The following diagram shows a kinetic model of a catalytic reaction:

<MultiModal>![The image contains nodes labeled with multiple letters: **A**, **B**, **C**, **D**, **E**, **F**, **G**, **S**. Node S has the molecular formula [H][Rh]([P](C1=CC=CC=C1)(C2=CC=CC=C2)C3=CC=CC=C3)([P](C4=CC=CC=C4)(C5=CC=CC=C5)C6=CC=CC=C6)[C]=O, which is the only annotation for this node. There is a two-way arrow between nodes **A** and **S**, labeled $K_1$ on the **S** to **A** side, and $CO$ on the **A** to **S** side. There is a two-way arrow between nodes **A** and **B**, labeled C=C[R] on the **A** to **B** side, and $K_2$ on the **B** to **A** side; there is a two-way arrow between nodes **B** and **C**, labeled $K_3$ on the **B** to **C** side; there is also a two-way arrow between nodes **C** and **D**, labeled $K_4$ on the **C** to **D** side. There is also a two-way arrow between nodes **D** and **E**, pointing from **D** to **E** and labeled "$H_2$, $k_1$", and the reverse arrow from **E** to **D** is labeled "$k_{-1}$". There is a one-way arrow between nodes **E** and **S**, pointing from **E** to **S** and labeled "$k_2$ fast". There is a two-way arrow between **F** and **D**, labeled $CO$ on the **D** to **F** side, and $K_5$ on the **F** to **D** side; and there is a two-way arrow between **G** and **S**, labeled $\mathrm { P P h }_3$ on the **S** to **G** side, and $K_6$ on the **G** to **F** side.](/media/uploads/66eb0ee8-0ded-4055-a49e-67445b37e8cc.png)</MultiModal>

The respective conversion processes in the catalytic cycle are: **S** reversibly combines with $\mathrm{CO}$ to generate **A**, and the equilibrium constant of the reaction is $K_1$; **A** reacts with olefin ($\mathrm{R-CH=CH_2}$, denoted as $\mathrm{Aly}$) to obtain **B**, and the equilibrium constant is $K_2$; **B** can be isomerized to **C**, and **C** can be isomerized to **D**, and the equilibrium constants for isomerization are $K_3$ and $K_4$, respectively; **D** reversibly combines with $\mathrm{CO}$ to obtain **F**, and the equilibrium constant is $K_5$. On the other hand, **D** can also react with $\mathrm{H_2}$ to obtain **E**, the forward reaction rate constant is $k_1$, and the reverse reaction rate constant is $k_{-1}$; **E** rapidly eliminates a molecule of product **P** (not shown in the figure), and simultaneously obtains **S**, the rate constant for this step is $k_2$; **S** can also reversibly react with $\mathrm { P P h }_3$ to obtain catalytically inactive **G**, and its equilibrium constant is $K_ 6$.

It is known that the concentrations of the three species **A**, **B**, and **C** in the catalytic cycle are negligible relative to other $\mathrm{Rh}$-containing species. The following options are descriptions of the reaction rate ($r=d[\mathbf{P}]/dt$). Please select the correct option by combining reasonable approximations and derivations.
下图展示了一个催化反应的动力学模型：

<MultiModal>![该图像包含多个字母标记的节点：**A**、**B**、**C**、**D**、**E**、**F**、**G**、**S**。节点S处有分子式[H][Rh]([P](C1=CC=CC=C1)(C2=CC=CC=C2)C3=CC=CC=C3)([P](C4=CC=CC=C4)(C5=CC=CC=C5)C6=CC=CC=C6)[C]=O，这是该节点的唯一标注。节点**A**和**S**之间有一条双向箭头，**S**至**A**侧标注$K_1$，**A**至**S**侧标注$CO$。节点**A**和**B**之间有一条双向箭头，**A**至**B**侧标注C=C[R]，**B**至**A**侧标注$K_2$；**B**和**C**之间有一条双向箭头，**B**至**C**侧标注$K_3$；**C**和**D**之间同样有一条双向箭头，**C**至**D**侧标注$K_4$。节点**D**与**E**之间也有一条双向箭头，箭头方向从**D**指向**E**，并标注为“$H_2$, $k_1$”，反向的箭头从**E**指向**D**，标注为“$k_{-1}$”。节点**E**与**S**之间有一条单向箭头，从**E**指向**S**的方向标注为“$k_ 2$ fast”。**F**和**D**之间有一条双向箭头，**D**至**F**侧标注$CO$，**F**至**D**侧标注$K_5$；而**G**和**S**之间有一条双向箭头，**S**至**G**侧标注$\mathrm { P P h }_3$，**G**至**F**侧标注$K_6$。](/media/uploads/66eb0ee8-0ded-4055-a49e-67445b37e8cc.png)</MultiModal>

