The artificial nitrogen fixation process is one of the lifelines of mankind, and the industrial preparation of ammonia is the most important reaction in it.

1.  At $700 K$, in a $1.0  {  \ m  } ^ { 3 }$ container, ${1.0\  mol}$ $\mathrm { N_2 }$ and $3 . 0{\ mol } \mathrm { \ H_2 }$ react to produce ammonia. The thermodynamic data are shown in Table 1:

| - | $\rm N_2(g)$ |$\rm H_2(g)$ |$\rm NH_3(g)$ |
| ----- | ----- | ----- | ----- |
|${\Delta_fH_m^\circ}/kJ·mol^{-1}$| -- | -- | -45.9 |
| ${S_m^\circ}/J·mol^{-1}·K^{-1}$ | 191.6| 130.7| 192.8|

2.  Cool the ammonia produced after the system in question 1 reaches equilibrium from ${700\ K}$ to ${50\ K}$ (always maintaining a pressure of ${1\ bar}$). The relevant thermodynamic data of ammonia are shown in Table 2:

| | ${\Delta_{vap}H_m^\circ}$ |${\Delta_{vap}S_m^\circ}$ |${\Delta_{fus}H_m^\circ}$ |${\Delta_{fus}S_m^\circ}$ |${C_{p,m}(g)}$ |${C_{p,m}(l)}$ |${C_{p,m}(s)}$ |
| ----- | ----- | ----- | ----- | ----- | ----- | ----- | ----- |
|$\rm NH_3(g)$| 23.5 | 98.3 | 5.66|32.0|35.1|78.5|78.5 (approximate)|

The unit of heat capacity in the table is $ J·mol^{-1}K^{-1}$ , the unit of enthalpy change is $kJ·mol^{-1}$ , and the unit of entropy change is $J·mol^{-1}K^{-1}$.

Which of the following options is correct:
人工固氮过程是人类的生命线之一，而工业制备氨气则是其中最重要的反应。

1.在$700 K$下， $1.0  {  \ m  } ^ { 3 }$ 的容器中，${1.0\  mol}$ $\mathrm { N_2 }$ 和 $3 . 0{\ mol } \mathrm { \ H_2 }$ 反应生成氨。热力学数据如表1所示：

| - | $\rm N_2(g)$ |$\rm H_2(g)$ |$\rm NH_3(g)$ |
| ----- | ----- | ----- | ----- |
|${\Delta_fH_m^\circ}/kJ·mol^{-1}$| -- | -- | -45.9 |
| ${S_m^\circ}/J·mol^{-1}·K^{-1}$ | 191.6| 130.7| 192.8|

2.将第1题中体系平衡后产生的氨气，从${700\ K}$ 降温至${50\ K}$(始终保持${1\ bar}$压强)，氨的有关热力学数据如表2所示：

| | ${\Delta_{vap}H_m^\circ}$ |${\Delta_{vap}S_m^\circ}$ |${\Delta_{fus}H_m^\circ}$ |${\Delta_{fus}S_m^\circ}$ |${C_{p,m}(g)}$ |${C_{p,m}(l)}$ |${C_{p,m}(s)}$ |
| ----- | ----- | ----- | ----- | ----- | ----- | ----- | ----- |
|$\rm NH_3(g)$| 23.5 | 98.3 | 5.66|32.0|35.1|78.5|78.5（近似）|

表中热容单位为 $ J·mol^{-1}K^{-1}$ ，焓变单位为 $kJ·mol^{-1}$ ，熵变单位为 $J·mol^{-1}K^{-1}$。

下列选项正确的是：
In question 1, the $K^\circ$ for ammonia synthesis is $4.58×10^{-4}$
The conversion rate of ammonia synthesis in question 1 is $0.54\%$
Due to the presence of hydrogen bonds, solid ammonia and water exhibit similar properties upon melting.
In problem 2, the enthalpy change of the entire process is $\Delta H=-60.2 kJ$
The melting point of ammonia decreases as pressure increases.
All of the above options are incorrect.
第1题中合成氨的$K^\circ=4.58×10^{-4}$
第1题中合成氨的转化率为$0.54\%$
由于氢键的存在，在融化时固态氨和水会表现出类似的性质
第2题中整个过程的焓变为$\Delta H=-60.2 kJ$
氨的熔点随压强升高而降低
以上选项均不正确
Problem 1:

$$\Delta _rG_m^\circ(700K)=\Delta _rH_m^\circ(700K)-T·\Delta S_m^\circ(700K)=-45.9kJ·mol^{-1}×2-700K×(192.8×2-191.6×1-130.7×3)J·mol^{-1}·K^{-1}=46.87kJ·mol^{-1} $$

