$\mu$-oxo-bis[bis(diethyldithiocarbamato)oxomolybdenum(V)] (Complex **A**) is a molybdenum complex containing sulfur ligands. Studying the properties of such substances helps us understand the structure and performance of various molybdenum enzymes in biological systems. Experimental results show that the solution of **A** does not obey the Lambert-Beer law, indicating the existence of the following disproportionation equilibrium:  

$\mathrm {[MoO(S_2CNEt_2)_2]_2O}$ (**A**) $\rightleftharpoons \mathrm {MoO(S_2CNEt_2)_2} $(**B**) $+\mathrm {MoO_2(S_2CNEt_2)_2}$(**C**)  

Solutions of **A** were prepared with initial concentrations of $c _ { 1 } = 2 . 0 7 \times 1 0 ^ { - 4 } \mathrm { m o l { \cdot } L ^ { - 1 } }$ and $c _ { 2 } = 1 . 5 5 \times 1 0 ^ { - 4 } \mathrm { m o l { \cdot } L ^ { - 1 } }$, and allowed to reach disproportionation equilibrium. The absorbances were measured at $\lambda = 5 1 0 \mathrm { n m }$ (the absorption band of the bridging oxygen) in a $b = 1 . 0 0 \ \mathrm { c m }$ cuvette, yielding $A _ { 1 } = 0 . 6 1 6$ and $A _ { 2 } = 0 . 3 8 5$, respectively. It is known that at $\lambda = 5 1 0 \mathrm { n m }$, the molar absorptivity of **B** is $\varepsilon _ { \mathrm { B } } { = } 7 1 5 \mathrm { L } { \cdot } \mathrm { m o l ^ { - 1 } } { \cdot } \mathrm { c m } ^ { - 1 }$, and its contribution to absorbance cannot be ignored. In contrast, the molar absorptivity of **C** is very small, and its effect on absorbance can be neglected. Please calculate the equilibrium constant $K$ of the disproportionation reaction and the true molar absorptivity $\varepsilon _ { \mathrm { A } }$ of **A**.
$\mu$-氧-双[双-(二乙基二硫代氨基甲酸)氧钼(V)](配合物 **A**)是一种含硫配体的钼络合物，研究此类物质的性质有助于我们了解生物体内各种钼酶的结构和性能。实验结果表明，**A**的溶液不遵守朗伯-比尔定律，预示了下面的歧化平衡的存在:

$\mathrm {[MoO(S_2CNEt_2)_2]_2O}$ (**A**) $\rightleftharpoons \mathrm {MoO(S_2CNEt_2)_2} $(**B**) $+\mathrm {MoO_2(S_2CNEt_2)_2}$(**C**)

将**A**配成初始浓度为 $c _ { 1 } = 2 . 0 7 \times 1 0 ^ { - 4 } \mathrm { m o l { \cdot } L ^ { - 1 } }$ 和 $c _ { 2 } = 1 . 5 5 \times 1 0 ^ { - 4 } \mathrm { m o l { \cdot } L ^ { - 1 } }$ 的溶液，并使之达到歧化平衡，在 $\lambda = 5 1 0 \mathrm { n m }$ 处(桥联氧的吸收带)、 $b = 1 . 0 0 \ \mathrm { c m }$ 吸收池中分别测得其吸光度分别为 $\begin{array} { r } { A _ { 1 } = 0 . 6 1 6 } \end{array}$ 和 $A _ { 2 } = 0 . 3 8 5$ 。已知 $\lambda = 5 1 0 \mathrm { n m }$ 下**B**的摩尔吸光系数为 $\varepsilon _ { \mathrm { B } } { = } 7 1 5 \mathrm { L } { \cdot } \mathrm { m o l ^ { - 1 } } { \cdot } \mathrm { c m } ^ { - 1 }$ 对吸光度的贡献不可忽略；而**C**的摩尔吸光系数很小，对吸光度的影响可以忽略。请计算歧化反应的平衡常数 $K$ 以及**A**真正的摩尔吸光系数 $\varepsilon _ { \mathrm { A } }$ 。
$K = 2.5 \times 1 0 ^ { - 3 } \ \mathrm { m o l \cdot L ^ { - 1 } } $

$\varepsilon _ { \mathrm { A } } = 2 . 0 \times 1 0 ^ { 4 } \ \mathrm { L } { \cdot } \mathrm { m o l } ^ { - 1 } { \cdot } \mathrm { c m } ^ { - 1 } $
$K = 4.0 \times 1 0 ^ { - 3 } \ \mathrm { m o l \cdot L ^ { - 1 } } $

$\varepsilon _ { \mathrm { A } } = 6.1 \times 1 0 ^ { 3 } \ \mathrm { L } { \cdot } \mathrm { m o l } ^ { - 1 } { \cdot } \mathrm { c m } ^ { - 1 } $
$K = 1 . 8 \times 1 0 ^ { - 3 } \ \mathrm { m o l \cdot L ^ { - 1 } } $