催化循环中各转化过程分别为：**S** 与 $\mathrm{CO}$ 可逆结合生成 **A**，反应的平衡常数为 $K_1$；**A** 与烯烃（$\mathrm{R-CH=CH_2}$，记作 $\mathrm{Aly}$）反应得到 **B**，平衡常数为 $K_2$；**B** 可异构化为 **C**，**C** 又可异构化为 **D**，异构化的平衡常数分别为 $K_3$ 和 $K_4$；**D** 与 $\mathrm{CO}$ 可逆结合得到 **F**，平衡常数为 $K_5$，另一方面，**D** 也可与 $\mathrm{H_2}$ 反应得到 **E**，正反应速率常数为 $k_1$，逆反应的速率常数为 $k_{-1}$；**E** 快速脱除一分子产物 **P** （在图中未示出），同时得到 **S**，此步速率常数为 $k_ 2$；**S** 还会与 $\mathrm { P P h }_3$ 可逆反应得到无催化活性的 **G**，其平衡常数为 $K_ 6$。

已知催化循环过程中 **A**、**B**、**C** 三个物种的浓度相对其他含 $\mathrm{Rh}$ 物种可忽略。以下选项是对反应速率（$r=d[\mathbf{P}]/dt$）的描述，请结合合理的近似推导，选择正确选项。
The reaction rate's order with respect to the total concentration $[\mathrm{Rh}]$ of $\mathrm{Rh}$-containing species is related to the concentrations of other species.
When the partial pressure of $\mathrm{CO}$ is sufficiently high, the reaction rate is -1 order with respect to $[\mathrm{CO}]$, 1 order with respect to $[\mathrm{H}_2]$, and depends on the values of $K_1$, $K_2$, $K_3$, and $K_4$.
When the partial pressure of $\mathrm{H}_2$ is sufficiently high, the reaction rate is 0th order with respect to $[\mathrm{CO}]$, 0th order with respect to $[\mathrm{H}_2]$, and independent of the value of $k_1$, but dependent on the value of $k_2$.
When the partial pressure of $\mathrm{H}_2$ is sufficiently low, the reaction rate is first order with respect to $[\mathrm{H}_2]$, and among all the equilibrium constants and rate constants mentioned in the problem, there exists a value that is independent of the reaction rate.
When $[\mathrm{PPh}_3]$ is sufficiently large, the reaction order with respect to $[\mathrm{PPh}_3]$ is -1, and it is independent of $[\mathrm{H}_2]$ and $[\mathrm{CO}]$.
As the partial pressure of $\mathrm{Aly}$ continuously increases, the reaction rate will reach a maximum value that is related to $[\mathrm{CO}]$, $[\mathrm{Rh}]$, $[\mathrm{H_2}]$, and $[\mathrm{PPh}_3]$.
反应速率关于含$\mathrm{Rh}$物种总浓度$[\mathrm{Rh}]$的级数与其他物种浓度有关。
当$\mathrm{CO}$分压足够大时，反应速率关于$[\mathrm{CO}]$的级数是-1级，关于$[\mathrm{H}_2]$的级数是1级，且与$K_1$、$K_2$、$K_3$、$K_4$的值有关。
当$\mathrm{H}_2$分压足够大时，反应速率关于$[\mathrm{CO}]$的级数是0级，关于$[\mathrm{H}_2]$的级数是0级，且与$k_1$的值无关，与$k_2$的值有关。
当$\mathrm{H}_2$分压足够小时，反应速率关于$[\mathrm{H}_2]$的级数是1级，且在题目中提到的所有平衡常数和速率常数中，存在与反应速率无关的值。
当$[\mathrm{PPh}_3]$足够大时，反应速率关于$[\mathrm{PPh}_3]$的级数为-1级，且与$[\mathrm{H}_2]$和$[\mathrm{CO}]$无关。
随着$\mathrm{Aly}$分压不断增加，反应速率会达到一个与$[\mathrm{CO}]$、$[\mathrm{Rh}]$、$[\mathrm{H_2}]$、$[\mathrm{PPh}_3]$均相关的最大值。
From the figure, we can obtain the equilibrium equations and reaction expressions including stoichiometric numbers:
$$
\mathbf{S}+\mathrm{CO} \rightleftharpoons\mathbf{A}
$$
$$
\mathbf{A}+\mathrm{Aly} \rightleftharpoons \mathbf{B} \rightleftharpoons \mathbf{C} \rightleftharpoons \mathbf{D}
$$
$$
\mathbf{D}+\mathrm{H_2} \rightleftharpoons \mathbf{E} \to \mathbf{E}+\mathbf{S}
$$
$$
\mathbf{D}+\mathrm{CO} \rightleftharpoons \mathbf{F}
$$
$$
\mathbf{S}+\mathrm{{PPh}_3} \rightleftharpoons \mathbf{G}
$$

