Place $0.100 \mathrm{~mol} ~\mathrm{H}_{2}$ and $0.100 \mathrm{~mol} ~\mathrm{Cl}_{2}$ in a pre-evacuated special $5 \mathrm{~L}$ closed container, heat to $2000 \mathrm{~K}$, the system reaches equilibrium, and the total pressure is $6.70 ~\mathrm{bar}(1~ \mathrm{bar}=100~ \mathrm{kPa})$. The following reaction relationships exist in the system:

$\begin{array}{ll}\mathrm{Cl}_{2}(\mathrm{g})=2 \mathrm{Cl}(\mathrm{g}) & K_{p 1}=0.576 \\ \mathrm{H}_{2}(\mathrm{g})+\mathrm{Cl}_{2}(\mathrm{g})=2 \mathrm{HCl}(\mathrm{g}) & K_{p 2} \\ \mathrm{HCl}(\mathrm{g})=\mathrm{H}(\mathrm{g})+\mathrm{Cl}(\mathrm{g}) & K_{p 3}=2.23 \times 10^{-6} \\ \mathrm{H}_{2}(\mathrm{g})=2 \mathrm{H}(\mathrm{g}) & K_{p 4}\end{array}$

Which of the following statements is incorrect:

$1$. The total pressure of the system at $2000 \mathrm{~K}$ when $\mathrm{H}_{2}(\mathrm{g})$ and $\mathrm{Cl}_{2}(\mathrm{g})$ are not decomposed is $6.65\mathrm{~bar}$.

$2$. The equilibrium partial pressure of $\mathrm{Cl}_{2}(\mathrm{g})$ is $0.0651\mathrm{~bar}$.

$3$. There are $3$ independent reactions in the system.

$4$. $K_{p4}=4.08\times10^{-7}$.
将 $0.100 \mathrm{~mol} ~\mathrm{H}_{2}$ 和 $0.100 \mathrm{~mol} ~\mathrm{Cl}_{2}$ 置于预先抽真空的特制 $5 \mathrm{~L}$ 密闭容器, 加热至 $2000 \mathrm{~K}$, 体系达到平衡, 总压强为 $6.70 ~\mathrm{bar}(1~ \mathrm{bar}=100~ \mathrm{kPa})$ 。体系中存在如下反应关系:

$\begin{array}{ll}\mathrm{Cl}_{2}(\mathrm{g})=2 \mathrm{Cl}(\mathrm{g}) & K_{p 1}=0.576 \\ \mathrm{H}_{2}(\mathrm{g})+\mathrm{Cl}_{2}(\mathrm{g})=2 \mathrm{HCl}(\mathrm{g}) & K_{p 2} \\ \mathrm{HCl}(\mathrm{g})=\mathrm{H}(\mathrm{g})+\mathrm{Cl}(\mathrm{g}) & K_{p 3}=2.23 \times 10^{-6} \\ \mathrm{H}_{2}(\mathrm{g})=2 \mathrm{H}(\mathrm{g}) & K_{p 4}\end{array}$

下列说法中，错误说法为：

$1$. $2000 \mathrm{~K}$ 体系中 $\mathrm{H}_{2}(\mathrm{g})$ 和 $\mathrm{Cl}_{2}(\mathrm{g})$ 未分解时容器的总压为$6.65\mathrm{~bar}$。

$2$. $\mathrm{Cl}_{2}(\mathrm{g})$ 平衡分压为$0.0651\mathrm{~bar}$。

$3$. 该体系中有$3$个独立反应。

$4$. $K_{p4}=4.08\times10^{-7}$。
1
2
3
4
1，2
1，3
1，4
2，3
2，4
3，4
1，2，3
1，3，4
2，3，4
1，2，4
1，2，3，4
No incorrect options.
1
2
3
4
1，2
1，3
1，4
2，3
2，4
3，4
1，2，3
1，3，4
2，3，4
1，2，4
1，2，3，4
无错误选项。
To simplify the processing, the following symbols are used in the calculations, and in the equilibrium expression, all partial pressure terms are divided by the standard partial pressure by default:

$\mathrm{Cl_2(g)}$ initial partial pressure $x_0$; $\mathrm{Cl_2(g)}$ equilibrium partial pressure $x_1$; $\mathrm{Cl(g)}$ equilibrium partial pressure $x_2$; $\mathrm{H_2(g)}$ initial partial pressure $y_0$; $\mathrm{H_2(g)}$ equilibrium partial pressure $y_1$; $\mathrm{H(g)}$ equilibrium partial pressure $y_2$;
$\mathrm{HCl(g)}$ equilibrium partial pressure $z$;
Total system pressure $p_t$;
From the ideal gas equation of state, we can obtain:
$p_{0}=\frac{n_{0} R T} { V}=0.2 \times 8.314 \times 2000 \div 5 \times 10^{3} \mathrm{~Pa}=6.65 \mathrm{~bar}$ 

<Checkpoint>[$p_{0}=6.65 \mathrm{~bar}$](1)</Checkpoint>

$1$ Correct.

