北京化工大学 Model of Mid-term Examination of <Inorganic Chemistry) bilingual course) Course code 70 T 课程代码 Class No.: Name and ID: Items(题号) 二三四 五六 Total score(总分) Score(得分) -、1,2x3V4x56x789x1011x12V 二、1A;2C,3A;4D5D,D:6B:7D8B9C,10D11B12B13B;14C 三、 1、左；CT,Cu2;降低：蓝色。2、75.97：75.97：H0g.3、4.46;稍稍变小。4、大：大： 小：小。5、BaS0:CaC0,)+2NH+2H,0=Ca2+2NH.H,0+l,C0.6、Fe3;Sn2+。 7、标准氢电极：甘汞电极。8、硝酸四氨，二水合铬(I):Co(H20(en)2CC2.9、9.37x103, 1.32x103. 10、Cu2*与cn形成的配合物：熵。 四、 1.7PbO2 +2 MnBr2 14 HNO3 =7 Pb(NO3)+2Br2 +2 HMnO4+6H2O 2、CI03+6r+6H=CI+32+3H0 五、 Fe,NE,AP.NH,4,0FeO,Ao,↓Fe(OH,)↓ Ni(NH)2+ 第1页
北京化工大学 Model of Mid-term Examination of《Inorganic Chemistry》 （bilingual course） Course code 课程代码 C H M 2 1 7 0 T Class No.: Name and ID: Items (题号) 一 二 三 四 五 六 Total score(总分) Score(得分) 一、1 √; 2 x; 3 √; 4 x; 5 √; 6 x; 7 √; 8 √; 9 x; 10 √; 11 x; 12 √ 二、1 A; 2 C; 3 A; 4 D; 5 D, D; 6 B; 7D; 8 B; 9 C; 10 D; 11 B; 12 B; 13 B; 14 C 三、 1、左; Cl- , Cu2+ ; 降低；蓝色。 2、75.97； 75.97； H2O(g). 3、4.46; 稍稍变小。4、大；大； 小；小。 5、BaSO4; CaCO3(s) + 2NH4 + + 2H2O = Ca2+ + 2NH3.H2O + H2CO3。 6、 Fe3+; Sn2+。 7、标准氢电极；甘汞电极。 8、硝酸四氨⋅二水合铬（II）； [Co(H2O)(en)2Cl]Cl2。 9、 9.37x10-13 ； 1.32x10-3。 10、 Cu2+与en形成的配合物；熵。 四、 1、 7 PbO2 + 2 MnBr2 + 14 HNO3 = 7 Pb(NO3)2 + 2 Br2 + 2 HMnO4 + 6H2O 2、 ClO3 - + 6 I- + 6H+ = Cl- + 3 I2 + 3 H2O 五、 1. 第 1 页
2.2Hf(aq)+2e→H2-Ag+r-e→AgI(+) ()Ag(s)l AgBr(s)IH'(aq)Ha(g),Pt(+) 六 1.answer CaCO3 (s)=Ca"(aq)+CO(aq) First initial 0.05 0.25 Second initial 0.0 0.20 At equilibria 0.20+x x0.20+x）=Kp°(CaC0)=2.8x109=x≈1.4x10molr M(CaC0g)=(0.05-x)Fw(CaC0)x200=(0.05-x)x100x200/1000=1.0g 2、Answer: Ag'+e=Ag E(Ag'lAg) Fe+e=Fe2+E(Fe/Fe2) AgBr=Ag*+Brsp (AgBr)=[Ag] Ag+Br=AgBr+e Since E=E(+)-E(-)=E(Fe/Fe2)-E(Ag"/Ag) S0,0.700=0.771-0.0592/1lg(1/MAg7 0.700=0.771-0.0592/1lg(/Kp°(AgBr) Kp°(AgBr)=5.04x10 3、Answer: CaS04=Ca2++S042 BaSOa=Ba2+SO 0.01 X 0.01y Kp°(CaS0)=0.01x=9.1x109 Kp°(BaS0)=0.01y=1.8x100 x=9.1x104 y=1.8x10 so,BaSO precipitates first. When CaSOstarts to precipitate. Ba21=Kp9(BaS04)/[S042=1.8x100/9.1x104=1.98x107mol. 4、Answer: 2HCI + 1/202(g) C12(g+ H20(g) Initial(mol）) 0.02 0.08x0.2 0 0 At equilibria 0.02-2x 0.016-x/2 第2页