<Checkpoint>[$$\Delta _rG_m^\circ(700K)=46.87kJ·mol^{-1} $$](1)</Checkpoint>

$$\Delta _rG_m^\circ = -RT ln(K^\circ) $$ 

$$-\Delta _rG_m^\circ / (RT) = -46.87kJ·mol^{-1} / (8.314J·mol^{-1} ·K^{-1} × 700K) ≈ -8.054$$

$$K^\circ = e^{-8.054}≈ 3.180 × 10^{-4}$$

<Checkpoint>[$$K^\circ =3.180 × 10^{-4}$$](1)</Checkpoint>

Equation: $\mathrm{N_2(g) + 3H_2(g)= 2NH_3(g)}$

According to $pV=nRT$, the initial $\mathrm{N_2(g)}$ is calculated to be $0.05820bar$, and $\mathrm{H_2(g)}$ is $0.1746bar$

Let the amount of $\mathrm{NH_3(g)}$ produced be $2x$, then the amount of $\mathrm{N_2(g)}$ is $0.05820-x$, and the amount of $\mathrm{H_2(g)}$ is $0.1746-3x$, in units of $bar$

List the equation $$3.180 × 10^{-4}=\frac{4x^2·(p^\circ)^2}{(0.05820-x)(0.1746-3x)^3}$$

Solving for $x$ yields $x=1.56×10^{-4}$ 

<Checkpoint>[$x=1.56×10^{-4}$ ](1)</Checkpoint>

The conversion rate is $0.27\%$ 

<Checkpoint>[The conversion rate is $0.27\%$ ](1)</Checkpoint>

Due to the presence of hydrogen bonds, solid ammonia and water exhibit different properties upon melting, because hydrogen atoms in solid ammonia participate in hydrogen bond formation, and a large number of hydrogen bonds make the structure more compact and denser.

<Checkpoint>[Due to the presence of hydrogen bonds, solid ammonia and water exhibit different properties upon melting, because hydrogen atoms in solid ammonia participate in hydrogen bond formation, and a large number of hydrogen bonds make the structure more compact and denser](1)</Checkpoint>

Problem 2:
Now calculate the melting point and boiling point based on the enthalpy change and entropy change, construct a cycle, and finally calculate the enthalpy change of the entire process.

$$T_b=\frac{\Delta_{vap}H_m^\circ}{\Delta_{vap}S_m^\circ}=\frac{23.5kJ·mol^{-1} }{98.3J·mol^{-1}·K^{-1} }=239 K$$

<Checkpoint>[$$T_b=\frac{\Delta_{vap}H_m^\circ}{\Delta_{vap}S_m^\circ}=\frac{23.5kJ·mol^{-1} }{98.3J·mol^{-1}·K^{-1} }=239 K$$](1)</Checkpoint>

$$T_f=\frac{\Delta_{fus}H_m^\circ}{\Delta_{fus}S_m^\circ}=\frac{5.66kJ·mol^{-1}}{32.0J·mol^{-1}·K^{-1}}=177 K$$

<Checkpoint>[$$T_f=\frac{\Delta_{fus}H_m^\circ}{\Delta_{fus}S_m^\circ}=\frac{5.66kJ·mol^{-1}}{32.0J·mol^{-1}·K^{-1}}=177 K$$](1)</Checkpoint>

The entire process can be expressed as $$\Delta H= \Delta H_1 (5.4 mmol,g,700K\rightarrow239K,1 bar)+ \Delta H_2 (5.4 mmol,g\rightarrow l,239K,1 bar)+\Delta H_3 (5.4 mmol,l,239K\rightarrow 177K,1 bar)+\Delta H_4 (5.4 mmol,l\rightarrow s,177K,1 bar)+\Delta H_5 (5.4 mmol,s,177K\rightarrow 50K,1 bar)$$

$$\Delta H=5.4×10^{-3}mol×[(239-700)K×35.1J·mol^{-1}·K^{-1}+(177-239)K× 78.5J·mol^{-1}·K^{-1}+(50-177)K× 78.5J·mol^{-1}·K^{-1}-23.5kJ·mol^{-1}-5.66 kJ·mol^{-1}] =-320J$$

<Checkpoint>[$$\Delta H = -320J$$](1)</Checkpoint>

According to the Clapeyron equation:

$$\frac{\Delta P}{\Delta T}=\frac{\Delta H}{T\Delta V}$$

For the solidification process $$\Delta H<0，T>0，\Delta V<0$$

Therefore $$\frac{\Delta P}{\Delta T}>0$$
The melting point of ammonia increases with increasing pressure.

