$NH_{4}HS$ is a very unstable compound and easily decomposes into $NH_{3}$ and $H_{2}S$. Put $30.00\,g\;NH_{4}HS$ and $0.100mol\;\mathrm{N_{2}}$ into a container with variable volume, then heat to $308.0K$ and adjust the external pressure to $P=2.000 \times 10^5~\mathrm{Pa}$. The decomposition of $NH_{3}$ and $H_{2}S$ can be ignored at this temperature. In this problem, assume that the gases separately obey their own van der Waals equation of state, ignore the volume change caused by mixing, and $p^\circ=101325~\mathrm{Pa}$. Analyze the deviation calculated using the van der Waals equation and the ideal gas equation. Select the correct one.

| | $\Delta_{f}H_{298}^{\circ}, \,kJ/mol$ | $S_{298}^{\circ}, \,J/(mol\cdot K)$ |
|:---------------:|:------:|:------:|
| $NH_{4}HS(s)$|-156.9 | 113.4  |
|   $NH_{3}(g)$   |-46.11 |192.45 |
|   $H_{2}S(g)$   | -20.63|205.79|

|  |$10^3a/(Pa\cdot m^6\cdot mol^{-2})$|$10^6b/(m^3\cdot mol^{-1})$|
|:-----------:|:------:|:-----:|
| $NH_3(g)$ | 422.5 | 37.07 |
| $H_2S(g)$ | 449.0 | 42.87 |
| $N_2(g)$   | 140.8  | 39.13 |
$NH_{4}HS$是一种很不稳定的化合物，很容易分解为 $NH_{3}$ 和 $H_{2}S$。在一个容积可变的容器中放入$30.00\,g\;NH_{4}HS$ 和 $0.100mol\;\mathrm{N_{2}}$，再加热到 $308.0K$ 并调节外压至 $P=2.000 \times 10^5~\mathrm{Pa}$， $NH_{3}$ 和 $H_{2}S$ 的分解在此温度下可以忽略。本题中，假设气体分别服从各自的范德华状态方程，忽略混合引起的体积变化，$p^\circ=101325~\mathrm{Pa}$。分析使用范德华方程和理想气体方程计算的偏差。选择正确的一项。

| | $\Delta_{f}H_{298}^{\circ}, \,kJ/mol$ | $S_{298}^{\circ}, \,J/(mol\cdot K)$ |
|:---------------:|:------:|:------:|
| $NH_{4}HS(s)$|-156.9 | 113.4  |
|   $NH_{3}(g)$   |-46.11 |192.45 |
|   $H_{2}S(g)$   | -20.63|205.79|

|  |$10^3a/(Pa\cdot m^6\cdot mol^{-2})$|$10^6b/(m^3\cdot mol^{-1})$|
|:-----------:|:------:|:-----:|
| $NH_3(g)$ | 422.5 | 37.07 |
| $H_2S(g)$ | 449.0 | 42.87 |
| $N_2(g)$   | 140.8  | 39.13 |
If the ideal gas equation is used, the absolute error in calculating the remaining solid mass is on the order of $10^{-1}g$.
If using the ideal gas equation, the absolute error in calculating the remaining solid mass is on the order of $10^{-2}g$.
If using the ideal gas law, the absolute error in calculating the remaining solid mass is on the order of $10^{-3}g$.
Remaining solid mass $25.55 g$
Remaining solid mass $30.00 g$
Equilibrium volume $3.51 L$
Equilibrium volume $1.28 L$
Total amount of gaseous substances produced by the reaction $0.0855mol$
All other options are incorrect.
如果使用理想气体方程，计算剩余固体质量的误差绝对值为 $10^{-1}g$ 数量级
如果使用理想气体方程，计算剩余固体质量的误差绝对值为 $10^{-2}g$ 数量级
如果使用理想气体方程，计算剩余固体质量的误差绝对值为 $10^{-3}g$ 数量级
剩余固体质量 $25.55 g$
剩余固体质量 $30.00 g$
平衡时体积 $3.51 L$
平衡时体积 $1.28 L$
反应生成气体总物质的量 $0.0855mol$
其他选项均不正确
At 298K, the enthalpy change of the reaction $\Delta_{r}H_{298}^{\circ}=(-46.11)+(-20.63)-(-156.9)=90.16\;kJ/mol$, entropy change $\Delta_{r}S_{298}^{\circ}=192.45+205.79-113.4=284.84\;J/(mol\cdot K)$

<Checkpoint>[$\Delta_{r}H_{298}^{\circ}=90.16\;kJ/mol, \Delta_{r}S_{298}^{\circ}=284.8\;J/(mol\cdot K)$](0.5)</Checkpoint>

