Extractant $\mathrm{L}$ (structure shown in Figure 1) can extract $\mathrm{Pd}\left(\mathrm{NO}_{3}\right)_{2}$ from the aqueous phase to the organic phase $\left(\mathrm{CHCl}_{3}\right)$:
$$
2 \mathrm{Pd}^{2+}(\mathrm{aq})+4 \mathrm{NO}_{3}^{-}(\mathrm{aq})+\mathrm{L}(\mathrm{o}) \rightleftharpoons \mathrm{Pd}_{2}\left(\mathrm{NO}_{3}\right)_{4} \mathrm{L}(\mathrm{o})
$$
In the above equation, aq represents the aqueous phase, and o represents the organic phase. $\mathrm{Pd}^{2+}$ and $\mathrm{NO_3^-}$ exist only in the aqueous phase, and $\mathrm{L}$ and $\mathrm{Pd}_{2} \mathrm{(NO_3)_4} \mathrm{L}$ exist only in the organic phase. At the beginning of the reaction, the volumes of the aqueous phase and the organic phase are the same, and the concentrations of each species are controlled as follows:
$$
c_{0}\left(\mathrm{NO}_{3}^{-}\right)=1.00 \mathrm{~mol} / \mathrm{L} ; c_{0}(\mathrm{~L})=2.0 \times 10^{-3} \mathrm{~mol} / \mathrm{L} ; c_{0}\left(\mathrm{Pd}^{2+}\right)=1.0 \times 10^{2} \mathrm{mg} / \mathrm{L};
$$
At equilibrium, it is measured that:
$$
c\left(\mathrm{Pd}^{2+}\right): c\left(\mathrm{Pd}_{2}\left(\mathrm{NO}_{3}\right)_{4} \mathrm{L}\right)=0.30
$$

<MultiModal>![Fig .1, the molecule in the figure is represented by SMILES as: CCN(C1=CC=CC=C1)C(=S)C2=NC(=CC=C2)C3=NC(=CC=C3)C(=S)N(CC)C4=CC=CC=C4](/media/uploads/2ec65003-9c0d-4e86-9a72-22c6acd19cf9.png)</MultiModal>

Which of the following options are correct:
萃取剂 $\mathrm{L}$(结构如图1) 能将$\mathrm{Pd}\left(\mathrm{NO}_{3}\right)_{2}$ 从水相萃取至有机相 $\left(\mathrm{CHCl}_{3}\right)$ :
$$
2 \mathrm{Pd}^{2+}(\mathrm{aq})+4 \mathrm{NO}_{3}^{-}(\mathrm{aq})+\mathrm{L}(\mathrm{o}) \rightleftharpoons \mathrm{Pd}_{2}\left(\mathrm{NO}_{3}\right)_{4} \mathrm{L}(\mathrm{o})
$$
上式中 aq 表示水相, $\mathrm{o}$ 表示有机相。 $\mathrm{Pd}^{2+} 、 \mathrm{NO_3^-}$ 仅存在于水相, $\mathrm{L}$ 和 $\mathrm{Pd}_{2} \mathrm{(NO_3)_4} \mathrm{L}$ 仅存在于有机相。反应开始时, 水相和有机相体积相同, 控制各物种浓度如下:
$$
c_{0}\left(\mathrm{NO}_{3}^{-}\right)=1.00 \mathrm{~mol} / \mathrm{L} ; c_{0}(\mathrm{~L})=2.0 \times 10^{-3} \mathrm{~mol} / \mathrm{L} ; c_{0}\left(\mathrm{Pd}^{2+}\right)=1.0 \times 10^{2} \mathrm{mg} / \mathrm{L};
$$
平衡时, 测得:
$$
c\left(\mathrm{Pd}^{2+}\right): c\left(\mathrm{Pd}_{2}\left(\mathrm{NO}_{3}\right)_{4} \mathrm{L}\right)=0.30
$$

<MultiModal>![Fig .1，图中分子以SMILES表示为：CCN(C1=CC=CC=C1)C(=S)C2=NC(=CC=C2)C3=NC(=CC=C3)C(=S)N(CC)C4=CC=CC=C4](/media/uploads/2ec65003-9c0d-4e86-9a72-22c6acd19cf9.png)</MultiModal>