From the reaction equilibrium, we know that

$x_{2}{ }^{2}=0.576 x_{1} \cdots(1)$,

$x_{2} y_{2}=2.23 \times 10^{-6} z \cdots(2)$

<Checkpoint>[$x_{2}{ }^{2}=0.576 x_{1} $, $x_{2} y_{2}=2.23 \times 10^{-6} z $](1)</Checkpoint>

From the conservation of mass, we know that

$2 x_{1}+x_{2}=2 y_{1}+y_{2} \cdots(3)$

From the change in pressure, we know that

$p_{\mathrm{t}}-p_{0}=0.5 x_{2}+0.5 y_{2} \cdots(4)$ 

From the total system pressure, we know that

$x_{1}+x_{2}+y_{1}+y_{2}+z=p_{\mathrm{t}} \cdots(5)$

From $(1)$, we know that $x_{1}=x_{2}^{2} / 0.576 \cdots(6)$

From $(4)$, we know that $y_{2}=2\left(p_{\mathrm{t}}-p_{0}\right)-x_{2} \cdots(7)$

Substituting $(7)$ into $(2)$, we know that $z=\frac{x_{2}\left[2\left(p_{\mathrm{t}}-p_{0}\right)-x_{2}\right] }{\left(2.23 \times 10^{-6}\right) }\cdots(8)$ 

<Checkpoint>[$z=\frac{x_{2}\left[2\left(p_{\mathrm{t}}-p_{0}\right)-x_{2}\right] }{\left(2.23 \times 10^{-6}\right) }$](1)</Checkpoint>

Substituting $(6)$ and $(7)$ into $(3)$, we know that $y_{1}=x_{2}{ }^{2} / 0.576+x_{2}-\left(p_{\mathrm{t}}-p_{0}\right) \cdots(9)$

<Checkpoint>[$y_{1}=x_{2}{ }^{2} / 0.576+x_{2}-\left(p_{\mathrm{t}}-p_{0}\right) $](1)</Checkpoint>

Substituting $(6)(7)(8)(9)$ into $(5)$ yields the equation:

$\frac{2 x_{2}^{2}} {0.576}+\left(p_{\mathrm{t}}-p_{0}\right)+\frac{x_{2}\left[2\left(p_{\mathrm{t}}-p_{0}\right)-x_{2}\right] }{\left(2.23 \times 10^{-6}\right)}=p_{\mathrm{t}}$

Substituting the data, we get: $x_{2}=0.0998 \mathrm{~bar}$

<Checkpoint>[$x_{2}=0.0998 \mathrm{~bar}$](1)</Checkpoint>

Then $x_{1}=0.0173 \mathrm{~bar}，y_{1}=0.0672 \mathrm{~bar},，y_{2}=1.48 \times 10^{-4} \mathrm{~bar}，z=6.5
\mathrm{~bar}$。

<Checkpoint>[The equilibrium partial pressure of $\mathrm{Cl}_{2}(\mathrm{~g})$ is $0.0173\mathrm{~bar}$.](1)</Checkpoint>

$2$ Incorrect.

$K_{p 2}=\frac{z^{2} }{x_{1} y_{1}}=3.63 \times 10^{5}$

<Checkpoint>[$K_{p 2}=\frac{z^{2} }{x_{1} y_{1}}=3.63 \times 10^{5}$](1)</Checkpoint>

$K_{p 4}=\frac{K_{p 2} K_{p 3}{ }^{2}}{K_{p 1}}=3.13 \times 10^{-6}$

<Checkpoint>[$K_{p 4}=\frac{K_{p 2} K_{p 3}{ }^{2}}{K_{p 1}}=3.13 \times 10^{-6}$](1)</Checkpoint>

$4$ Incorrect.

$3$. In this system, reaction $1+$ reaction $4-2\times$ reaction $3=$ reaction $2$, there are $3$ independent reactions.

<Checkpoint>[There are $3$ independent reactions.](1)</Checkpoint>

$3$ Correct.