$\varepsilon _ { \mathrm { A } } = 2 . 5 \times 1 0 ^ { 4 } \ \mathrm { L } { \cdot } \mathrm { m o l } ^ { - 1 } { \cdot } \mathrm { c m } ^ { - 1 } $
$K = 4.6 \times 1 0 ^ { - 4 } \ \mathrm { m o l \cdot L ^ { - 1 } } $

$\varepsilon _ { \mathrm { A } } = 1.2 \times 1 0 ^ { 4 } \ \mathrm { L } { \cdot } \mathrm { m o l } ^ { - 1 } { \cdot } \mathrm { c m } ^ { - 1 } $
$K = 1 . 8 \times 1 0 ^ { - 4 } \ \mathrm { m o l \cdot L ^ { - 1 } } $

$\varepsilon _ { \mathrm { A } } = 2 . 5 \times 1 0 ^ { 4 } \ \mathrm { L } { \cdot } \mathrm { m o l } ^ { - 1 } { \cdot } \mathrm { c m } ^ { - 1 } $
$K = 4.0 \times 1 0 ^ { - 4 } \ \mathrm { m o l \cdot L ^ { - 1 } } $

$\varepsilon _ { \mathrm { A } } = 6.1 \times 1 0 ^ { 3 } \ \mathrm { L } { \cdot } \mathrm { m o l } ^ { - 1 } { \cdot } \mathrm { c m } ^ { - 1 } $
$K = 2.5 \times 1 0 ^ { - 4 } \ \mathrm { m o l \cdot L ^ { - 1 } } $

$\varepsilon _ { \mathrm { A } } = 2 . 0 \times 1 0 ^ { 4 } \ \mathrm { L } { \cdot } \mathrm { m o l } ^ { - 1 } { \cdot } \mathrm { c m } ^ { - 1 } $
$K = 4.6 \times 1 0 ^ { - 3 } \ \mathrm { m o l \cdot L ^ { - 1 } } $

$\varepsilon _ { \mathrm { A } } = 1.2 \times 1 0 ^ { 4 } \ \mathrm { L } { \cdot } \mathrm { m o l } ^ { - 1 } { \cdot } \mathrm { c m } ^ { - 1 } $
$K = 2.5 \times 1 0 ^ { - 3 } \ \mathrm { m o l \cdot L ^ { - 1 } } $

$\varepsilon _ { \mathrm { A } } = 2 . 0 \times 1 0 ^ { 4 } \ \mathrm { L } { \cdot } \mathrm { m o l } ^ { - 1 } { \cdot } \mathrm { c m } ^ { - 1 } $
$K = 4.0 \times 1 0 ^ { - 3 } \ \mathrm { m o l \cdot L ^ { - 1 } } $

$\varepsilon _ { \mathrm { A } } = 6.1 \times 1 0 ^ { 3 } \ \mathrm { L } { \cdot } \mathrm { m o l } ^ { - 1 } { \cdot } \mathrm { c m } ^ { - 1 } $
$K = 1 . 8 \times 1 0 ^ { - 3 } \ \mathrm { m o l \cdot L ^ { - 1 } } $

$\varepsilon _ { \mathrm { A } } = 2 . 5 \times 1 0 ^ { 4 } \ \mathrm { L } { \cdot } \mathrm { m o l } ^ { - 1 } { \cdot } \mathrm { c m } ^ { - 1 } $
$K = 4.6 \times 1 0 ^ { - 4 } \ \mathrm { m o l \cdot L ^ { - 1 } } $

$\varepsilon _ { \mathrm { A } } = 1.2 \times 1 0 ^ { 4 } \ \mathrm { L } { \cdot } \mathrm { m o l } ^ { - 1 } { \cdot } \mathrm { c m } ^ { - 1 } $
$K = 1 . 8 \times 1 0 ^ { - 4 } \ \mathrm { m o l \cdot L ^ { - 1 } } $

$\varepsilon _ { \mathrm { A } } = 2 . 5 \times 1 0 ^ { 4 } \ \mathrm { L } { \cdot } \mathrm { m o l } ^ { - 1 } { \cdot } \mathrm { c m } ^ { - 1 } $
$K = 4.0 \times 1 0 ^ { - 4 } \ \mathrm { m o l \cdot L ^ { - 1 } } $

$\varepsilon _ { \mathrm { A } } = 6.1 \times 1 0 ^ { 3 } \ \mathrm { L } { \cdot } \mathrm { m o l } ^ { - 1 } { \cdot } \mathrm { c m } ^ { - 1 } $
$K = 2.5 \times 1 0 ^ { - 4 } \ \mathrm { m o l \cdot L ^ { - 1 } } $

