普通化学复习题(一)09 -计算题答案
四、计算题:
⊙⊙⊙1. 解:△r G m (T) = △r H m (298K) - T △r S m (298K)
⊙---1)△r G m (298K) = (402 -298×189.6×103) kJ·mol 1 =345.5 kJ·mol 1>0, 说明该反应在25℃时不能自发进行。
⊙⊙⊙⊙2)因为该反应的△r S m >0,温度升高,△r S m 基本不变,T △r S m 增大,△r G m 数
值将减小,从而有利于反应的进行。
⊙⊙⊙3)令 △r G m (T) = △r H m (298K) - T △r S m (298K) = 0
⊙⊙ 则 T 转变= △r H m ÷ △r S m = 402×103 ÷ 189.6 = 2120 K
∴ 反应逆向进行所需的最高温度为2120 K。
⊙⊙⊙2. 解:△r H m =∑νB △f H m (B),将有关数据代入右式可得在298.15K 时反应的△r H m ⊙⊙⊙-⊙⊙为:△r H m =△f H m (NH4+) + △f H m (OH) - △f H m (NH3) - △f H m (H2O)
--= (-132.51 - 229.99 + 80.29 + 285.83)kJ ·mol 1=3.62kJ ·mol 1
298.15K 时反应的标准摩尔熵变为:
⊙⊙⊙-⊙⊙△r S m = S m (NH4+) + S m (OH) - S m (NH3) - S m (H2O)
-- = (-113.4 - 10.75 - 113.4 - 69.91) J·K 1·mol 1
-- = -78.56 J·K 1·mol 1
⊙⊙⊙由公式 △r G m (T) = △r H m (T) - T △r S m (T) 可求得298.15 K时反应的
⊙⊙⊙△r G m = △r H m (298K) - T △r S m (298K) =
-- = [3.62×103 —298.15×(-78.56)] J·mol 1 = 27.04 kJ·mol 1
⊙⊙⊙△r G m =-RTln Kb = - 2.303 lg Kb
⊙⊙∴ lg Kb = -△r G m ∕(2.303RT) = -27.04×103÷(2.303×8.314×298.15)
⊙- = -4.74 K b = 1.80×105
3. 解:设最初容器中有1 mol的NO 2,则
1) 2 NO2(g) == 2 NO(g) + O 2 (g)
开始时物质的量∕mol 1 0 0
平衡时物质的量∕mol 1-0.56 0.56 0.28
依题意,该过程为定压定温,n 总 = ( 1-0.56+0.56+0.28) =1.28 mol 各物种对应的物质的量分数(摩尔分数)为:
x (NO2) =(1-0.56)÷ 1.28 = 0.34 x (NO)=0.56/1.28= 0.44
x (O2) = 0.28/1.28=0.22 p (B) = x (B)p总,p 总=p ,
⊙⊙⊙⊙ [p (NO)/p]2·[p (O2)/p] (0.44p总/p) 2·(0.22p总/p)
⊙K =——————————————=————————————————
⊙⊙[p (NO2) /p]2 (0.34p总/p) 2
= 0.442×0.22÷0.342=0.37
2) 2 NO2(g) == 2 NO(g) + O 2 (g)
开始时物质的量∕mol 1 0 0
平衡时物质的量∕mol 1-0.80 0.80 0.40
n 总=(1-0.80+0.80+0.40)mol = 1.40 mol
∴ x (NO2) =(1-0.80)÷ 1.40 = 0.14 x (NO)=0.80 /1.40= 0.57
x (O2) = 0.40/1.40=0.29
⊙⊙⊙⊙ [p (NO)/p]2·[p (O2)/p] (0.57p总/p) 2·(0.29p总/p) ⊙⊙K =————————————=—————————————=4.8 p总/p=0.37
⊙⊙[p (NO2) /p]2 (0.14p总/p) 2
∴ p 总= 0.37p ÷4.8 = 7.7 kPa ⊙⊙
4. 解:1)pH =8.00, pOH =pK w —pH =14.00-8.00=6.00 ,