Then the Gibbs free energy change can be calculated as $\Delta_{r}G_{298}^{\circ}=\Delta_{r}H_{298}^{\circ}-T\Delta_{r}S_{298}^{\circ}=90.16-298.0\times284.84\times10^{-3}=5.27768\;kJ/mol$

<Checkpoint>[$\Delta_{r}G_{298}^{\circ}=5.278\;kJ/mol$](0.5)</Checkpoint>

From the formula $\Delta_{r}G_{298}^{\circ}=-RT\ln K_a^{298K}$ solve for $K_a^{298K}=0.11881563281$

Assume that the change in $\Delta_{r}H^{\circ}$ at 298K and 308K can be ignored, that is $\Delta_{r}H_{308}^{0}=\Delta_{r}H_{298}^{\circ}=90.16\;kJ/mol$

<Checkpoint>[$\Delta_{r}H_{308}^{\circ}=90.16\;kJ/mol$](0.5)</Checkpoint>

To calculate $K_p$, first calculate $K_a$. From the formula $d\ln K_a=\dfrac{\Delta_{r}H^{\circ}}{RT^2}$ we get $\ln\dfrac{K_{a}^{T_2}}{K_{a}^{T_1}}=-\dfrac{\Delta_{r}H^{\circ}}{R}(\dfrac1{T_2}-\dfrac1{T_1})$, solve for $K_a^{308K}=0.38725431767$

<Checkpoint>[$K_a^{308K}=0.3873$](0.5)</Checkpoint>

Let the total volume be $V$, and the amount of substance converted by the reaction be $n_{gas}$. Substituting the van der Waals equation for the three gases respectively and rearranging, we have the system of equations $\begin{cases}(P(N_2) + \frac{1.408\times10^{-3}}{V^2})(V-3.913\times10^{-6}) = 2.560712\times10^2 \\ (P(NH_3) + \frac{0.4225\times n_{gas}^2}{V^2})(V-3.707\times10^{-5}\times n_{gas}) = n_{gas}\times2.560712\times10^3 \\ (P(H_2S) + \frac{0.4490\times n_{gas}^2}{V^2})(V-4.287\times10^{-5}\times n_{gas}) = n_{gas}\times2.560712\times10^3\end{cases}$

And the equilibrium constant satisfies $K_a^{308K}=P(NH_3)\times P(H_2S)/{p^\circ}^2$, and the total pressure satisfies $P=P(N_2)+P(NH_3)+P(H_2S)$.

Solve for $\begin{cases}P(N_2)=7.39225036\times10^4Pa \\ P(NH_3)=6.30422915\times10^4Pa \\ P(H_2S)=6.30352049\times10^4Pa \\ n_{gas}=0.0855127459mol \\ V=3.4624675L \end{cases}$

<Checkpoint>[Equilibrium volume $V=3.46L$, options F and G are incorrect](1)</Checkpoint>

The total amount of gaseous substance produced is $n=2n_{gas}=0.171mol$.

<Checkpoint>[The total amount of gaseous substance produced is $n=2n_{gas}=0.171mol$, option H is incorrect](1)</Checkpoint>

The mass of the remaining solid is $m=30.00-0.0855127459\times51.11244=25.62923490595\,g$

<Checkpoint>[The mass of the remaining solid is $m=25.63\,g$, options D and E are incorrect](1)</Checkpoint>

Using the ideal gas law to derive the system of equations $\begin{cases} P(N_2)=\dfrac{0.1\times8.314\times308}{V} \\ P(NH_3)=\dfrac{n_{gas}\times8.314\times308}{V} \\ P(H_2S)=\dfrac{n_{gas}\times8.314\times308}{V}\\ 2.000\times10^5=P(N_2)+P(NH_3)+P(H_2S) \\ 0.38725431767=P(NH_3)\times P(H_2S)/{101325}^2 \end{cases}$

Solve for $\begin{cases}P(N_2)=7.39225038\times10^4Pa \\ P(NH_3)=P(H_2S)=6.30387481\times10^4Pa \\ n_{gas}=0.0852768018mol \\ V=3.46404930L \end{cases}$

The mass of the remaining solid is $m=30.00-0.0852768018\times51.11244=25.64129799568\,g$

Deviation $m_{ideal}-m_{Van}=25.64129799568\,g-25.62923490595\,g=0.0120630897\,g$

<Checkpoint>[The absolute error is about $0.01g$, which is on the order of $10^{-2}g$, option B is correct, options A and C are incorrect](1)</Checkpoint>
298K时，反应的焓变 $\Delta_{r}H_{298}^{\circ}=(-46.11)+(-20.63)-(-156.9)=90.16\;kJ/mol$，熵变 $\Delta_{r}S_{298}^{\circ}=192.45+205.79-113.4=284.84\;J/(mol\cdot K)$ 

