It is known that for $\mathrm{H_2S}$, $K_{\mathrm{a1}} = 1.1 \times 10^{-7}$, $K_{\mathrm{a2}} = 1.3 \times 10^{-13}$; $K_{\mathrm{sp}}(\mathrm{CuS}) = 6.3 \times 10^{-36}$, $K_{\mathrm{sp}}(\mathrm{Cu_2S}) = 2.5 \times 10^{-48}$. $\varphi_{\mathrm{A}}^{\theta}(\mathrm{S}/\mathrm{H_2S}) = 0.142\ \mathrm{V}$, $\varphi_{\mathrm{B}}^{\theta}(\mathrm{S}/\mathrm{S}^{2-}) = -0.476\ \mathrm{~V~}$; $\varphi_{\mathrm{A}}^{\theta}(\mathrm{NO_3}^{-}/\mathrm{NO}) = 0.942\ \mathrm{V}$, $\varphi^{\theta}(\mathrm{Cu}^{2+}/\mathrm{Cu}) = 0.342\ \mathrm{V}$, $\varphi^{\theta}(\mathrm{Cu}^{+}/\mathrm{Cu}) = 0.521\ \mathrm{V}$.  

Based on the above conditions, write the ionic equation for the dissolution of $\mathrm{Cu_2S}$ in dilute nitric acid to form a blue solution and a yellow precipitate, and calculate the standard equilibrium constant of this reaction using Hess's law. Select the option closest to your calculated equilibrium constant.
已知 $\mathrm{ H _2S }$ 的 $K _ { \mathrm { a 1 } } = 1 . 1 { \times } 1 0 ^ { - 7 }$ ， $K _ { \mathrm{a 2} } = 1 . 3 { \times } 1 0 ^ { - 1 3 }$ ；  $K _ { \mathrm { s p } } ( \mathrm { C u S } ) = 6 . 3 { \times } 1 0 ^ { - 3 6 } $， $K _ { \mathrm { s p } } ( \mathrm { C u _ { 2 } S } ) = 2 . 5 { \times } 1 0 ^ { - 4 8 }$。 $\varphi _ { \mathrm { A } } ^ { \theta } ( \mathrm { S } / \mathrm {H _ 2S } ) = 0 . 1 4 2 \ \mathrm { V }$，$\varphi _ { \mathrm { B } } ^ { \theta } ( \mathrm { S } / \mathrm { S } ^ { 2 - } ) = - 0 . 4 7 6\ \mathrm { V }$ ；$\varphi _ { \mathrm { A } } ^ { \theta } ( \mathrm { N O } _ { 3}^{-}/  \mathrm { N O } ) = 0 . 9 4 2\ \mathrm { V } $，$\varphi ^ { \theta } ( \mathrm { C u } ^ { 2 + } / \mathrm { C u } )= 0 . 3 4 2 \mathrm { \ V }$ ， $\varphi ^ { \theta } ( \mathrm { C u } ^ { + } / \mathrm { C u } ) = 0 . 5 2 1\ \mathrm { V  }$。

根据以上条件，写出 $\mathrm { C u _ { 2 } S }$ 溶于稀硝酸生成蓝色溶液与黄色沉淀的离子方程式，利用Hess 定律等计算该反应的标准平衡常数。选择最接近你计算的平衡常数的选项。
All other options are incorrect
$K^{\theta}=1.1\times 10^{80}$
$K^{\theta}=2.0 \times 10^{130}$
$K^{\theta}=2.4 \times 10^{8}$
$K^{\theta}=5.1 \times 10^{43}$
其他选项均不正确
$K^{\theta}=1.1\times 10^{80}$
$K^{\theta}=2.0 \times 10^{130}$
$K^{\theta}=2.4 \times 10^{8}$
$K^{\theta}=5.1 \times 10^{43}$
First, calculate the standard potential for the reduction of cupric ion to cuprous ion:

$\varphi _ { \mathrm { B } } ^ { \theta } ( \mathrm { Cu^{2+} } / \mathrm { Cu } ^ { + }）=2 \varphi ^ { \theta } ( \mathrm { C u } ^ { 2 + } / \mathrm { C u })-\varphi ^ { \theta } ( \mathrm { C u } ^ { + } / \mathrm { C u } )=0.163 \ \mathrm{V}$

