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无机化学课后习题.答案内容(天大第四版)

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-_ 第2章 化学反应的方向、速率和限度 习题参考答案

1.解：mrH = 3347.6 kJ·mol1;mrS = 216.64 J·mol1·K1;mrG = 3283.0 kJ·mol1 ＜ 0 该反应在298.15K及标准态下可自发向右进行。 2.解： mrG = 113.4 kJ·mol1 ＞ 0 该反应在常温(298.15 K)、标准态下不能自发进行。 （2）mrH = 146.0 kJ·mol1;mrS = 110.45 J·mol1·K1;mrG = 68.7 kJ·mol1 ＞ 0 该反应在700 K、标准态下不能自发进行。 3.解：mrH = 70.81 kJ·mol1 ;mrS = 43.2 J·mol1·K1;mrG = 43.9 kJ·mol1 （2）由以上计算可知： mrH(298.15 K) = 70.81 kJ·mol1;mrS(298.15 K) = 43.2 J·mol1·K1

mrG = mrH  T ·mrS ≤ 0

T ≥K) (298.15K) (298.15mrmrSH = 1639 K

4.解：（1）cK = O)H( )(CH )(H (CO) 2432cccc pK = O)H( )(CH )(H (CO) 2432pppp K = pppppppp / O)H( /)(CH / )(H / (CO) 2432 （2）cK =  )(NH )(H )(N 3232212ccc pK =  )(NH )(H )(N 3232212ppp K = pppppp/ )(NH /)(H /)(N 3232212 （3）cK =)(CO 2c pK =)(CO 2p K =pp/)(CO 2

（4）cK =3232 )(H O)(H cc pK = 3232 )(H O)(H pp

K= 3232 /)(H /O)(H pppp 5.解：设mrH、mrS基本上不随温度变化。 mrG = mrH  T ·mrS

mrG(298.15 K) = 233.60 kJ·mol1

mrG(298.15 K) = 243.03 kJ·mol1 -_ Klg(298.15 K) = 40.92, 故 K(298.15 K) = 8.31040

Klg(373.15 K) = 34.02,故 K(373.15 K) = 1.01034

6.解：（1） mrG=2mfG(NH3, g) = 32.90 kJ·mol1 ＜0 该反应在298.15 K、标准态下能自发进行。 （2） Klg(298.15 K) = 5.76, K(298.15 K) = 5.8105

7. 解：（1） mrG(l) = 2mfG(NO, g) = 173.1 kJ·mol1 1lgK= RTG 303.2)1(mf = 30.32, 故 1K= 4.81031

（2）mrG(2) = 2mfG(N2O, g) =208.4 kJ·mol1 2lgK= RTG 303.2)2(mf = 36.50, 故 2K= 3.21037

（3）mrG(3) = 2mfG(NH3, g) = 32.90 kJ·mol1 3lgK= 5.76, 故 3K= 5.8105

由以上计算看出：选择合成氨固氮反应最好。

8.解：mrG = mfG(CO2, g)  mfG(CO, g) mfG(NO, g) = 343.94 kJ·mol1< 0，所以该反应从理论上讲是可行的。 9.解: mrH(298.15 K) = mfH(NO, g) = 90.25 kJ·mol1 mrS(298.15 K) = 12.39 J·mol1·K1

mrG(1573.15K)≈mrH(298.15 K) 1573.15mrS(298.15 K)

= 70759 J ·mol1 Klg(1573.15 K) = 2.349, K(1573.15 K) = 4.48103

10. 解： H2(g) + I2(g) 2HI(g) 平衡分压／kPa 2905.74 χ 2905.74 χ 2χ

22)74.2905()2(xx= 55.3

χ= 2290.12 p (HI) = 2χkPa = 4580.24 kPa

n = pVRT = 3.15 mol 11.解：p (CO) = 1.01105 Pa, p (H2O) = 2.02105 Pa p (CO2) = 1.01105 Pa, p (H2) = 0.34105 Pa CO(g) + H2O(g)  CO2(g) + H2(g) 起始分压／105 Pa 1.01 2.02 1.01 0.34 J = 0.168, pK= 1＞0.168 = J,故反应正向进行。

