Given: At 693 K, the initial amounts of Xe and F₂ in the container are 2 mol and 1 mol, respectively.  

Reaction equations:  

$Xe(g) + F_2(g) \rightleftharpoons XeF_2(g) \quad K_{p1} = 360.3$  

$Xe(g) + 2F_2(g) \rightleftharpoons XeF_4(g) \quad K_{p2} = 1978.8$  

$Xe(g) + 3F_2(g) \rightleftharpoons XeF_6(g) \quad K_{p3} = 35.99$  

The volume of the container is variable, and the total pressure is maintained at 1 bar. For simplicity and convenience in calculations, the following symbolic conventions must be used. In the equilibrium expressions, all partial pressure terms are divided by the standard partial pressure by default, and the units in the problem are all in bar.  

| Total system pressure | Initial partial pressure of $Xe(g)$ | Initial partial pressure of $F_2(g)$ | Equilibrium partial pressure of $Xe(g)$ | Equilibrium partial pressure of $F_2(g)$ | Equilibrium partial pressure of $XeF_2(g)$ | Equilibrium partial pressure of $XeF_4(g)$ | Equilibrium partial pressure of $XeF_6(g)$ |  
|-----------------------|-----------------------------------|------------------------------------|----------------------------------|-----------------------------------|-------------------------------------|--------------------------------------|--------------------------------------|  
| $p_1$                 | $x_0$                             | $x_1$                              | $y_0$                            | $y_1$                             | $y_2$                               | $y_3$                                | $y_4$                                |  

Then $\frac{y_4}{y_0+100y_1}=$:
已知：693 K 下，初始情况下容器内 Xe 与 F₂ 的物质量分别为 2 mol 与 1 mol。

反应方程式：

$Xe(g) + F_2(g) \rightleftharpoons XeF_2(g) \quad K_{p1} = 360.3$

$Xe(g) + 2F_2(g) \rightleftharpoons XeF_4(g) \quad K_{p2} = 1978.8$

$Xe(g) + 3F_2(g) \rightleftharpoons XeF_6(g) \quad K_{p3} = 35.99$

容器体积可变，总压维持在 1 bar。为使处理过程简洁方便，计算过程中务必使用以下约定符号。 在平衡表达式中默认各分压项均除以标准分压，题中单位均为bar。

| 体系总压 | $Xe(g)$ 起始分压 | $F_2(g)$ 起始分压 | $Xe(g)$ 平衡分压 | $F_2(g)$ 平衡分压 | $XeF_2(g)$ 平衡分压 | $XeF_4(g)$ 平衡分压 | $XeF_6(g)$ 平衡分压 |
|----------|------------------|-------------------|------------------|-------------------|---------------------|----------------------|----------------------|
| $p_1$    | $x_0$            | $x_1$             | $y_0$            | $y_1$             | $y_2$               | $y_3$                | $y_4$                |

则 $\frac{y_4}{y_0+100y_1}=$：（保留一位有效数字）
$4 \times 10^{-7}$
$8 \times 10^{-7}$
$2 \times 10^{-6}$
$1 \times 10^{-7}$
$8 \times 10^{-6}$
$4 \times 10^{-6}$
$8\times10^{-8}$
None of the above options are correct
$4\times 10^{-7} $
$8\times 10^{-7}$
$2\times 10^{-6}$
$1\times 10^{-7}$
$8\times 10^{-6}$
$4\times 10^{-6}$
$8\times10^{-8}$
以上选项均错误
From the quantitative relationship:
$$
y_0 + y_1 + y_2 + y_3 + y_4 = 1 \quad (1)
$$

<Checkpoint>[$y_0 + y_1 + y_2 + y_3 + y_4 = 1 \quad (1)$](1)</Checkpoint>

From the molar relationship:
$$
x_0 = 2x_1
$$

From the atomic conservation relationship:
$$ 
y_0 + y_2 + y_3 + y_4 = 2y_1 + 2y_2 + 4y_3 + 6y_4 
$$