---c (OH) =1.0×106mol ·dm 3
---pH =10.0,pOH =14.00-10.00=4.00, c (OH) =1.0×104 mol ·dm 3
当两溶液等体积混合后,浓度减半,即
-----c (OH) =(1.0×106+1.0×104)÷2 = 5.05×105 mol ·dm 3
pOH =4.30,pH =9.70
--1) pH =2.00,c (H+) = 1.0×102 mol ·dm 3
--pH =13.00, pOH =1.00, c (OH) =0.10 mol·dm 3
当两溶液等体积混合后,浓度减半,且发生了强酸强碱中和反应,
-由于c (OH) >c (H+) ,故碱过量,则
-----c (OH) =(1.0×101+1.0×102)÷2 = 4.5×102 mol ·dm 3
pOH =1.35,pH =12.65
-5. 解:1)根据溶度积规则,可分别计算出生成Pb(OH)2和Cr(OH)3沉淀所需OH 的
最低浓度。
-Pb(OH)2(s) == Pb 2+ (aq) + 2 OH(aq)
⊙-K sp (Pb(OH)2) =c (Pb2+) ·c 2 (OH)
-⊙--c (OH) =[Ksp (Pb(OH)2) ÷ c (Pb2+)]1/2=(1.2×1015÷3.0×102) 1/2
=2.0×10-7 mol·dm-3
-Cr(OH)3(s) == Cr 3+ (aq) + 3 OH(aq)
⊙-K sp (Cr (OH)3) =c (Cr3+) ·c 3 (OH)
-⊙--C(OH) =[Ksp (Cr (OH)3) ÷ c (Cr3+)]1/3=(6.3×1031÷2.0×102) 1/3
- =3.1×1010 mol·dm -3
-∵ 在Pb 2+ 和 Cr 3+共存的条件下,Cr(OH)3沉淀的c (OH) 小于Pb (OH)2沉淀的
-c (OH) ,
∴ Cr(OH)3沉淀先析出。
-2)当Cr 3+离子完全沉淀时所需的c (OH) 为:
----c (OH) =(6.3×1031÷1.0×105) 1/3 =4.0×109 mol·dm -3
⊙----c (H+) = K w ÷ c (OH) =1.0×1014÷4.0×109=2.5×106 mol·dm -3
--pH =5.6, 而Pb 2+离子开始沉淀时的c (OH) =2.0×107 mol·dm -3 , pH=7.3
∴ 要分离这两种离子,溶液的pH 应控制在5.6~7.3范围内。
6. 解:1)锌在酸中置换氢: Zn + 2 H+ = Zn 2+ + H 2
-在标准状态下,c (H+) =c (Zn2+) =1.0 mol·dm 3
⊙⊙⊙E MF =E MF =E (H +/H2)- E (Zn2+/Zn) =0-(-0.76)=0.76 V >0
∴ 所以上述反应能自发进行,即锌在酸中能置换氢。
--2)锌在碱中置换氢: Zn + 2 OH = ZnO 22 + H 2
----在标准状态下,c (OH) =1.0 mol·dm 3,c (H+) =1.0×1014 mol·dm 3
--- 2 H2O + 2 e = H 2 + 2 OH 或 2 H+ + 2 e = H 2
⊙⊙⊙E(H+/H2) =E (H +/H2)+(0.0592÷2)lg[c2(H+) ÷(p(H2)/p)] (当p(H2) =p )
- =0+(0.0592÷2)lg(1.0×1014)2 =-0.83 V
---⊙已知:ZnO 22 + 2 H2O + 2 e = Zn 2+ + 4 OH E = -1.22 V
∴ E MF =-0.83 - (-1.22) =0.39 V > 0
即锌在碱中置换氢的反应是自发的。
7. (1)将下列半反应组成原电池:
﹣⊙ Cu + + e = Cu E (Cu+/Cu) = 0.52 V; ⊙
[Cu(CN)2] + e = Cu + 2 CN E 2 = ﹣0.896 V;
⊙-⊙⊙- E (Cu(CN)2∕Cu) = E (Cu+/Cu) - 0.0592 lg Kf (Cu(CN)2)
⊙-⊙⊙- lg Kf (Cu(CN)2) = [E(Cu+/Cu) - E (Cu(CN)2∕Cu)]÷ 0.0592
= (0.52 + 0.896) ÷0.0592 = 23.92