12.解：（1） NH4HS(s)  NH3(g) + H2S(g) 平衡分压/kPa x x -_ K= / S)(H / )(NH 23pppp= 0.070

则 x= 0.26100 kPa = 26 kPa 平衡时该气体混合物的总压为52 kPa （2）T不变，K不变。 NH4HS(s)  NH3(g) + H2S(g) 平衡分压/kPa 25.3+y y

K= / / ) 25.3pypy（ = 0.070

y= 17 kPa

13.解：（1） PCl5(g)  PCl3(g) + Cl2(g)

平衡浓度／(mol·L1) 0.250.070.0 0.250.0 0.250.0

cK = ) PCl()Cl()PCl(523ccc = 0.62mol· L1， (PCl5) = 71%

PCl5(g)  PCl3(g) + Cl2(g) 平衡分压 0.20VRT 0.5VRT 0.5VRT

K=pppppp/ )(PCl /) (Cl /)(PCl 523

= 27.2 （2） PCl5(g)  PCl3(g) + Cl2(g)

新平衡浓度／(mol·L1) 0.10 + y 0.25 y 0.25 +y210.0

cK= )10.0()30.0)(25.0(yyymol·L1 = 0.62mol· L1 (T不变，cK不变)

y=0.01 mol·L1，(PCl5) = 68% （3） PCl5(g)  PCl3(g) + Cl2(g) 平衡浓度／(mol·L1) z35.0 z 0.050 +z

cK= zzz35.0)050.0(= 0.62 mol·L1

z= 0.24 mol·L1，(PCl5) = 68%

比较(2)、(3)结果，说明最终浓度及转化率只与始、终态有关，与加入过程无关。

14.解： N2(g) + 3H2(g)  2NH3(g) 平衡浓度／(mol·L1) 1.0 0.50 0.50

cK= 32223)H()N()NH(ccc

=21)L · mol(0.2 若使N2的平衡浓度增加到1.2mol· L1，设需从容器中取走x摩尔的H2。 N2(g) + 3H2(g) 2NH3(g) 新平衡浓度／(mol·L1) 1.2 0.50+(30.2) x 0.5020.20 -_ cK=2132)L · mol()02350.0(2.1)20.0250.0(x=21)L · mol(0.2

x=0.94

15. 解：（1）α（CO）=61.5%;（2）α（CO）=86.5%; (3)说明增加反应物中某一物质浓度可提高另一物质的转化率;增加反应物浓度，平衡向生成物方向移动。

16.解： 2NO(g) + O2(g) 2NO2(g) 平衡分压／kPa 10179.2 = 21.8 286  79.2／2 = 246 79.2

K（673K）=pppppp/)(O/(NO)/)(NO2222= 5.36

mrG= KRTlg303.2，mrG(673 K) =  9.39 kJ·mol1

17.解： mrG(298.15 K) = 95278.54 J·mol1 mrG(298.15 K) = mrH(298.15 K)  298.15 K·mrS(298.15 K)

mrS(298.15 K) = 9.97 J·mol1·K1，mrG(500 K) ≈97292 J·mol1

Klg(500 K) = 0.16， 故 )K500(K=1.41010

或者 12lnKK≈ RH)K15.298(mr2112TTTT，K(500 K) = 1.41010 18.解：因mrG(298.15 K) = mrG(1) +mrG(2) = 213.0 kJ·mol1 <0, 说明该耦合反应在上述条件可自发进行。

第3章 酸碱反应和沉淀反应 习题参考答案 解：（1）pH=-lg c(H+)=12.00 （2）0.050mol·L-1HOAc溶液中， HOAc H+ + OAc- c平/(mol·L-1) 0.050-x x x

5108.1x0.050xx)HOAc()OAc()H(Kc

cc

a

c(H+) = 9.5×10-4mol·L-1 pH = -lg c(H+) = 3.02

2．解：（1）pH = 1.00 c(H+) = 0.10mol·L-1 pH = 2.00 c(H+) = 0.010mol·L-1 等体积混合后：c(H+) =（0.10mol·L-1+0.010mol·L-1）/2 = 0.055 mol·L-1 pH = -lgc(H+) = 1.26 （2）pH = 2.00 c(H+) = 0.010mol·L-1 pH = 13.00 pOH = 14.00-13.00 = 1.00, c(OH-) = 0.10mol·L-1 等体积混合后：-1-10.010molL(H)0.0050molL2c