The chemical equation mentioned in the question is:

$3\mathrm{Cu_{2}S}+4\mathrm{NO_{3}^{-}}+16\mathrm{H}^{+}=6\mathrm{Cu^{2+}}+3\mathrm{S}+4\mathrm{NO}+8\mathrm{H_{2}O}$

<Checkpoint>[The chemical equation is: $3\mathrm{Cu_{2}S}+4\mathrm{NO_{3}^{-}}+16\mathrm{H}^{+}=6\mathrm{Cu^{2+}}+3\mathrm{S}+4\mathrm{NO}+8\mathrm{H_{2}O}$](1)</Checkpoint>

Decompose the reaction to separate the precipitation dissolution equilibrium and the redox reaction:

<Checkpoint>[Decompose the reaction to separate the precipitation dissolution equilibrium and the redox reaction](1)</Checkpoint>

$3\mathrm{Cu_{2}S}=6\mathrm{Cu^{+}}+3\mathrm{S^{2-}} \ \mathrm{K_{1}={K_{sp}(Cu_{2}S)}^{3}}=1.56\times 10^{-143}$

<Checkpoint>[$3\mathrm{Cu_{2}S}=6\mathrm{Cu^{+}}+3\mathrm{S^{2-}} \ \mathrm{K_{1}={K_{sp}(Cu_{2}S)}^{3}}=1.56\times 10^{-143}$](1)</Checkpoint>

$\mathrm{3S^{2-}+2NO_{3}^{-}+8H^{+}=3S+2NO+4H_{2}O} \ \mathrm{K_{2}}$

<Checkpoint>[$\mathrm{3S^{2-}+2NO_{3}^{-}+8H^{+}=3S+2NO+4H_{2}O} \ \mathrm{K_{2}=exp(\frac{6F}{RT}\times(\varphi _ { \mathrm { A } } ^ { \theta } ( \mathrm { N O } _ { 3}^{-}/  \mathrm { N O } )-\varphi _ { \mathrm { B } } ^ { \theta } ( \mathrm { S } / \mathrm { S } ^ { 2 - }))=6.64\times10^{143}}$](2)</Checkpoint>

$\mathrm{6Cu^{+}+2NO_{3}^{-}+8H^{+}=6Cu^{2+}+2NO+4H_{2}O} \ \mathrm{K_{3}}$

<Checkpoint>[$\mathrm{6Cu^{+}+2NO_{3}^{-}+8H^{+}=6Cu^{2+}+2NO+4H_{2}O} \ \mathrm{K_{3}=exp(\frac{6F}{RT}\times(\varphi _ { \mathrm { A } } ^ { \theta } ( \mathrm { N O } _ { 3}^{-}/  \mathrm { N O } )-\varphi _ { \mathrm { B } } ^ { \theta } ( \mathrm { Cu^{2+} } / \mathrm { Cu } ^ { + }))=1.03\times10^{79}}$](2)</Checkpoint>

Where, for $\mathrm{K_{2}}$ and $\mathrm{K_{3}}$, we have

$\mathrm{K_{2}=exp(\frac{6F}{RT}\times(\varphi _ { \mathrm { A } } ^ { \theta } ( \mathrm { N O } _ { 3}^{-}/  \mathrm { N O } )-\varphi _ { \mathrm { B } } ^ { \theta } ( \mathrm { S } / \mathrm { S } ^ { 2 - }))=6.64\times10^{143}}$

$\mathrm{K_{3}=exp(\frac{6F}{RT}\times(\varphi _ { \mathrm { A } } ^ { \theta } ( \mathrm { N O } _ { 3}^{-}/  \mathrm { N O } )-\varphi _ { \mathrm { B } } ^ { \theta } ( \mathrm { Cu^{2+} } / \mathrm { Cu } ^ { + }))=1.03\times10^{79}}$

$\mathrm{K^{\theta}=K_{1}K_{2}K_{3}=1.1\times10^{80}}$

<Checkpoint>[$\mathrm{K^{\theta}=K_{1}K_{2}K_{3}=1.1\times10^{80}}$](1)</Checkpoint>
首先计算二价铜还原到一价铜的标准电势：