以下选项正确的是：
All other options are incorrect.
The equilibrium constant for the extraction reaction is $\mathrm{1.7×10^{7}}$
The equilibrium constant for the extraction reaction is $\mathrm{8.2×10^{-4}}$.
The equilibrium constant for the extraction reaction is $\mathrm{2.1×10^{3}}$
Extraction rate was 77%
Extraction rate: 23%
Increasing the initial $\mathrm{NO_3^-}$ concentration cannot increase the extraction rate at equilibrium.
Increasing the initial concentration of $\mathrm{L}$ does not increase the extraction rate at equilibrium.
Increasing the initial concentration of $\mathrm{Pd^{2+}}$ can increase the extraction rate at equilibrium.
Extending the reaction time can improve the extraction rate at equilibrium.
其他选项均不正确
萃取反应的平衡常数为$\mathrm{1.7×10^{7}}$
萃取反应的平衡常数为$\mathrm{8.2×10^{-4}}$
萃取反应的平衡常数为$\mathrm{2.1×10^{3}}$
萃取率为77%
萃取率为23%
提高$\mathrm{NO_3^-}$初始浓度无法提高平衡时的萃取率
提高$\mathrm{L}$初始浓度无法提高平衡时的萃取率
提高$\mathrm{Pd^{2+}}$初始浓度可以提高平衡时的萃取率
延长反应时间可以提高平衡时的萃取率
Extraction rate $E=2 \times 1 /(0.30+2 \times 1)=87 \%$

<Checkpoint>[Extraction rate is 87%](1)</Checkpoint>

Options E and F are incorrect.

Convert the initial concentration of $\mathrm{Pd}^{2+}$ to molar concentration:

$c_{0}\left(\mathrm{Pd}^{2+}\right)=1.0 \times 10^{2} \mathrm{mg} / \mathrm{L} / 106.4 \mathrm{~g} / \mathrm{mol}=9.4 \times 10^{-4} \mathrm{~mol} / \mathrm{L}$

Therefore, at equilibrium:

$\mathrm{c}\left(\mathrm{Pd}^{2+}\right)=(1-0.87) \times 9.4 \times 10^{-4} \mathrm{~mol} / \mathrm{L}=1.2 \times 10^{-4} \mathrm{~mol} / \mathrm{L}$

<Checkpoint>[Calculated equilibrium $\mathrm{c}\left(\mathrm{Pd}^{2+}\right)=1.2 \times 10^{-4} \mathrm{~mol} / \mathrm{L}$](0.5)</Checkpoint>

$\mathrm{c}\left(\mathrm{Pd}_{2}\left(\mathrm{NO}_{3}\right)_{4} \mathrm{L}\right)=0.87 \times 9.4 \times 10^{-4} \mathrm{~mol} / \mathrm{L} \div 2=4.1 \times 10^{-4} \mathrm{~mol} / \mathrm{L}$

<Checkpoint>[Calculated equilibrium $\mathrm{c}\left(\mathrm{Pd}_{2}\left(\mathrm{NO}_{3}\right)_{4} \mathrm{L}\right)=4.1 \times 10^{-4} \mathrm{~mol} / \mathrm{L}$](0.5)</Checkpoint>

The concentration of $\mathrm{L}$ at equilibrium is:

$\mathrm{c}(\mathrm{L})=2.0 \times 10^{-3} \mathrm{~mol} / \mathrm{L}-4.1 \times 10^{-4} \mathrm{~mol} / \mathrm{L}=1.6 \times 10^{-3} \mathrm{~mol} / \mathrm{L}$

<Checkpoint>[Calculated equilibrium $\mathrm{c}(\mathrm{L})=1.6 \times 10^{-3} \mathrm{~mol} / \mathrm{L}$](0.5)</Checkpoint>

The nitrate ion is in large excess in the system, and its equilibrium concentration is:

$c\left(\mathrm{NO}_{3}^{-}\right)=c_{0}\left(\mathrm{NO}_{3}^{-}\right)=1.00 \mathrm{~mol} / \mathrm{L}$

