Perdeuterated pyridine ($py-d_5$, denoted directly as $py$ or $Py$ in responses) can catalyze the pyrolysis of the vanadium compound $[\mathbf{L}(O^iPr)V=O](\mathbf{A})$ (where $\mathbf{L}^{2-}$ represents the 2,6-pyridinedicarboxylate anion). It is known that this reaction proceeds quantitatively and yields the following products (unbalanced):

$ Py + \mathbf{A} \to \mathbf{P} + ^iPrOH + CH_3COCH_3$

The structures of $\mathbf{A}$ and $\mathbf{P}$ are as follows:

$\mathbf{A}$:

<MultiModal>![O=[V]12(OC(C)C)[N]3=C(C(O2)=O)C=CC=C3C(O1)=O
](/media/uploads/11443316-06e5-419e-b532-28f1331dd044.png)</MultiModal>

$\mathbf{P}$:

<MultiModal>![O=C(O1)C2=CC=CC(C(O3)=O)=[N]2[V]13(C4=NC=CC=C4)(C5=NC=CC=C5)OC(C)C
](/media/uploads/ded414c2-f784-4a0a-b25d-1fd877b354e0.png)</MultiModal>

The experimental team proposed three possible reaction pathways for the pyrolysis:

$\mathbf{Path I}$

$Py + \mathbf{A} \xrightleftharpoons[]{K_{eq}} \mathbf{B} \xrightarrow[]{k_1} \mathbf{C} \xrightarrow[]{fast} \mathbf{P}$

$\mathbf{Path II}$

$Py + \mathbf{A} \xrightleftharpoons[]{K_{eq}} \mathbf{B} \xrightarrow[Py]{k_1} \mathbf{C} \xrightarrow[]{fast} \mathbf{P}$

$\mathbf{Path III}$

Both $ Py + \mathbf{A} \xrightleftharpoons[]{K_{eq}} \mathbf{B} \xrightarrow[Py]{k_1} \mathbf{C} \xrightarrow[]{fast} \mathbf{P}$

And $ \mathbf{A} \xrightarrow[Py]{k_2} \mathbf{D} \xrightarrow[]{fast} \mathbf{P}$

The structures of $\mathbf{B}$, $\mathbf{C}$, and $\mathbf{D}$ are as follows:

$\mathbf{B}$:

<MultiModal>![O=[V]12(OC(C)C)(C3=NC=CC=C3)[N]4=C(C(O2)=O)C=CC=C4C(O1)=O
](/media/uploads/16998d1b-570d-4786-ad95-07e35abc2aa7.png)</MultiModal>

$\mathbf{C}$:

<MultiModal>![O[V]12(/O=C(C)/C)(C3=NC=CC=C3)[N]4=C(C(O2)=O)C=CC=C4C(O1)=O
](/media/uploads/78a937cf-f225-423c-bc4e-d1ac6aecb672.png)</MultiModal>

$\mathbf{D}$:

<MultiModal>![O[V]12(/O=C(C)/C)[N]3=C(C(O2)=O)C=CC=C3C(O1)=O
](/media/uploads/8c53d21a-4ec2-4c53-a377-929c66c4a33e.png)</MultiModal>

It is known that in $Path II$ and $Path III$, one molecule of $py-d_5$ participates in the proton transfer process during the formation of $\mathbf{C}$ or $\mathbf{D}$, while $\mathbf{C}$ and $\mathbf{D}$ are highly reactive, so their concentrations can be neglected.

Experimental measurements at $340K$ show the following variation curve of $k_{obs}$ with $[py]$:

<MultiModal>![This image is a scatter plot of the apparent rate constant $k_{obs}$ versus the concentration of perdeuterated pyridine $[pyridine]$. The x-axis represents $[pyridine]$ in units of $M$, with a range of $0-15$, and the y-axis represents $k_{obs}$ in units of $10^{-4}s$, with a range of $0-6$. The estimated coordinates of the data points are $(0,0), (0.33,0.59), (0.49,0.75), (0.98,1.47), (2.95,1.98), (5.93,2.98), (8.93,4.07), (12.40,4.96)$.](/media/uploads/5bba2295-5120-45b3-abc9-f22717a0a331.png)</MultiModal>

Based on the above information, which of the following statements are **incorrect**?