$\varphi _ { \mathrm { B } } ^ { \theta } ( \mathrm { Cu^{2+} } / \mathrm { Cu } ^ { + }）=2 \varphi ^ { \theta } ( \mathrm { C u } ^ { 2 + } / \mathrm { C u })-\varphi ^ { \theta } ( \mathrm { C u } ^ { + } / \mathrm { C u } )=0.163 \ \mathrm{V}$


题中所述化学方程式为：  

$3\mathrm{Cu_{2}S}+4\mathrm{NO_{3}^{-}}+16\mathrm{H}^{+}=6\mathrm{Cu^{2+}}+3\mathrm{S}+4\mathrm{NO}+8\mathrm{H_{2}O}$

<Checkpoint>[化学方程式为：    $3\mathrm{Cu_{2}S}+4\mathrm{NO_{3}^{-}}+16\mathrm{H}^{+}=6\mathrm{Cu^{2+}}+3\mathrm{S}+4\mathrm{NO}+8\mathrm{H_{2}O}$](1)</Checkpoint>


将该反应进行分解，使得沉淀溶解平衡和氧化还原反应被分开：

<Checkpoint>[将该反应进行分解，使得沉淀溶解平衡和氧化还原反应被分开](1)</Checkpoint>

$3\mathrm{Cu_{2}S}=6\mathrm{Cu^{+}}+3\mathrm{S^{2-}} \ \mathrm{K_{1}={K_{sp}(Cu_{2}S)}^{3}}=1.56\times 10^{-143}$

<Checkpoint>[$3\mathrm{Cu_{2}S}=6\mathrm{Cu^{+}}+3\mathrm{S^{2-}} \ \mathrm{K_{1}={K_{sp}(Cu_{2}S)}^{3}}=1.56\times 10^{-143}$](1)</Checkpoint>

$\mathrm{3S^{2-}+2NO_{3}^{-}+8H^{+}=3S+2NO+4H_{2}O} \ \mathrm{K_{2}}$

<Checkpoint>[$\mathrm{3S^{2-}+2NO_{3}^{-}+8H^{+}=3S+2NO+4H_{2}O} \ \mathrm{K_{2}=exp(\frac{6F}{RT}\times(\varphi _ { \mathrm { A } } ^ { \theta } ( \mathrm { N O } _ { 3}^{-}/  \mathrm { N O } )-\varphi _ { \mathrm { B } } ^ { \theta } ( \mathrm { S } / \mathrm { S } ^ { 2 - }))=6.64\times10^{143}}$](2)</Checkpoint>

$\mathrm{6Cu^{+}+2NO_{3}^{-}+8H^{+}=6Cu^{2+}+2NO+4H_{2}O} \ \mathrm{K_{3}}$

<Checkpoint>[$\mathrm{6Cu^{+}+2NO_{3}^{-}+8H^{+}=6Cu^{2+}+2NO+4H_{2}O} \ \mathrm{K_{3}=exp(\frac{6F}{RT}\times(\varphi _ { \mathrm { A } } ^ { \theta } ( \mathrm { N O } _ { 3}^{-}/  \mathrm { N O } )-\varphi _ { \mathrm { B } } ^ { \theta } ( \mathrm { Cu^{2+} } / \mathrm { Cu } ^ { + }))=1.03\times10^{79}}$](2)</Checkpoint>


其中，对于$\mathrm{K_{2}}$和$\mathrm{K_{3}}$，有


$\mathrm{K_{2}=exp(\frac{6F}{RT}\times(\varphi _ { \mathrm { A } } ^ { \theta } ( \mathrm { N O } _ { 3}^{-}/  \mathrm { N O } )-\varphi _ { \mathrm { B } } ^ { \theta } ( \mathrm { S } / \mathrm { S } ^ { 2 - }))=6.64\times10^{143}}$


$\mathrm{K_{3}=exp(\frac{6F}{RT}\times(\varphi _ { \mathrm { A } } ^ { \theta } ( \mathrm { N O } _ { 3}^{-}/  \mathrm { N O } )-\varphi _ { \mathrm { B } } ^ { \theta } ( \mathrm { Cu^{2+} } / \mathrm { Cu } ^ { + }))=1.03\times10^{79}}$


$\mathrm{K^{\theta}=K_{1}K_{2}K_{3}=1.1\times10^{80}}$

<Checkpoint>[$\mathrm{K^{\theta}=K_{1}K_{2}K_{3}=1.1\times10^{80}}$](1)</Checkpoint>