Equilibrium constant $K=c\left(\mathrm{Pd}_{2}\left(\mathrm{NO}_{3}\right)_{4} \mathrm{L}\right) / c^{2}\left(\mathrm{Pd}^{2+}\right) c^{4}\left(\mathrm{NO}_{3}{ }^{-}\right) c(\mathrm{~L})=1.7 \times 10^{7}$

<Checkpoint>[Calculated equilibrium constant as $\mathrm{1.7 \times 10^{7}}$](1)</Checkpoint>

Option B is correct, C and D are incorrect

Increasing the extraction rate is actually trying to find a way to shift the equilibrium to the right. Increasing any species on the left side of the equation other than $\mathrm{Pd^{2+}}$ can increase the extraction rate. Options G, H, and I are incorrect. Extending the reaction time after equilibrium will not change the concentration of each species. Option J is incorrect.
萃取率 $E=2 \times 1 /(0.30+2 \times 1)=87 \%$

<Checkpoint>[萃取率为87%](1)</Checkpoint>

选项E、F错误。

将 $\mathrm{Pd}^{2+}$ 的初始浓度转化为摩尔浓度为:

$c_{0}\left(\mathrm{Pd}^{2+}\right)=1.0 \times 10^{2} \mathrm{mg} / \mathrm{L} / 106.4 \mathrm{~g} / \mathrm{mol}=9.4 \times 10^{-4} \mathrm{~mol} / \mathrm{L}$

故平衡时, 有:

$\mathrm{c}\left(\mathrm{Pd}^{2+}\right)=(1-0.87) \times 9.4 \times 10^{-4} \mathrm{~mol} / \mathrm{L}=1.2 \times 10^{-4} \mathrm{~mol} / \mathrm{L}$

<Checkpoint>[算得平衡时$\mathrm{c}\left(\mathrm{Pd}^{2+}\right)=1.2 \times 10^{-4} \mathrm{~mol} / \mathrm{L}$](0.5)</Checkpoint>

$\mathrm{c}\left(\mathrm{Pd}_{2}\left(\mathrm{NO}_{3}\right)_{4} \mathrm{L}\right)=0.87 \times 9.4 \times 10^{-4} \mathrm{~mol} / \mathrm{L} \div 2=4.1 \times 10^{-4} \mathrm{~mol} / \mathrm{L}$

<Checkpoint>[算得平衡时$\mathrm{c}\left(\mathrm{Pd}_{2}\left(\mathrm{NO}_{3}\right)_{4} \mathrm{L}\right)=4.1 \times 10^{-4} \mathrm{~mol} / \mathrm{L}$](0.5)</Checkpoint>

平衡时 $\mathrm{L}$ 的浓度为:

$\mathrm{c}(\mathrm{L})=2.0 \times 10^{-3} \mathrm{~mol} / \mathrm{L}-4.1 \times 10^{-4} \mathrm{~mol} / \mathrm{L}=1.6 \times 10^{-3} \mathrm{~mol} / \mathrm{L}$ 

<Checkpoint>[算得平衡时$\mathrm{c}(\mathrm{L})=1.6 \times 10^{-3} \mathrm{~mol} / \mathrm{L}$](0.5)</Checkpoint>

体系中硝酸根离子大过量, 其平衡浓度为:

$c\left(\mathrm{NO}_{3}^{-}\right)=c_{0}\left(\mathrm{NO}_{3}^{-}\right)=1.00 \mathrm{~mol} / \mathrm{L}$

平衡常数 $K=c\left(\mathrm{Pd}_{2}\left(\mathrm{NO}_{3}\right)_{4} \mathrm{~L}\right) / c^{2}\left(\mathrm{Pd}^{2+}\right) c^{4}\left(\mathrm{NO}_{3}{ }^{-}\right) c(\mathrm{~L})=1.7 \times 10^{7}$

<Checkpoint>[算得平衡常数为$\mathrm{1.7 \times 10^{7}}$](1)</Checkpoint>

选项B正确，C、D错误

提高萃取率实际上就是想办法让平衡右移，提高$\mathrm{Pd^{2+}}$以外的方程左侧物种均可提高萃取率，选项G、H、I错误。平衡后延长反应时间不会改变各物种浓度，选项J错误。