<Checkpoint>[$\Delta_{r}H_{298}^{\circ}=90.16\;kJ/mol, \Delta_{r}S_{298}^{\circ}=284.8\;J/(mol\cdot K)$](0.5)</Checkpoint>

则可以计算吉布斯自由能变 $\Delta_{r}G_{298}^{\circ}=\Delta_{r}H_{298}^{\circ}-T\Delta_{r}S_{298}^{\circ}=90.16-298.0\times284.84\times10^{-3}=5.27768\;kJ/mol$ 

<Checkpoint>[$\Delta_{r}G_{298}^{\circ}=5.278\;kJ/mol$](0.5)</Checkpoint>

由公式 $\Delta_{r}G_{298}^{\circ}=-RT\ln K_a^{298K}$ 解出 $K_a^{298K}=0.11881563281$ 

 假设 $\Delta_{r}H^{\circ}$ 在298K和308K时的改变可以忽略不计，即$\Delta_{r}H_{308}^{0}=\Delta_{r}H_{298}^{\circ}=90.16\;kJ/mol$ 

<Checkpoint>[$\Delta_{r}H_{308}^{\circ}=90.16\;kJ/mol$](0.5)</Checkpoint>

为计算 $K_p$ 先计算 $K_a$，由公式 $d\ln K_a=\dfrac{\Delta_{r}H^{\circ}}{RT^2}$ 得到 $\ln\dfrac{K_{a}^{T_2}}{K_{a}^{T_1}}=-\dfrac{\Delta_{r}H^{\circ}}{R}(\dfrac1{T_2}-\dfrac1{T_1})$，解出 $K_a^{308K}=0.38725431767$

<Checkpoint>[$K_a^{308K}=0.3873$](0.5)</Checkpoint>

设总体积 $V$，反应转化物质的量 $n_{gas}$。对三种气体分别代入范德华方程并整理有方程组$\begin{cases}(P(N_2) + \frac{1.408\times10^{-3}}{V^2})(V-3.913\times10^{-6}) = 2.560712\times10^2 \\ (P(NH_3) + \frac{0.4225\times n_{gas}^2}{V^2})(V-3.707\times10^{-5}\times n_{gas}) = n_{gas}\times2.560712\times10^3 \\ (P(H_2S) + \frac{0.4490\times n_{gas}^2}{V^2})(V-4.287\times10^{-5}\times n_{gas}) = n_{gas}\times2.560712\times10^3\end{cases}$ 

且平衡常数满足 $K_a^{308K}=P(NH_3)\times P(H_2S)/{p^\circ}^2$ ，总压力满足 $P=P(N_2)+P(NH_3)+P(H_2S)$ 。

解出 $\begin{cases}P(N_2)=7.39225036\times10^4Pa \\ P(NH_3)=6.30422915\times10^4Pa \\ P(H_2S)=6.30352049\times10^4Pa \\ n_{gas}=0.0855127459mol \\ V=3.4624675L \end{cases}$

<Checkpoint>[平衡体积 $V=3.46L$，选项F和G错误](1)</Checkpoint>

生成气体总物质的量 $n=2n_{gas}=0.171mol$。

<Checkpoint>[生成气体总物质的量 $n=2n_{gas}=0.171mol$，选项H错误](1)</Checkpoint>

剩下固体的质量 $m=30.00-0.0855127459\times51.11244=25.62923490595\,g$ 

<Checkpoint>[剩下固体的质量$m=25.63\,g$，选项D和E错误](1)</Checkpoint>

用理想气体得出方程组 $\begin{cases} P(N_2)=\dfrac{0.1\times8.314\times308}{V} \\ P(NH_3)=\dfrac{n_{gas}\times8.314\times308}{V} \\ P(H_2S)=\dfrac{n_{gas}\times8.314\times308}{V}\\ 2.000\times10^5=P(N_2)+P(NH_3)+P(H_2S) \\ 0.38725431767=P(NH_3)\times P(H_2S)/{101325}^2 \end{cases}$ 

解出 $\begin{cases}P(N_2)=7.39225038\times10^4Pa \\ P(NH_3)=P(H_2S)=6.30387481\times10^4Pa \\ n_{gas}=0.0852768018mol \\ V=3.46404930L \end{cases}$ 

剩下固体的质量 $m=30.00-0.0852768018\times51.11244=25.64129799568\,g$ 

偏差 $m_{ideal}-m_{Van}=25.64129799568\,g-25.62923490595\,g=0.0120630897\,g$

<Checkpoint>[绝对误差约 $0.01g$，属于$10^{-2}g$数量级，选项B正确，选项A和C错误](1)</Checkpoint>