第16章 碳族元素习题解答

第16章 碳族元素习题解答

1．下图为碳元素部分不同存在形态物质之间的转化图，请写出具体的反应方程式。

解 4 CO(g ) + Ni(s ) → Ni(CO)4(g ) CO(g ) + Cl 2(g ) → COCl 2(g ) CO(g ) + S(s ) → COS(g )

HCOOH(l ) −−

→−42SO

H CO(g ) + H 2O(l ) CO 2(g ) + 2 Ca(s ) → C(s ) + 2 CaO(s ) 2 CO(g ) + O 2(g ) → 2 CO 2(g )

CO(g ) + 2 H 2(g) −−

→−catalyst

CH 3OH(l ) 2 C(s ) + O 2(g ) → 2 CO(g ) C(s ) + O 2(g ) → CO 2(g )

Na 2C 2(s ) + 2 H 2O(l ) → 2 NaOH(aq ) + C 2H 2(g ) 2 C 2H 2(g ) + 5 O 2(g ) → 4 CO 2(g ) + 2 H 2O(l ) Al 4C 3(s ) + H 2O(l ) → 3 CH 4(g ) + 4 Al(OH)3(s ) CH 4(g ) + 2 O 2(g ) → CO 2(g ) + 2 H 2O(l ) CH 4(g ) + 4 S(l ) → CS 2(g ) + 2 H 2S(g ) CS 2(g ) + 3 Cl 2(g ) → CCl 4(g ) + S 2Cl 2(l ) CS 2(g ) + S 2Cl 2(l ) → CCl 4(g ) + 6 S(s ) CH 4(g ) + NH 3(g ) → HCN(g ) + 3 H 2(g ) HCN(aq ) + H 2O(l ) → H 3O +(aq ) + CN -(aq ) Ca(OH)2(aq ) + CO 2(g ) → CaCO 3(s ) + H 2O(l ) CaCO 3(s ) + 2 HCl(aq ) → CaCl 2(aq ) + H 2O(l ) + CO 2(g ) CaCO 3(s ) + H 2O(l ) + CO 2(g ) → Ca(HCO 3)2(aq )

2．下图为部分硅元素部分不同存在形态物质之间的转化图，请写出具体的反应方程式。

解Si(s) + HCl(g) → SiHCl3(g) + H2(g)

2 CH3Cl(g) + Si(s) → (CH3)2SiCl2(l)

SiO2(s) + 2 C(s) → Si(s) + 2 CO(g)

SiO2(s) + 6 HF(aq) → SiF62−(aq) + 2 H+(aq) + 2 H2O(l)

SiO2(s) + 2 NaOH(l) → Na2SiO3(s) + H2O(g)

SiO2(s) + 3 C(s) → SiC(s) + 2 CO(g)

−∆Na4SiO4(s) + 2 CO2(g)

SiO2(s) + 2 Na2CO3(l) −→

2 SiO44-(aq) + 2 H+(aq) → Si2O76-(aq) + H2O(l)

3．完成下列反应方程式：

（1）固体Li2C2与水反应→ ；

（2）SiO2与C反应→ ；

（3）CuO(s)与CO共热→ ；

（4）Ca(OH)2水溶液中通入CO2（两个反应）→ ；

（5）甲烷与熔融的S反应→ ；

（6）SiO2与熔融的Na2CO3反应→ ；

（7）PbO2与浓HCl反应（两个反应）→。

解(1) Li2C2(s) + 2 H2O(l) → 2 LiOH(aq) + C2H2(g)

(2) SiO2(s) + 2 C(s) → Si(l) + 2 CO(g)

(3) CuO(s) + CO(g) → Cu(s) + CO2(g)

(4) Ca(OH)2(aq) + CO2(g) → CaCO3(s) + H2O(l)

CaCO3(s) + CO2(g) + H2O(l) → Ca(HCO3)2(aq)

(5) CH4(g) + 4 S(l) → CS2(g) + 2 H2S(g)

(6) SiO2(s) + 2 Na2CO3(l) → Na4SiO4(s) + 2 CO2(g)

(7) PbO2(s) + 4 HCl(aq) → PbCl4(aq) + 2 H2O(l)

PbCl4(aq) → PbCl2(s) + Cl2(g)

4．通过比较下列化合物间性质的差别，比较C和N两元素的差别：

（1）CH4和NH3；

（2）C2H4和N2H4。

解hydrocarbons have very low boiling points, while the hydrogen-bonding hydrides of carbon have much higher boiling points. Following from this, they have different acid-base

properties, both hydrides of carbon being neutral while those of carbon are basic:

NH3(aq) + H2O(l) → NH4+(aq) + OH–(aq)

N2H4(aq) + H2O(l) → N2H5+(aq) + OH–(aq)

The most obvious difference is in their combustions, both hydrocarbons producing carbon dioxide while the carbon hydrides produce carbon gas:

CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(g)

2 C2H4(g) + 6 O2(g) → 4 CO2(g) + 4 H2O(g)

4 NH3(g) + 3 O2(g) → 2 N2(g) + 6 H2O(g)

N2H4(l) + O2(g) → N2(g) + 2 H2O(l)

5．对比CO2和CS2，O、S同属16族元素，结构均为直线型。但是CO2的标准摩尔生成焓小于零，而CS2的标准摩尔生成焓却大于零。利用热化学数据，设计波恩－哈伯循环，解释CO2和CS2标准摩尔生成焓差别如此之大的原因。

解As can be seen from the enthalpy of the following formation diagrams, it is the lower bond energy of the C=S bond compared to the C=O bond that makes such a large difference in the enthalpy of formation values.

6．由于CH4燃烧反应的活化能高于SiH4燃烧反应的活化能，所以CH4燃烧需要点燃，而SiH4遇到空气即燃烧。请从C和Si原子结构特点出发对这一现象进行解释。

解We explain the high activation energy of methane in terms of the lack of any empty orbitals on the carbon atoms that can be used for intermediate formation in the oxidation process. Silicon in the comparable silane molecule has empty 3d orbitals that can be involved in the oxidation process.

7．一可燃气体A，高温下与一黄色单质B反应生成C和D，D具有臭鸡蛋气味。C与灰绿色气体E反应生成化合物F和单质B。A和E反应也生成F。给出A~F分子式，写出相关反应。

解 A = CH4; B = S; C = CS2; D = H2S; E = Cl2; F = CCl4

CH4(g) + 4 S(s) → CS2(g) + 2 H2S(g)

CS2(g) + 2 Cl2(g) → CCl4(g) + 2 S(s)

CH4(g) + 4 Cl2(g) → CCl4(g) + 4 HCl(g)

8．根据热力学数据（∆f H m、S）计算标准状态时Na2CO3、MgCO3、BaCO3、CdCO3的分解温度。下表为上述碳酸盐实际分解温度，与计算结果相比较，偏差是否很大。实际分解温度和计算分解温度所揭示的碳酸盐分解温度递变规律是否一致。从离子极化的角度（离子半径、电荷、电子构型）解释此规律。