<Checkpoint>[Equilibrium exists: $ \mathbf{S}+\mathrm{CO} \rightleftharpoons\mathbf{A} $](0.4)</Checkpoint>

<Checkpoint>[Equilibrium exists: $ \mathbf{A}+\mathrm{Aly} \rightleftharpoons \mathbf{B} \rightleftharpoons \mathbf{C} \rightleftharpoons \mathbf{D} $](0.4)</Checkpoint>

<Checkpoint>[Equilibrium exists: $ \mathbf{D}+\mathrm{H_2} \rightleftharpoons \mathbf{E} \to \mathbf{E}+\mathbf{S} $](0.4)</Checkpoint>

<Checkpoint>[Equilibrium exists: $ \mathbf{D}+\mathrm{CO} \rightleftharpoons \mathbf{F} $](0.4)</Checkpoint>

<Checkpoint>[Equilibrium exists: $ \mathbf{S}+\mathrm{{PPh}_3} \rightleftharpoons \mathbf{G} $](0.4)</Checkpoint>

Based on the speculated structure and reaction process, and according to the steady-state hypothesis, give the expressions of [**D**], [**F**], and [**G**] with respect to [**S**]:

$[\mathbf{D}]=K_{1} K_{2} K_{3} K_{4}[\mathbf{S}][\mathrm{CO}][\mathrm{Aly}]$

$[\mathbf{F}]=K_{1} K_{2} K_{3} K_{4} K_{5}[\mathbf{S}][\mathrm{CO}]^{2}[\mathrm{Aly}] $

$[\mathbf{G}]=K_{6}[\mathbf{S}][\mathrm{PPh}_3]$

Where $[\mathrm{Aly}]$ is the concentration of the alkene species; $[\mathrm{PPh}_3]$ is the concentration of the triphenylphosphine species.

<Checkpoint>[The expression of [**D**] with respect to [**S**] is $[\mathbf{D}]=K_{1} K_{2} K_{3} K_{4}[\mathbf{S}][\mathrm{CO}][\mathrm{Aly}]$](1)</Checkpoint>

<Checkpoint>[The expression of [**F**] with respect to [**S**] is $[\mathbf{F}]=K_{1} K_{2} K_{3} K_{4} K_{5}[\mathbf{S}][\mathrm{CO}]^{2}[\mathrm{Aly}] $](1)</Checkpoint>

<Checkpoint>[The expression of [**G**] with respect to [**S**] is $[\mathbf{G}]=K_{6}[\mathbf{S}][\mathrm{PPh}_3]$](1)</Checkpoint>