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第四版无机化学 章课后答案

第四版无机化学 章课后答案

第21章 p区金属思考题21-1如何制备无水AlCl3？能否用加热脱水的方法从AlCl3·6H2O中制取无水AlCl3？解：制备无水AlCl3，可采用干法，即用Al和Cl2在加热的条件下反应制得。

2Al + 3Cl2 === 2AlCl3也可在HCl气氛下，由AlCl3·6H2O加热脱水得到。

但不能直接由AlCl3·6H2O 加热脱水得到无水AlCl3，因为铝离子的强烈水解性，在加热过程中，发生水解，不能得到无水AlCl3。

21-2矾的特点是什么？哪些金属离子容易成矾？解：矾在组成上可分为两类：M I2SO4·M II SO4·6H2O（M I=K+、Rb+、Cs+和NH4+等，M II=Mg2+、Fe2+等）；M I M III(SO4)2·12H2O（M I=Na+、K+、Rb+、Cs+和NH4+等，M II=Al3+、Fe3+、Cr3+、Co3+、Ga3+、V3+等），都为硫酸的复盐，形成复盐后，稳定性增强，在水中的溶解性减小。

21-3 若在SnCl2溶液中含有少量的Sn4+，如何除去它？若在SnCl4溶液中含有少量Sn2+，又如何除去？解：(1) SnCl2中含有少量的Sn4+，可加入锡粒将其除去。

发生的化学反应是：SnCl4＋Sn === 2SnCl2(2) 在SnCl4中含有少量的Sn2+，可利用Sn2+的还原性，用H2O2将其氧化为Sn4+。

21-4比较Sb、Bi的硫化物和氧化物的性质，并联系第16章内容，总结归纳砷分族元素的氧化物及其水合物、硫化物的酸碱性、氧化还原性的递变规律。

解：砷分族元素的氧化物及其水合物的酸碱性和氧化还原性的递变规律与其对应的硫化物相似。

即：从上到下，酸性减弱，碱性增强；+V氧化值的氧化性从上到下增强，+III 氧化值的还原性从上到下减弱；同一元素+V氧化值的酸性比+III氧化值的强。

21-5 说明三氯化铝在气态及在水溶液中的存在形式。

(精选)无机化学简明教程(天津大学)课后习题参考答案

(精选)无机化学简明教程(天津大学)课后习题参考答案

第1章 化学反应中的质量关系和能量关系 习题参考答案1．解：1.00吨氨气可制取2.47吨硝酸。

2．解：氯气质量为2.9×103g 3．解：一瓶氧气可用天数33111-1222()(13.210-1.0110)kPa 32L9.6d 101.325kPa 400L d n p p V n p V -⨯⨯⨯===⨯⨯４．解：pV MpVT nR mR== = 318 K 44.9=℃ ５．解：根据道尔顿分压定律ii n p p n=p (N 2) = 7.6104 Pap (O 2) = 2.0104 Pa p (Ar) =1103 Pa６．解：（1）2(CO )n = 0.114mol; 2(CO )p = 42.87 10 Pa ⨯（2）222(N )(O )(CO )p p p p =--43.7910Pa =⨯ （3）4224(O )(CO ) 2.6710Pa0.2869.3310Pan p n p ⨯===⨯ 7.解：（1）p (H 2) =95.43 kPa （2）m (H 2) =pVMRT= 0.194 g 8.解：（1） = 5.0 mol（2） = 2.5 mol结论: 反应进度()的值与选用反应式中的哪个物质的量的变化来进行计算无关，但与反应式的写法有关。