$\varepsilon _ { \mathrm { A } } = 2 . 0 \times 1 0 ^ { 4 } \ \mathrm { L } { \cdot } \mathrm { m o l } ^ { - 1 } { \cdot } \mathrm { c m } ^ { - 1 } $
$K = 4.6 \times 1 0 ^ { - 3 } \ \mathrm { m o l \cdot L ^ { - 1 } } $

$\varepsilon _ { \mathrm { A } } = 1.2 \times 1 0 ^ { 4 } \ \mathrm { L } { \cdot } \mathrm { m o l } ^ { - 1 } { \cdot } \mathrm { c m } ^ { - 1 } $
Let the equilibrium concentrations of **B** in the two sets of experiments be $x _ { 1 }$ and $x _ { 2 }$, respectively. Then, the following equations are established:  

$K=x_{1}^2/(c_1-x_1)$  

$K=x_{2}^2/(c_2-x_2)$  

$A_1=\varepsilon _ { \mathrm { A } } b(c_1−𝑥_1)+\varepsilon _ { \mathrm { B } } bx_1$  

$A_2=\varepsilon _ { \mathrm { A } } b(c_2−x_2)+\varepsilon _ { \mathrm { B } } bx_2 $  

<Checkpoint>[$K=x_{1}^2/(c_1-x_1)$](1)</Checkpoint>  

<Checkpoint>[$K=x_{2}^2/(c_2-x_2)$](1)</Checkpoint>  

<Checkpoint>[$A_1=\varepsilon _ { \mathrm { A } } b(c_1−𝑥_1)+\varepsilon _ { \mathrm { B } } bx_1$](1)</Checkpoint>  

<Checkpoint>[$A_2=\varepsilon _ { \mathrm { A } } b(c_2−x_2)+\varepsilon _ { \mathrm { B } } bx_2 $](1)</Checkpoint>  

The solution yields:  

$K = 1 . 8 \times 1 0 ^ { - 3 } \ \mathrm { m o l \cdot L ^ { - 1 } } $  

<Checkpoint>[$K = 1 . 8 \times 1 0 ^ { - 3 } \ \mathrm { m o l \cdot L ^ { - 1 } } $](2)</Checkpoint>  

$\varepsilon _ { \mathrm { A } } = 2 . 5 \times 1 0 ^ { 4 } \ \mathrm { L } { \cdot } \mathrm { m o l } ^ { - 1 } { \cdot } \mathrm { c m } ^ { - 1 } $  

<Checkpoint>[$\varepsilon _ { \mathrm { A } } = 2 . 5 \times 1 0 ^ { 4 } \ \mathrm { L } { \cdot } \mathrm { m o l } ^ { - 1 } { \cdot } \mathrm { c m } ^ { - 1 } $](2)</Checkpoint>
设**B**在两组实验中的平衡浓度分别为 $x _ { 1 }$ 和 $x _ { 2 }$，则联立以下方程：

$K=x_{1}^2/(c_1-x_1)$

$K=x_{2}^2/(c_2-x_2)$

$A_1=\varepsilon _ { \mathrm { A } } b(c_1−𝑥_1)+\varepsilon _ { \mathrm { B } } bx_1$

$A_2=\varepsilon _ { \mathrm { A } } b(c_2−x_2)+\varepsilon _ { \mathrm { B } } bx_2 $

<Checkpoint>[$K=x_{1}^2/(c_1-x_1)$](1)</Checkpoint>

<Checkpoint>[$K=x_{2}^2/(c_2-x_2)$](1)</Checkpoint>

<Checkpoint>[$A_1=\varepsilon _ { \mathrm { A } } b(c_1−𝑥_1)+\varepsilon _ { \mathrm { B } } bx_1$](1)</Checkpoint>

<Checkpoint>[$A_2=\varepsilon _ { \mathrm { A } } b(c_2−x_2)+\varepsilon _ { \mathrm { B } } bx_2 $](1)</Checkpoint>

解得： 

$K = 1 . 8 \times 1 0 ^ { - 3 } \ \mathrm { m o l \cdot L ^ { - 1 } } $

<Checkpoint>[$K = 1 . 8 \times 1 0 ^ { - 3 } \ \mathrm { m o l \cdot L ^ { - 1 } } $](2)</Checkpoint>

$\varepsilon _ { \mathrm { A } } = 2 . 5 \times 1 0 ^ { 4 } \ \mathrm { L } { \cdot } \mathrm { m o l } ^ { - 1 } { \cdot } \mathrm { c m } ^ { - 1 } $

<Checkpoint>[$\varepsilon _ { \mathrm { A } } = 2 . 5 \times 1 0 ^ { 4 } \ \mathrm { L } { \cdot } \mathrm { m o l } ^ { - 1 } { \cdot } \mathrm { c m } ^ { - 1 } $](2)</Checkpoint>