<Checkpoint>[The melting point of ammonia increases with increasing pressure](1)</Checkpoint>

The final result is F
第1题：

$$\Delta _rG_m^\circ(700K)=\Delta _rH_m^\circ(700K)-T·\Delta S_m^\circ(700K)=-45.9kJ·mol^{-1}×2-700K×(192.8×2-191.6×1-130.7×3)J·mol^{-1}·K^{-1}=46.87kJ·mol^{-1} $$

<Checkpoint>[$$\Delta _rG_m^\circ(700K)=46.87kJ·mol^{-1} $$](1)</Checkpoint>

$$\Delta _rG_m^\circ = -RT ln(K^\circ) $$ 

$$-\Delta _rG_m^\circ / (RT) = -46.87kJ·mol^{-1} / (8.314J·mol^{-1} ·K^{-1} × 700K) ≈ -8.054$$

$$K^\circ = e^{-8.054}≈ 3.180 × 10^{-4}$$

<Checkpoint>[$$K^\circ =3.180 × 10^{-4}$$](1)</Checkpoint>

方程：$\mathrm{N_2(g) + 3H_2(g)= 2NH_3(g)}$

根据$pV=nRT$，计算出开始时$\mathrm{N_2(g)}$为$0.05820bar$,$\mathrm{H_2(g)}$为$0.1746bar$

将产生$\mathrm{NH_3(g)}$的量设为$2x$,则$\mathrm{N_2(g)}$的量为$0.05820-x$,$\mathrm{H_2(g)}$的量为$0.1746-3x$,单位为$bar$

列方程$$3.180 × 10^{-4}=\frac{4x^2·(p^\circ)^2}{(0.05820-x)(0.1746-3x)^3}$$

解得$x=1.56×10^{-4}$ 

<Checkpoint>[$x=1.56×10^{-4}$ ](1)</Checkpoint>

转化率为$0.27\%$ 

<Checkpoint>[转化率为$0.27\%$ ](1)</Checkpoint>

由于氢键的存在，在融化时固态氨和水会表现出不同的性质，因为固态氨氢原子参与氢键形成，大量氢键使得结构更加紧密，密度更大

<Checkpoint>[由于氢键的存在，在融化时固态氨和水会表现出不同的性质，因为固态氨氢原子参与氢键形成，大量氢键使得结构更加紧密，密度更大](1)</Checkpoint>

第2题：
现根据焓变和熵变计算熔点和沸点，在构建循环，最后计算整个过程焓变。

$$T_b=\frac{\Delta_{vap}H_m^\circ}{\Delta_{vap}S_m^\circ}=\frac{23.5kJ·mol^{-1} }{98.3J·mol^{-1}·K^{-1} }=239 K$$

<Checkpoint>[$$T_b=\frac{\Delta_{vap}H_m^\circ}{\Delta_{vap}S_m^\circ}=\frac{23.5kJ·mol^{-1} }{98.3J·mol^{-1}·K^{-1} }=239 K$$](1)</Checkpoint>

$$T_f=\frac{\Delta_{fus}H_m^\circ}{\Delta_{fus}S_m^\circ}=\frac{5.66kJ·mol^{-1}}{32.0J·mol^{-1}·K^{-1}}=177 K$$

<Checkpoint>[$$T_f=\frac{\Delta_{fus}H_m^\circ}{\Delta_{fus}S_m^\circ}=\frac{5.66kJ·mol^{-1}}{32.0J·mol^{-1}·K^{-1}}=177 K$$](1)</Checkpoint>

整个过程可以表述为$$\Delta H= \Delta H_1 (5.4 mmol,g,700K\rightarrow239K,1 bar)+ \Delta H_2 (5.4 mmol,g\rightarrow l,239K,1 bar)+\Delta H_3 (5.4 mmol,l,239K\rightarrow 177K,1 bar)+\Delta H_4 (5.4 mmol,l\rightarrow s,177K,1 bar)+\Delta H_5 (5.4 mmol,s,177K\rightarrow 50K,1 bar)$$

$$\Delta H=5.4×10^{-3}mol×[(239-700)K×35.1J·mol^{-1}·K^{-1}+(177-239)K× 78.5J·mol^{-1}·K^{-1}+(50-177)K× 78.5J·mol^{-1}·K^{-1}-23.5kJ·mol^{-1}-5.66 kJ·mol^{-1}] =-320J$$

<Checkpoint>[$$\Delta H = -320J$$](1)</Checkpoint>

根据Clapeyon方程：

$$\frac{\Delta P}{\Delta T}=\frac{\Delta H}{T\Delta V}$$

凝固过程$$\Delta H<0，T>0，\Delta V<0$$

所以$$\frac{\Delta P}{\Delta T}>0$$
氨的熔点随压强升高而升高

<Checkpoint>[氨的熔点随压强升高而升高](1)</Checkpoint>

最后结果选F