2. 2H+ (aq) + 2e → H2 (-); Ag + I- - e → AgI (+) (-) Ag (s) ⎢AgBr(s) ⎢⎢H+ (aq) H2(g), Pt (+) 六、 1. answer CaCO3 (s) = Ca2+(aq) + CO3 2-(aq) First initial 0.05 0.25 Second initial 0.0 0.20 At equilibria x 0.20 + x x(0.20 + x) = Ksp Θ (CaCO3) = 2.8x10-9 ⇒ x ≈ 1.4x10-8mol.l-1. M(CaCO3) = (0.05 – x) Fw(CaCO3) x 200 = (0.05-x) x100 x200/1000 = 1.0g 2、Answer: Ag+ + e = Ag EΘ(Ag+ /Ag) Fe3+ + e = Fe2+ EΘ(Fe3+/Fe2+) AgBr = Ag+ + Br- ⇒ Ksp Θ (AgBr) = [Ag+ ] Ag + Br- = AgBr + e Since E = E(+) – E(-) = EΘ(Fe3+/Fe2+) - EΘ(Ag+ /Ag) So, 0.700 = 0.771 – 0.0592/1 lg(1/[Ag+ ] 0.700 = 0.771 – 0.0592/1 lg(1/ Ksp Θ (AgBr)) Ksp Θ (AgBr) = 5.04x10-13 3、Answer: CaSO4 = Ca2+ + SO4 2- 0.01 x Ksp Θ (CaSO4) = 0.01x = 9.1 x10-6 x = 9.1 x10-4 BaSO4 = Ba2+ + SO4 2- 0.01 y Ksp Θ (BaSO4) = 0.01y = 1.8 x10-10 y = 1.8 x10-8 so, BaSO4 precipitates first. When CaSO4 starts to precipitate, [Ba2+] = Ksp Θ (BaSO4) / [SO4 2-] = 1.8x10-10/ 9.1x10-4 = 1.98x10-7mol.l-1 4、Answer: 2HCl + 1/2O2(g) Cl2(g)+ H2O(g) Initial(mol) 0.02 0.08x0.2 0 0 At equilibria 0.02 – 2x 0.016 – x/2 x x 第 2 页
W0.064+0.02-2x+0.016-2+x+x)=4.0% so,x=0.039,the tital of the substances is 0.0998mol PN2)=0.08x0.8/0.0998x100=64.0kPa PHC)=(0.02-2x0.0394)/0.0998x100=12.2kP PH0)=PC1,)=0.0394/0.0998X100=3.9kPa P(O2)=(0.016-0.5x0.039)/0.098x100=14.1kPa Ke={P(C2p]PH20)p]}/{PHCI/pP(O2p]2}=(3.91/100) (3.91/100/{12.2/100}2(14.1/100)2=0.27 5、Answer Ni(s)+4CO(g)=Ni(CO)4(l) △Hm9KJ.molK-) 0 -110.5 605.0 →△，Hm(KJ.moK)=-605.0-4-110.5)=-163.0 S(J.morK) 29.9 197.9420.0 -△SnmJ.molK=420.0-4x197.9-29.9=-401.5 T=△，Hme1△Sne-163.0x103/401.5=405.9K 第3页
x/(0.064 + 0.02 – 2x + 0.016 – x/2 + x + x) = 4.0% so, x = 0.039, the tital of the substances is 0.0998mol P(N2) = 0.08x0.8/0.0998 x 100 = 64.0 kPa P(HCl) = (0.02 - 2x0.0394)/0.0998 x 100= 12.2 kPa P(H2O) = P(Cl2) = 0.0394/0.0998 x 100= 3.9 kPa P(O2) = (0.016 -0.5x0.039)/0.098 x100 = 14.1kPa KΘ = {[P(Cl2)/pΘ][P(H2O )/pΘ]} / {[P(HCl)/pΘ] 2 [P(O2)/pΘ] 1/2} = (3.91/100) (3.91/100)/{12.2/100}2 (14.1/100)1/2} = 0.27 5、Answer: Ni (s) + 4CO(g) = Ni(CO)4(l) ΔfHm Θ(KJ.mol-1.K-1) 0 -110.5 -605.0 ⇒ ΔrHm Θ(KJ.mol-1.K-1) = -605.0 – 4(-110.5) = -163.0 Sm Θ(J.mol-1.K-1) 29.9 197.9 420.0 ⇒ ΔrSm Θ(J.mol-1.K-1) = 420.0 – 4x197.9 – 29.9 = -401.5 T = ΔrHm Θ/ ΔrSm Θ = 163.0x103 /401.5 = 405.9 K 第 3 页