1. For all three reaction pathways, the reaction order with respect to $[\mathbf{A}]$ is 1.

2. $\mathbf{Path II}$ has a reaction order of 2 with respect to $[py]$.

3. When $[py] \gg 1/K_{eq}$, the variation of $k_{obs}$ with $[py]$ at $340K$ best fits $\mathbf{Path I}$.
五氘代吡啶（$py-d_5$，作答时直接用$py$或$Py$表示）可催化钒化合物$[\mathbf{L}(O^iPr)V=O](\mathbf{A})$的热解（其中$\mathbf{L}^{2-}$表示2,6-吡啶二甲酸根）。已知该反应可定量进行并生成如下产物（未配平）：

$ Py + \mathbf{A} \to \mathbf{P} + ^iPrOH + CH_3COCH_3$

$\mathbf{A}$和$\mathbf{P}$的结构分别如下：

$\mathbf{A}$：

<MultiModal>![O=[V]12(OC(C)C)[N]3=C(C(O2)=O)C=CC=C3C(O1)=O
](/media/uploads/11443316-06e5-419e-b532-28f1331dd044.png)</MultiModal>

$\mathbf{P}$：

<MultiModal>![O=C(O1)C2=CC=CC(C(O3)=O)=[N]2[V]13(C4=NC=CC=C4)(C5=NC=CC=C5)OC(C)C
](/media/uploads/ded414c2-f784-4a0a-b25d-1fd877b354e0.png)</MultiModal>

实验人员对热解反应提出了三种可能的反应历程：

$\mathbf{Path I}$

$Py + \mathbf{A} \xrightleftharpoons[]{K_{eq}} \mathbf{B} \xrightarrow[]{k_1} \mathbf{C} \xrightarrow[]{fast} \mathbf{P}$

$\mathbf{Path II}$

$Py + \mathbf{A} \xrightleftharpoons[]{K_{eq}} \mathbf{B} \xrightarrow[Py]{k_1} \mathbf{C} \xrightarrow[]{fast} \mathbf{P}$

$\mathbf{Path III}$

既有 $ Py + \mathbf{A} \xrightleftharpoons[]{K_{eq}} \mathbf{B} \xrightarrow[Py]{k_1} \mathbf{C} \xrightarrow[]{fast} \mathbf{P}$

又有 $ \mathbf{A} \xrightarrow[Py]{k_2} \mathbf{D} \xrightarrow[]{fast} \mathbf{P}$

$\mathbf{B}$、$\mathbf{C}$和$\mathbf{D}$的结构分别如下：

$\mathbf{B}$：

<MultiModal>![O=[V]12(OC(C)C)(C3=NC=CC=C3)[N]4=C(C(O2)=O)C=CC=C4C(O1)=O
](/media/uploads/16998d1b-570d-4786-ad95-07e35abc2aa7.png)</MultiModal>

$\mathbf{C}$：

<MultiModal>![O[V]12(/O=C(C)/C)(C3=NC=CC=C3)[N]4=C(C(O2)=O)C=CC=C4C(O1)=O
](/media/uploads/78a937cf-f225-423c-bc4e-d1ac6aecb672.png)</MultiModal>

$\mathbf{D}$：

<MultiModal>![O[V]12(/O=C(C)/C)[N]3=C(C(O2)=O)C=CC=C3C(O1)=O
](/media/uploads/8c53d21a-4ec2-4c53-a377-929c66c4a33e.png)</MultiModal>

已知在$Path II$与$Path III$中生成$\mathbf{C}$或$\mathbf{D}$一步有一分子$py-d_5$参与质子转移过程，而$\mathbf{C}$与$\mathbf{D}$十分活泼，故其浓度可以忽略不计。

实验测得在$340K$下，$k_{obs}$随$[py]$的变化曲线如下：

<MultiModal>![这张图片为表观速率常数$k_{obs}$与五氘代吡啶浓度$[pyridine]$的散点图。横轴是以$M$为单位的$[pyridine]$，坐标范围是$0-15$，纵轴是以$10^{-4}s$为单位的$k_{obs}$，坐标范围是$0-6$。各散点的估读坐标为$(0,0), (0.33,0.59), (0.49,0.75), (0.98,1.47), (2.95,1.98), (5.93,2.98), (8.93,4.07), (12.40,4.96)。$](/media/uploads/5bba2295-5120-45b3-abc9-f22717a0a331.png)</MultiModal>