Applying steady-state approximation to [**E**]:
$$
d[\mathbf{E}]/dt=k_1[\mathbf{D}][\mathrm{H}_2]-
(k_{-1}+k_2)[\mathbf{E}]=0
$$

Then:
$$
[\mathbf{E}]=k_1[\mathbf{D}][\mathrm{H}_2]/(k_{-1}+k_2)
$$

<Checkpoint>[Applying steady-state approximation to [**E**]: $[\mathbf{E}]=k_1[\mathbf{D}][\mathrm{H}_2]/(k_{-1}+k_2)$](1)</Checkpoint>

Substituting into the material balance relationship:
$$
[\mathrm{Rh}]=[\mathbf{S}]+[\mathbf{D}]+[\mathbf{E}]+[\mathbf{F}]+[\mathbf{G}]
$$

The expression for $[\mathbf{S}]$ is obtained as:
$$
[\mathbf{S}]=[\mathrm{Rh}]/\lbrace1+K_6[\mathrm{PPh}_3]+[1+K_5[\mathrm{CO}]+k_1[\mathrm{H}_2]/(k_{-1}+k_2)]K_1K_2K_3K_4[\mathrm{CO}][\mathrm{Aly}]\rbrace
$$

<Checkpoint>[The expression for $[\mathbf{S}]$ is $ [\mathbf{S}]=[\mathrm{Rh}]/\lbrace1+K_6[\mathrm{PPh}_3]+[1+K_5[\mathrm{CO}]+k_1[\mathrm{H}_2]/(k_{-1}+k_2)]K_1K_2K_3K_4[\mathrm{CO}][\mathrm{Aly}]\rbrace $](1)</Checkpoint>

Substituting into the reaction rate expression:
$$
r=d[\mathbf{P}]/dt=k_2[\mathbf{E}]=k_1k_2[\mathbf{D}][\mathrm{H}_2]/(k_{-1}+k_2)=k_1k_2K_{1} K_{2} K_{3} K_{4}[\mathbf{S}][\mathrm{CO}][\mathrm{Aly}][\mathrm{H}_2]/(k_{-1}+k_2)
$$

Substituting $[\mathbf{S}]$, the reaction rate expression is obtained:
$$
r=k_1k_2K_{1} K_{2} K_{3} K_{4}[\mathrm{Rh}][\mathrm{CO}][\mathrm{Aly}][\mathrm{H}_2]/(k_{-1}+k_2)\lbrace1+K_6[\mathrm{PPh}_3]+[1+K_5[\mathrm{CO}]+k_1[\mathrm{H}_2]/(k_{-1}+k_2)]K_1K_2K_3K_4[\mathrm{CO}][\mathrm{Aly}]\rbrace
$$

<Checkpoint>[Reaction rate expression: $ r=k_1k_2K_{1} K_{2} K_{3} K_{4}[\mathrm{Rh}][\mathrm{CO}][\mathrm{Aly}][\mathrm{H}_2]/(k_{-1}+k_2)\lbrace1+K_6[\mathrm{PPh}_3]+[1+K_5[\mathrm{CO}]+k_1[\mathrm{H}_2]/(k_{-1}+k_2)]K_1K_2K_3K_4[\mathrm{CO}][\mathrm{Aly}]\rbrace $](1)</Checkpoint>

Judgment based on the reaction rate expression:

Regardless of the concentration values of other species, the exponent of $[\mathrm{Rh}]$ is always 1, so the reaction order is independent of other species, therefore option A is incorrect;

<Checkpoint>[The exponent of $[\mathrm{Rh}]$ is always 1, so the reaction order is independent of other species](1)</Checkpoint>

When $[\mathrm{CO}]$ is large enough, we have:
$$
r=k_1k_2[\mathrm{Rh}][\mathrm{H}_2]/K_5(k_{-1}+k_2)[\mathrm{CO}]
$$

<Checkpoint>[When $[\mathrm{CO}]$ is large enough, we have: $ r=k_1k_2[\mathrm{Rh}][\mathrm{H}_2]/K_5(k_{-1}+k_2)[\mathrm{CO}] $](1)</Checkpoint>