9.解：∆U = Q p p ∆V = 0.771 kJ 10.解： （1）V 1 = 38.3⨯10-3m 3= 38.3L（2） T 2 = nRpV 2= 320 K （3）W =(p V ) = 502 J（4） U = Q + W = -758 J （5） H = Q p = -1260 J11.解：NH 3(g) +45O 2(g) 298.15K−−−−→标准态NO(g) + 23H 2O(g) m r H ∆= 226.2 kJ ·mol 1 12.解：m r H ∆= Q p = 89.5 kJ m r U ∆= mr H ∆ nRT =96.9 kJ13.解：（1）C (s) + O 2 (g) → CO 2 (g)m r H ∆ =m f H ∆(CO 2, g) = 393.509 kJ ·mol 121CO 2(g) + 21C(s) → CO(g) m r H ∆ = 86.229 kJ ·mol 1CO(g) +31Fe 2O 3(s) → 32Fe(s) + CO 2(g)m r H ∆ =8.3 kJ ·mol 1各反应m r H ∆之和m r H ∆= 315.6 kJ·mol 1。

物化课后习题答案@天大第四版共86页

物化课后习题答案@天大第四版共86页

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物化课后习题答案@天大第四版
36、如果我们国家的法律中只有某种 神灵， 而Байду номын сангаас是 殚精竭 虑将神 灵揉进 宪法， 总体上 来说， 法律就 会更好 。—— 马克·吐 温 37、纲纪废弃之日，便是暴政兴起之 时。— —威·皮 物特
38、若是没有公众舆论的支持，法律 是丝毫 没有力 量的。 ——菲 力普斯 39、一个判例造出另一个判例，它们 迅速累 聚，进 而变成 法律。 ——朱 尼厄斯
无机化学第四版第十四章习题答案