Combining with (1) yields:
$$
2y_2 + 4y_3 + 6y_4 = 1-3y_1 \quad (2)
$$

<Checkpoint>[$2y_2 + 4y_3 + 6y_4 = 1-3y_1 \quad (2)$](1)</Checkpoint>

Equilibrium constant expressions:
$$ 
y_2 = K_{p1} y_0 y_1 \quad (3)
$$

<Checkpoint>[$ y_2 = K_{p1} y_0 y_1 \quad (3) $](1)</Checkpoint>

$$ 
y_3 = K_{p2} y_0 y_1^2 \quad (4)
$$

<Checkpoint>[$ y_3 = K_{p2} y_0 y_1^2 \quad (4) $](1)</Checkpoint>

$$ 
y_4 = K_{p3} y_0 y_1^3 \quad (5)
$$

<Checkpoint>[$ y_4 = K_{p3} y_0 y_1^3 \quad (5) $](1)</Checkpoint>

From (1), (3), (4), and (5), $y_0$ can be expressed as a function of $y_1$. Substituting into (2) yields a univariate equation in $y_1$, ultimately solving to:

<Checkpoint>[$ y_1 = 2.633 \times 10^{-3} \text{ bar} $ ](1)</Checkpoint>

Substituting back into the equations yields:
$$ 
y_0 = 0.5082 \text{ bar} 
$$

<Checkpoint>[$ y_0 = 0.5082 \text{ bar} $ ](1)</Checkpoint>

$$
y_2 = 0.482 \text{ bar} 
$$

<Checkpoint>[$ y_2 = 0.482 \text{ bar} $ ](1)</Checkpoint>

$$ 
y_3 = 6.966 \times 10^{-3} \text{ bar} 
$$

<Checkpoint>[$ y_3 = 6.966 \times 10^{-3} \text{ bar} $ ](1)</Checkpoint>

$$
y_4 = 3.255 \times 10^{-7} \text{ bar} 
$$

<Checkpoint>[$ y_4 = 3.255 \times 10^{-7} \text{ bar} $ ](1)</Checkpoint>

Substituting and calculating gives $\frac{y_4}{y_0+100y_1}=4.2\times 10^{-7}$

<Checkpoint>[$\frac{y_4}{y_0+100y_1}=4.2\times 10^{-7}$ Option A is correct](1)</Checkpoint>
由数量关系：
$$
y_0 + y_1 + y_2 + y_3 + y_4 = 1 （1）
$$

<Checkpoint>[$y_0 + y_1 + y_2 + y_3 + y_4 = 1（1）$](1)</Checkpoint>

由物质的量关系：
$$
x_0 = 2x_1
$$

有原子守恒关系：
$$ 
2(y_0 + y_2 + y_3 + y_4) = 2y_1 + 2y_2 + 4y_3 + 6y_4 
$$

联立（1）得：
$$
2y_2 + 4y_3 + 6y_4 = 2-4y_1 （2）
$$

<Checkpoint>[$2y_2 + 4y_3 + 6y_4 = 1-3y_1 （2）$](1)</Checkpoint>

平衡常数表达式：
$$ 
y_2 = K_{p1} y_0 y_1 （3）
$$

<Checkpoint>[$  y_2 = K_{p1} y_0 y_1 （3） $](1)</Checkpoint>

$$ 
y_3 = K_{p2} y_0 y_1^2 （4）
$$

<Checkpoint>[$  y_3 = K_{p2} y_0 y_1^2 （4） $](1)</Checkpoint>

$$ 
y_4 = K_{p3} y_0 y_1^3 （5）
$$

<Checkpoint>[$  y_4 = K_{p3} y_0 y_1^3 （5） $](1)</Checkpoint>

由（1）（3）（4）（5）可以将$y_0$表示为$y_1$的函数，再代入（2）得到只有$y_1$的一元方程，最终解得：

<Checkpoint>[$  y_1 = 2.633 \times 10^{-3} \text{ bar} $ ](1)</Checkpoint>

再代入方程，解得：
$$ 
y_0 = 0.5082 \text{ bar} 
$$

<Checkpoint>[$  y_0 = 0.5082 \text{ bar}  $ ](1)</Checkpoint>

$$
y_2 = 0.482 \text{ bar} 
$$

<Checkpoint>[$ y_2 = 0.482 \text{ bar}  $ ](1)</Checkpoint>

$$ 
y_3 = 6.966 \times 10^{-3} \text{ bar} 
$$

<Checkpoint>[$  y_3 = 6.966 \times 10^{-3} \text{ bar}  $ ](1)</Checkpoint>

$$
y_4 = 3.255 \times 10^{-7} \text{ bar} 
$$

<Checkpoint>[$ y_4 = 3.255 \times 10^{-7} \text{ bar}  $ ](1)</Checkpoint>

代入计算得 $\frac{y_4}{y_0+100y_1}=4.2\times 10^{-7}$

<Checkpoint>[$\frac{y_4}{y_0+100y_1}=4.2\times 10^{-7}$ 选项A正确](1)</Checkpoint>