The reaction rate is independent of the values of $K_1$, $K_2$, $K_3$, and $K_4$, so option B is incorrect;

When $[\mathrm{H_2}]$ is large enough, we have:
$$
r=k_2[\mathrm{Rh}]
$$

<Checkpoint>[When $[\mathrm{H_2}]$ is large enough, we have: $ r=k_2[\mathrm{Rh}] $](1)</Checkpoint>

Therefore, option C is correct;

When $[\mathrm{H_2}]$ is small enough, we have:
$$
r=k_1k_2K_{1} K_{2} K_{3} K_{4}[\mathrm{Rh}][\mathrm{CO}][\mathrm{Aly}][\mathrm{H}_2]/(k_{-1}+k_2)\lbrace1+K_6[\mathrm{PPh}_3]+[1+K_5[\mathrm{CO}]]K_1K_2K_3K_4[\mathrm{CO}][\mathrm{Aly}]\rbrace
$$

<Checkpoint>[When $[\mathrm{H_2}]$ is small enough, we have: $$ r=k_1k_2K_{1} K_{2} K_{3} K_{4}[\mathrm{Rh}][\mathrm{CO}][\mathrm{Aly}][\mathrm{H}_2]/(k_{-1}+k_2)\lbrace1+K_6[\mathrm{PPh}_3]+[1+K_5[\mathrm{CO}]]K_1K_2K_3K_4[\mathrm{CO}][\mathrm{Aly}]\rbrace $$](1)</Checkpoint>

All rate constants and equilibrium constants in the question are related to the reaction rate, so option D is incorrect.

When $[\mathrm{PPh}_3]$ is large enough, we have:
$$
r=k_{1}k_{2}K_{1}K_{2}K_{3}K_{4}[\mathrm{Rh}][\mathrm{CO}][\mathrm{Aly}][\mathrm{H}_2]/(k_{-1}+k_{2})K_{6}[\mathrm{PPh}_3]
$$

<Checkpoint>[When $[\mathrm{PPh}_3]$ is large enough, we have: $$ r=k_{1}k_{2}K_{1}K_{2}K_{3}K_{4}[\mathrm{Rh}][\mathrm{CO}][\mathrm{Aly}][\mathrm{H}_2]/(k_{-1}+k_{2})K_{6}[\mathrm{PPh}_3] $$](1)</Checkpoint>

The rate is related to $[\mathrm{H}_2]$ and $[\mathrm{CO}]$, so option E is incorrect.

When $[\mathrm{Aly}]$ is large enough, we have:
$$
r=k_1k_2[\mathrm{Rh}][\mathrm{H}_2]/(k_{-1}+k_2)[1+K_5[\mathrm{CO}]+k_1[\mathrm{H}_2]/(k_{-1}+k_2)]
$$

<Checkpoint>[When $[\mathrm{Aly}]$ is large enough, we have: $$ r=k_1k_2[\mathrm{Rh}][\mathrm{H}_2]/(k_{-1}+k_2)[1+K_5[\mathrm{CO}]+k_1[\mathrm{H}_2]/(k_{-1}+k_2)] $$](1)</Checkpoint>

The rate is independent of $[\mathrm{PPh}_3]$, option F is incorrect.

In summary, option C should be selected.
由图，可以得到含化学计量数的平衡式和反应表达式：
$$
\mathbf{S}+\mathrm{CO} \rightleftharpoons\mathbf{A}
$$
$$
\mathbf{A}+\mathrm{Aly} \rightleftharpoons \mathbf{B} \rightleftharpoons \mathbf{C} \rightleftharpoons \mathbf{D}
$$
$$
\mathbf{D}+\mathrm{H_2} \rightleftharpoons \mathbf{E} \to \mathbf{E}+\mathbf{S}
$$
$$
\mathbf{D}+\mathrm{CO} \rightleftharpoons \mathbf{F}
$$
$$
\mathbf{S}+\mathrm{{PPh}_3} \rightleftharpoons \mathbf{G}
$$