根据以上信息，以下说法中**错误**的有哪些？

1. 对于三种反应路径，反应对$[\mathbf{A}]$的级数均为1。

2. $\mathbf{Path II}$对$[py]$的级数为2。

3. 当$[py] \gg 1/K_{eq}$时，最符合$340K$下$k_{obs}$随$[py]$变化关系的是$\mathbf{Path I}$
1
2
3
1,2
2,3
1,3
1,2,3
All three statements are correct
1
2
3
1,2
2,3
1,3
1,2,3
3种说法均正确
Let $C_{V(V)}$ denote the sum of the concentrations of all species containing only $V(V)$.

For $Path I$, the rate equation of the reaction can be expressed as:

$r=k_1[\mathbf{B}]$

By material balance:

$C_{V(V)}=[\mathbf{B}]+[\mathbf{A}]$

Additionally, the equilibrium relationship is:

$K_{eq}=[\mathbf{B}]/[[\mathbf{A}][py]]$

Thus, we have:

$[\mathbf{B}]=K_{eq}[py]C_{V(V)}/(K_{eq}[py]+1)$

$r=k_1K_{eq}[py]C_{V(V)}/(K_{eq}[py]+1)$

For $Path II$, the rate equation of the reaction can be expressed as:

$r=k_1[py][\mathbf{B}]$

The material balance and equilibrium relationship for $Path II$ are the same as for $Path I$, so we have:

$r=k_1K_{eq}[py]^2C_{V(V)}/(K_{eq}[py]+1)$

For $Path III$, the rate equation of the reaction can be expressed as:

$r=k_1[py][\mathbf{B}]+k_2[py][\mathbf{A}]$

From the material balance of $\mathbf{A}$ and the $K_{eq}$ relationship, we obtain:

$[\mathbf{B}]=K_{eq}[py]C_{V(V)}/(K_{eq}[py]+1)$

$[\mathbf{A}]=C_{V(V)}/(K_{eq}[py]+1)$

<Checkpoint>[$[\mathbf{B}]=K_{eq}[py]C_{V(V)}/(K_{eq}[py]+1)$](0.5)</Checkpoint>

<Checkpoint>[$[\mathbf{A}]=C_{V(V)}/(K_{eq}[py]+1)$](0.5)</Checkpoint>

$r=k_1K_{eq}[py]^2C_{V(V)}/(K_{eq}[py]+1)+k_2[py]C_{V(V)}/(K_{eq}[py]+1)$

Now, let’s analyze each statement:

1. The three reaction paths can be expressed in terms of $[\mathbf{A}]$ as:

$r_{Path I}=k_1K_{eq}[py][\mathbf{A}]$

$r_{Path II}=k_1K_{eq}[py]^2[\mathbf{A}]$

$r_{Path III}=(k_1K_{eq}[py]^2+k_2[py])[\mathbf{A}]$

Thus, for all three reaction paths, the reaction order with respect to $[\mathbf{A}]$ is 1. This is correct.

<Checkpoint>[For all three reaction paths, the reaction order with respect to $[\mathbf{A}]$ is 1. Statement 1 is correct.](1)</Checkpoint>

2. The rate equation for $Path II$ is:

$r=k_1K_{eq}[py]^2C_{V(V)}/(K_{eq}[py]+1)$

<Checkpoint>[$r=k_1K_{eq}[py]^2C_{V(V)}/(K_{eq}[py]+1)$](1)</Checkpoint>

It is evident that $[py]$ does not follow a simple integer-order relationship. Thus, the statement is incorrect.

<Checkpoint>[$[py]$ does not follow a simple integer-order relationship. Statement 2 is incorrect.](1)</Checkpoint>

3. When $[py] \gg 1/K_{eq}$, the apparent rate constants for each reaction path are:

$Path I: k_{obs}=k_1$

$Path II: k_{obs}=k_1[py]$

$Path III: k_{obs}=k_1[py]+k_2/K_{eq}$

<Checkpoint>[$Path III: k_{obs}=k_1[py]+k_2/K_{eq}$](0.5)</Checkpoint>

At high concentrations of $py-d_5$, plotting the tangent line of $k_{obs}$ reveals that $k_{obs}$ has a vertical intercept and exhibits a positive correlation with $[py]$ (approximately linear). Therefore, $Path III$ is the most plausible reaction pathway, and the statement is incorrect.