无机化学第四版第十四章习题答案

无机化学第四版第十四章习题答案（Inorganic chemistry, Fourth Edition, fourteenth chapter, exercises, answers）Fourteenth chapter halogen14-1 why do not use KF water solution when making fluorine by electrolysis? Why liquid hydrogen fluoride does not conduct electricity while potassium fluoride anhydrous hydrogen fluoride solution can conduct electricity?1 solution: because F2 reacts with water, there is no free ion in the liquid HF molecule, so it can not conduct electricity. In the anhydrous HF solution of KF, K+ and HF2- exist;14-2 what are the special characteristics of fluorine in this element? What are the characteristics of hydrogen fluoride and hydrofluoric acid?2, solution: (1) as the radius of the F is very small, so F2 can dissociation is very small, the F- hydration heat of other halogen ions.(2) hydrogen bonds exist between HF molecules, so the boiling point and heat of vaporization of HF molecules are especially high.(3) AgF is a compound easily soluble in water.(4) the product of F2 reaction with water is complex.(5) HF is a weak acid with increasing acidity and increasingconcentration.(6) HF can react with SiO2 or silicate to form gaseous SiF4;14-3 (1) the reaction tendency of Cl2 was generated by comparing KMnO4, K2Cr2O7 and MnO2 with hydrochloric acid (1mol.L-1) according to the electrode potential.(2) what is the minimum concentration of hydrochloric acid if MnO2 is reacted with hydrochloric acid to cause Cl2 to occur successfully?3, Xie: (1) according to the relationship between electrode potential, we can see the reaction trend:KMnO4>K2Cr2O7>MnO2;14-4 according to the potential map, the reaction equilibrium constant of Br2 in alkaline aqueous solution is reduced to Br- and BrO3- at 298K.4. Solution: by formula: -ZFE=-RTlnKGet: K=exp (ZFE/RT)=2.92 * 103814-5, three, fluorinated nitrogen NF3 (boiling point -129 degrees Celsius) does not show Lewis alkaline, and relatively low molecular mass compounds NH3 (boiling point -33 degrees Celsius) is a well known Lewis base. (a) to explain the reasonwhy their volatility is so great; (b) explain why they differ in alkalinity.5. Xie: (1) NH3 has a higher boiling point because of the hydrogen bond between the molecules.(2) the NF3 atom has a large radius of F atoms, which makes it difficult to match Lewis acid because of steric hindrance.In addition, the electronegativity of the F atom is larger, which weakens the electronegativity of the central atom N.14-6 from bittern making Br2 available chlorine oxidation method. But from a thermodynamic point of view, Br- can be oxidized to Br2 by O2. Why not use O2 to make Br2?14-7 pass Cl2 can be used in slaked lime to obtain bleaching powder, and hydrochloric acid can be added to Cl2 in the solution of bleaching powder. The two phenomena are explained by using electrode potential.7, Xie: because Cl2 pass into the hydrated lime is in alkaline medium, and because of, so Cl2 in alkaline conditions prone to disproportionation reaction.When the hydrochloric acid is added to the bleach solution, the following reaction can be carried to the right:HClO + Cl- + H+ = Cl2 + H2O14-8 which of the following oxides are anhydrides: OF2, Cl2O7,ClO2, Cl2O, Br2O and I2O5? If the anhydride is prepared, the reaction is obtained by the corresponding acid or other method of getting the anhydride.8 solution: Cl2O7 is the anhydride of HClO4. Cl2O, Br2O are HClO, HBrO anhydrides, respectively14-9 how do you identify the three salts of KClO, KClO3, and KClO4?9 、 solution: add a small amount of solid into the dried test tube, and then do the following experimentAdd dilute hydrochloric acid, that is, Cl2 gas release is KClO;KClO+2HCl=KCl+Cl2+H2OAdding concentrated hydrochloric acid has Cl2 and emits, and the solution turns yellow is KClO3;8KC1O3+24HCl (strong) =9Cl2 = +8KCl+60ClO2 (yellow) +12H2OThe other is KClO414-10 the reaction equation for preparing HIO4, KIO3, I2O5 and KIO4 was prepared with I2 as raw material.The solubility of 14-11 (1) I2 in water is very small. The concentration of I2 saturated solution is calculated from the following 2.5 reactions at 298K.I2 (s) + 2e- = 2I-; 0.535V = Phi thetaI2 (AQ) + 2e- = 2I-; 0.621V = Phi theta(2) 0.100mol I2 was dissolved in 1.00L 0.100mol.L-1 KI solution and I3- solution was obtained. The Kc value of the I3- generation reaction is 0.752, and the concentration of I2 in the I3- solution is calculated.14-11, (1) I2 (AQ) =I2 (s)K=exp (ZFE/RT)=812K=1/[I2 (AQ)]I2 (AQ) =1/812=0.00123mol/L(2) I2+I-=I3-;KC=; so [I2]=?The solution is x=0.0934mol/L.Explain the following phenomena: 14-12 by adding a small amount of NaClO in the electrode potential of starch potassium iodide solution, blue A solution, adding excessive NaClO, B get a colorless solution, then acidified and add a small amount of solid Na2SO3 in B solution, A blue reproduction, when the excessive amount of Na2SO3 blue and then faded become colorlesssolution C, A the solution then add NaIO3 solution and blue. Point out A, B, C why each kind of substance, and write out the reaction equation of each step.12, Xie: A:I2;B:NaIO3;C:NaI14-13 write out the equation for the reaction of iodate and excess H2O2, such as adding starch to the system. What do you see?13 solution: HIO3+3H2O2=3O2+HI+3H2O; if the starch is added to the system, the solution turns blue slowly and then fades.14-14 write out the molecular formula of three metal halides with covalent bonds, and explain the common properties of this type of halides.14 and Xie: (AlCl3) 2; (AlBr3) 2; (AlI3) 2; all molecules contain coordinate bonds14-15 what is a multi halide? What is the trend of the formation of Br3- and Cl3- ions in comparison with I3- ions?14-16 what are halogen compounds?(a) write ClF3, BrF3 and IF3 halogen halide complexes, central atom hybridization orbitals, molecular electronicconfigurations and molecular configurations.(b) is there any risk of explosion when the following compounds are exposed to BrF3? Explain why.SbF5; CH3OH; F2; S2Cl2(c) why are halogenated compounds often diamagnetic, covalent, and chemically active than halogen?14-17 laboratory has a calcium halide, soluble in water, try to use concentrated H2SO4 to determine the nature and name of this salt.17. Solution: different phenomena can be identified by the reaction of halide with concentrated sulfuric acid.14-18 please click the following example,The bromine, iodine, and halogen ions containing various conversion and transformation conditions as oxygen acid interaction diagram.。