52× 10-872<2.55×10-82PaP02.不可能发生
例2：求298K时，Fe生锈的逆反应发生的条件。 解： 2Fe (s.) + 3/2 O2 (g.) → Fe2O3 (s.) △fGm º(KJ/mol) 0 0 －741.0 ∴ △rGm º= －741.0 KJ/mol 若要逆向进行，需△rGm O = -741.0 —3.72㏑(PO2 / Pº) > 0 ㏑(PO2 / Pº) < －199.4 PO2 / Pº< 2.52×10－87 PO2 < 2.55×10－82 Pa 不可能发生
例3：求下述反应逆反应发生的条件Zn(s) + Cu2+(aq) = Zn2+(aq) + Cu(s)解: Zn(s) + Cu2+(aq) = Zn2+(aq) + Cu(s)(=01mol/dm301-xtX△,Gm(KJ/mol)064.77-153.891△,Gm= △,Gm(Zn2+)-△,Gm(Cu2+)=-153.89-64.77=-218.66(KJ/mol)△,Gm=△,G.+RTln=-218.66+8.314×298×10-31n（[Zn2+//|Cu2+D)>01n([Zn2+J/[Cu2+D)>(218.66/2.48)>0([Zn2+)/[Cu2+))>2.13×1038即若想使反应逆向进行，[Zn2+]>2.13×1038[Cu2+
例3：求下述反应逆反应发生的条件。 Zn(s) + Cu2+(aq) = Zn2+(aq) + Cu(s) 解： Zn(s) + Cu2+ (aq) = Zn2+ (aq) + Cu(s) t=0 1mol/dm3 0 t x 1-x △fGm º(KJ/mol) 0 64.77 －153.89 0 △rGm º= △fGm º(Zn2+ )－△fGm º(Cu2+ )=－153.89－64.77=－218.66 (KJ/mol) △rGm = △rGm º+ RT㏑ = －218.66 + 8.314×298×10－3㏑([Zn2+ ]/[Cu2+ ]) > 0 ㏑([Zn2+ ]/[Cu2+ ])>(218.66/2.48) > 0 ([Zn2+ ]/[Cu2+ ]) > 2.13 ×1038 即若想使反应逆向进行，[Zn2+ ] > 2.13×1038[Cu2+ ] + + C Cu C Zn 2 2
三点说明①当△G.数值较大时，可用△G.代替△G.估计/判断反应方向。[△,Gm>41.84KJ/mol②当△G㎡数值较小时，△，Gm<40KJ/mol，需用化学反应等温式计算出△Gm，用△G的符号判断方向。③当△G.=0时，反应达平衡，由化学反应等温式△,G.=-RTInQ=-RTInK°：一定温度下，一确定反应，其△G.是个定数：此时Q为一常数，称之为标准平衡常数，用K表示。显然同样类型的反应，相同温度时，K越大，表明反应正向进行的程度越大
三点说明 ① 当△rGm º数值较大时，可用△rGm º代替△rGm估计 / 判断 反应方向。 |△rGmº| > 41.84 KJ/mol ② 当△rGm º数值较小时，|△rGmº| < 40KJ/mol，需用化学反 应等温式计算出△rGm，用△rGm的符号判断方向。 ③ 当 △rGm = 0时，反应达平衡，由化学反应等温式 △rGm º=－RT㏑Q = －RT㏑Kº ∵一定温度下，一确定反应，其△rGm º是个定数 ∴此时Q为一常数，称之为标准平衡常数，用Kº表示。显然， 同样类型的反应，相同温度时，Kº越大，表明反应正向进行 的程度越大
2.Chemicalequlibrium2.1 Characteristics of an equilibrium2.2Equlibrium constants and equlibriumconstantexpressions: K, K, Kp2.3Transfer rate平衡转化率α=(C初一C平)/C初=△C/C初3. Shifts in chemicalequilibium3.1Effect of changes in concentrations3.2 Effectofchanges inpressure3.3 Effectof changes in temperature3.4Effectofcatalyst
2.Chemical equlibrium 2.1 Characteristics of an equilibrium 2.2 Equlibrium constants and equlibrium constant expressions：Kº, Kc , Kp 2.3 Transfer rate平衡转化率 α= (C初－C平) / C初 = △C / C初 3. Shifts in chemical equilibium 3.1 Effect of changes in concentrations 3.2 Effect of changes in pressure 3.3 Effect of changes in temperature —— 3.4 Effect of catalyst
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Each component concentration will not change withtime. Although different initial concentrations give differentequilibium concentrations, the quantity, [H]’eq/ [H,leql] eq,is constant.K,=[H]2e/ [H,leq[leqTheconstant is called equilibrium constant for theequilibrium and is reprensented bythe symbol Kc. Dynamic equilibrium. The forward reaction rate equalsto thereverse reaction rate10It is a conditional eguilibrium.When the concentration.pressure and temperature change, the equilibrium positionwill change
u Each component concentration will not change with time. Although different initial concentrations give different equilibium concentrations, the quantity, [HI]2 eq/ [H2 ] eq[I2 ] eq, is constant . Kc = [HI]2 eq/ [H2 ] eq[I2 ] eq The constant is called equilibrium constant for the equilibrium and is reprensented by the symbol Kc. • Dynamic equilibrium. The forward reaction rate equals to the reverse reaction rate. It is a conditional equilibrium. When the concentration, pressure and temperature change, the equilibrium position will change
2.2Equlibrium constants andEqulibrium constant expressions: K°, K,, KaA+bB=gG+hH2.2.1Experimental equilibrium constants(1)Concentration equilibrium constants K[G][H]K[A][B]C: mol/L
2.2 Equlibrium constants and Equlibrium constant expressions：Kº，Kc，Kp aA + bB = gG + hH 2.2.1 Experimental equilibrium constants (1) Concentration equilibrium constants Kc a b g h A B G H Kc = C: mol/L