Therefore, the correct option is I.
为使处理过程简洁方便，计算中使用如下约定符号，在平衡表达式中默认各分压项均除以标准分压：

$\mathrm{Cl_2(g)}$起始分压 $x_0$；$\mathrm{Cl_2(g)}$平衡分压 $x_1$；$\mathrm{Cl(g)}$平衡分压 $x_2$；$\mathrm{H_2(g)}$起始分压 $y_0$；$\mathrm{H_2(g)}$平衡分压 $y_1$；$\mathrm{H(g)}$平衡分压 $y_2$；
$\mathrm{HCl(g)}$平衡分压 $z$；
体系总压 $p_t$;
由理想气体状态方程可得：
$p_{0}=\frac{n_{0} R T} { V}=0.2 \times 8.314 \times 2000 \div 5 \times 10^{3} \mathrm{~Pa}=6.65 \mathrm{~bar}$ 

<Checkpoint>[$p_{0}=6.65 \mathrm{~bar}$ ](1)</Checkpoint>

$1$正确。

由反应平衡可知 

$x_{2}{ }^{2}=0.576 x_{1} \cdots(1)$,

$x_{2} y_{2}=2.23 \times 10^{-6} z \cdots(2)$

<Checkpoint>[$x_{2}{ }^{2}=0.576 x_{1} $, $x_{2} y_{2}=2.23 \times 10^{-6} z $](1)</Checkpoint>

由物料守恒可知

$2 x_{1}+x_{2}=2 y_{1}+y_{2} \cdots(3)$

由压强变化可知 

$p_{\mathrm{t}}-p_{0}=0.5 x_{2}+0.5 y_{2} \cdots(4)$ 

由体系总压可知 

$x_{1}+x_{2}+y_{1}+y_{2}+z=p_{\mathrm{t}} \cdots(5)$

由$(1)$可知 $x_{1}=x_{2}^{2} / 0.576 \cdots(6)$

由$(4)$可知 $y_{2}=2\left(p_{\mathrm{t}}-p_{0}\right)-x_{2} \cdots(7)$

将$(7)$代入$(2)$可知 $z=\frac{x_{2}\left[2\left(p_{\mathrm{t}}-p_{0}\right)-x_{2}\right] }{\left(2.23 \times 10^{-6}\right) }\cdots(8)$ 

<Checkpoint>[$z=\frac{x_{2}\left[2\left(p_{\mathrm{t}}-p_{0}\right)-x_{2}\right] }{\left(2.23 \times 10^{-6}\right) }$ ](1)</Checkpoint>

将$(6)$和$(7)$代入$(3)$可知 $y_{1}=x_{2}{ }^{2} / 0.576+x_{2}-\left(p_{\mathrm{t}}-p_{0}\right) \cdots(9)$

<Checkpoint>[$y_{1}=x_{2}{ }^{2} / 0.576+x_{2}-\left(p_{\mathrm{t}}-p_{0}\right) $](1)</Checkpoint>

将$(6)(7)(8)(9)$代入$(5)$得到方程:

$\frac{2 x_{2}^{2}} {0.576}+\left(p_{\mathrm{t}}-p_{0}\right)+\frac{x_{2}\left[2\left(p_{\mathrm{t}}-p_{0}\right)-x_{2}\right] }{\left(2.23 \times 10^{-6}\right)}=p_{\mathrm{t}}$

代入数据解得:$x_{2}=0.0998 \mathrm{~bar}$

<Checkpoint>[$x_{2}=0.0998 \mathrm{~bar}$](1)</Checkpoint>

则 $x_{1}=0.0173 \mathrm{~bar}，y_{1}=0.0672 \mathrm{~bar},，y_{2}=1.48 \times 10^{-4} \mathrm{~bar}，z=6.5
\mathrm{~bar}$。

<Checkpoint>[$\mathrm{Cl}_{2}(\mathrm{~g})$ 平衡分压为$0.0173\mathrm{~bar}$。](1)</Checkpoint>

$2$错误。

$K_{p 2}=\frac{z^{2} }{x_{1} y_{1}}=3.63 \times 10^{5}$

<Checkpoint>[$K_{p 2}=\frac{z^{2} }{x_{1} y_{1}}=3.63 \times 10^{5}$](1)</Checkpoint>

$K_{p 4}=\frac{K_{p 2} K_{p 3}{ }^{2}}{K_{p 1}}=3.13 \times 10^{-6}$

<Checkpoint>[$K_{p 4}=\frac{K_{p 2} K_{p 3}{ }^{2}}{K_{p 1}}=3.13 \times 10^{-6}$](1)</Checkpoint>

$4$错误。

$3$.该体系中,反应$1+$反应$4-2\times$反应$3=$反应$2$，有$3$个独立反应。

<Checkpoint>[有$3$个独立反应。](1)</Checkpoint>

$3$正确。

故正确选项为I。