<Checkpoint>[At high concentrations of $py-d_5$, plotting the tangent line of $k_{obs}$ reveals that $k_{obs}$ has a vertical intercept.](1)</Checkpoint>

<Checkpoint>[$Path III$ is the most plausible, so statement 3 is incorrect.](0.5)</Checkpoint>

Thus, the correct answer is E (2,3).
以$C_{V(V)}$表示仅含$V(V)$的物种浓度之和

对于$Path I$，反应的速率方程可以表示为：

$r=k_1[\mathbf{B}]$

由物料守恒：

$C_{V(V)}=[\mathbf{B}]+[\mathbf{A}]$

又有平衡关系：

$K_{eq}=[\mathbf{B}]/[[\mathbf{A}][py]]$

故有：

$[\mathbf{B}]=K_{eq}[py]C_{V(V)}/(K_{eq}[py]+1)$

$r=k_1K_{eq}[py]C_{V(V)}/(K_{eq}[py]+1)$

对于$Path II$，反应的速率方程可以表示为：

$r=k_1[py][\mathbf{B}]$

$Path II$的物料守恒和平衡关系与$Path I$相同，故有：

$r=k_1K_{eq}[py]^2C_{V(V)}/(K_{eq}[py]+1)$

对于$Path III$，反应的速率方程可以表示为：

$r=k_1[py][\mathbf{B}]+k_2[py][\mathbf{A}]$

由$\mathbf{A}$的物料守恒以及$K_{eq}$的关系式可得：

$[\mathbf{B}]=K_{eq}[py]C_{V(V)}/(K_{eq}[py]+1)$

$[\mathbf{A}]=C_{V(V)}/(K_{eq}[py]+1)$

<Checkpoint>[$[\mathbf{B}]=K_{eq}[py]C_{V(V)}/(K_{eq}[py]+1)$](0.5)</Checkpoint>

<Checkpoint>[$[\mathbf{A}]=C_{V(V)}/(K_{eq}[py]+1)$](0.5)</Checkpoint>

$r=k_1K_{eq}[py]^2C_{V(V)}/(K_{eq}[py]+1)+k_2[py]C_{V(V)}/(K_{eq}[py]+1)$

下面分析各说法：

1. 三种反应路径分别可以用$[\mathbf{A}]$表示成：

$r_{Path I}=k_1K_{eq}[py][\mathbf{A}]$

$r_{Path II}=k_1K_{eq}[py]^2[\mathbf{A}]$

$r_{Path III}=(k_1K_{eq}[py]^2+k_2[py])[\mathbf{A}]$

即对于三种反应路径，反应对$[\mathbf{A}]$的级数均为1，正确。

<Checkpoint>[对于三种反应路径，反应对$[\mathbf{A}]$的级数均为1，说法1正确](1)</Checkpoint>

2. $Path II$的速率方程为

$r=k_1K_{eq}[py]^2C_{V(V)}/(K_{eq}[py]+1)$

<Checkpoint>[$r=k_1K_{eq}[py]^2C_{V(V)}/(K_{eq}[py]+1)$](1)</Checkpoint>

可见$[py]$并非简单的整数级数关系，说法错误。

<Checkpoint>[$[py]$并非简单的整数级数关系，说法2错误](1)</Checkpoint>

3. 当$[py] \gg 1/K_{eq}$时，各反应路径的表观速率常数为：

$Path I: k_{obs}=k_1$

$Path II: k_{obs}=k_1[py]$

$Path III: k_{obs}=k_1[py]+k_2/K_{eq}$

<Checkpoint>[$Path III: k_{obs}=k_1[py]+k_2/K_{eq}$](0.5)</Checkpoint>

高浓度$py-d_5$下作$k_{obs}$的切线会发现$k_{obs}$具有纵轴截距，且$k_{obs}$与$[py]$呈正相关关系（近似直线），因此$Path III$为最合理的反应历程，说法错误。

<Checkpoint>[高浓度$py-d_5$下作$k_{obs}$的切线会发现$k_{obs}$具有纵轴截距](1)</Checkpoint>

<Checkpoint>[$Path III$最合适，说法3错误。](0.5)</Checkpoint>

因此正确答案选E（2,3）。