⊙- ∴K f (Cu(CN)2) = 8.3×1023
(2)将下列半反应组成原电池:
﹣⊙ Co 3+ + e = Co 2+ E (Co3+/Co2+) = 1.84 V;
﹣⊙ [Co(NH3) 6]3+ + e = [Co(NH3) 6]2+ E 4 = 0.10 V;
⊙E (Co(NH3) 63+∕Co(NH3) 62+)
⊙⊙⊙=E (Co3+/Co2+) + 0.0592 lg[Kf (Co(NH3) 62+) ÷ K f (Co(NH3) 63+)]
⊙将K f (Co(NH3)62+) = 1.3×105代入上式,得
⊙0.10 = 1.84 + 0.0592 lg 1.3×105 - 0.0592 lg Kf (Co(NH3) 63+)
⊙ ∴ K f (Co(NH3) 63+) = 3.2×1034
8. 解:(1)电池符号为
⊙--Pt ,Cl 2 (p) | Cl(aq) || MnO4(aq),Mn 2+ (aq),H + (aq) | Pt
(2 ) 电极反应为:
-⊙- 负极:2 Cl (aq) = Cl 2 (g , p) + 2 e
-- 正极:MnO 4(aq) + 8 H+ (aq) + 5 e = Mn 2+(aq) + 4 H2O ( l )
--电池反应:2 MnO4(aq)+10 Cl(aq)+16 H+(aq) = 2 Mn2+(aq)+5Cl 2(g)+8 H2O( l )
⊙⊙⊙ E MF = E (+)- E (-)= (1.51 - 1.36)V = 0.15 V
⊙⊙-(3)△r G m = -zFE MF = -10×96500×0.15 = -1.4×102 kJ·mol 1
⊙⊙ lg K = zFE MF ∕0.0592 = 10×0.15÷0.0592 = 24.52
⊙∴ K = 2.6×1025
----- (4)当 c (H+) =1.0×102 mol·dm 3,而c (Cl) =c (MnO4) =c (Mn2+) =1.0 mol·dm 3,
⊙p (Cl2) = p 时,
⊙ E MF = E MF +(0.0592÷10) lg [c (H+)]16 = 0.15+0.00592 lg (0.010)16=-0.040 V
-⊙- 或:E (MnO4/Mn2+) = E (MnO4/Mn2+) + (0.0592÷5) lg [c (H+)]8
- = 1.51 + (0.0592÷5) lg (102) 8 = 1.32 V
∴ E MF = 1.32 - 1.36 = - 0.040 V
- (5 ) △r G m = -zFE MF = -10×96500×(-0.040) = 38 kJ·mol 1
9. 解: 反应: Sn 2+ + Ni = Sn + Ni 2+
- 平衡时浓度/mol·dm 3 x 0.20-x
⊙⊙⊙⊙lg K = zFE MF ∕0.0592 = 2×[ E(Sn2+/Sn) - E (Ni2+/Ni)]÷ 0.0592
⊙= 2×[-0.14 - (-0.25)]÷0.0592 = 3.72 , K = 5.25×103
(0.20-x) ∕x = 5.25×103 0.20-x ≈ 0.20
-- ∴ x = 0.20÷5.25×103 = 3.8×105 mol ·dm 3
--c (Sn2+) = x = 3.8×105 mol ·dm 3
--c (Ni2+) = 0.20 - 3.8×105 = 0.20 mol·dm 3
-10. 解:设平衡时AgBr 的溶解度为x mol·dm 3 ,并假设溶液的体积不变,则有
--- AgBr (s) + 2 S2O 32 == [Ag(S2O 3) 2]3 + Br (aq)
-平衡时浓度/mol·dm 3 2.0-2x x x
⊙---⊙⊙- K = [c (Ag(S2O 3) 3) ·c (Br)]/c 2(S2O 32) =K sp (AgBr)·K sp (Ag(S2O 3) 23)
- x 2/(2.0-2x) 2 = 5.3×1013×2.89×103 = 14.45 -﹣-⊙
x /(2.0-2x) = 14.450.5 = 3.80 解得 x = 0.90 mol·dm 3
- 即 AgBr 的溶解度为0.90 mol·dm 3。 -