<Checkpoint>[存在平衡$ \mathbf{S}+\mathrm{CO} \rightleftharpoons\mathbf{A} $](0.4)</Checkpoint>

<Checkpoint>[存在平衡$ \mathbf{A}+\mathrm{Aly} \rightleftharpoons \mathbf{B} \rightleftharpoons \mathbf{C} \rightleftharpoons \mathbf{D} $](0.4)</Checkpoint>

<Checkpoint>[存在平衡$ \mathbf{D}+\mathrm{H_2} \rightleftharpoons \mathbf{E} \to \mathbf{E}+\mathbf{S} $](0.4)</Checkpoint>

<Checkpoint>[存在平衡$ \mathbf{D}+\mathrm{CO} \rightleftharpoons \mathbf{F} $](0.4)</Checkpoint>

<Checkpoint>[存在平衡$ \mathbf{S}+\mathrm{{PPh}_3} \rightleftharpoons \mathbf{G} $](0.4)</Checkpoint>

根据推测的结构和反应过程，根据平衡态假设，给出[**D**]、[**F**]、[**G**]关于[**S**]的表达式：

$[\mathbf{D}]=K_{1} K_{2} K_{3} K_{4}[\mathbf{S}][\mathrm{CO}][\mathrm{Aly}]$

$[\mathbf{F}]=K_{1} K_{2} K_{3} K_{4} K_{5}[\mathbf{S}][\mathrm{CO}]^{2}[\mathrm{Aly}] $

$[\mathbf{G}]=K_{6}[\mathbf{S}][\mathrm{PPh}_3]$

其中$[\mathrm{Aly}]$为烯烃物种浓度；$[\mathrm{PPh}_3]$为三苯基膦物种浓度。

<Checkpoint>[[**D**]关于[**S**]的表达式为$[\mathbf{D}]=K_{1} K_{2} K_{3} K_{4}[\mathbf{S}][\mathrm{CO}][\mathrm{Aly}]$](1)</Checkpoint>

<Checkpoint>[[**F**]关于[**S**]的表达式为$[\mathbf{F}]=K_{1} K_{2} K_{3} K_{4} K_{5}[\mathbf{S}][\mathrm{CO}]^{2}[\mathrm{Aly}] $](1)</Checkpoint>

<Checkpoint>[[**G**]关于[**S**]的表达式为$[\mathbf{G}]=K_{6}[\mathbf{S}][\mathrm{PPh}_3]$](1)</Checkpoint>

对[**E**]采取稳态近似：
$$
d[\mathbf{E}]/dt=k_1[\mathbf{D}][\mathrm{H}_2]-
(k_{-1}+k_2)[\mathbf{E}]=0
$$

则：
$$
[\mathbf{E}]=k_1[\mathbf{D}][\mathrm{H}_2]/(k_{-1}+k_2)
$$

<Checkpoint>[对[**E**]采取稳态近似:$[\mathbf{E}]=k_1[\mathbf{D}][\mathrm{H}_2]/(k_{-1}+k_2)$](1)</Checkpoint>

代入物料守恒关系式：
$$
[\mathrm{Rh}]=[\mathbf{S}]+[\mathbf{D}]+[\mathbf{E}]+[\mathbf{F}]+[\mathbf{G}]
$$

得到$[\mathbf{S}]$的表达式为：
$$
[\mathbf{S}]=[\mathrm{Rh}]/\lbrace1+K_6[\mathrm{PPh}_3]+[1+K_5[\mathrm{CO}]+k_1[\mathrm{H}_2]/(k_{-1}+k_2)]K_1K_2K_3K_4[\mathrm{CO}][\mathrm{Aly}]\rbrace
$$

<Checkpoint>[$[\mathbf{S}]$的表达式为$ [\mathbf{S}]=[\mathrm{Rh}]/\lbrace1+K_6[\mathrm{PPh}_3]+[1+K_5[\mathrm{CO}]+k_1[\mathrm{H}_2]/(k_{-1}+k_2)]K_1K_2K_3K_4[\mathrm{CO}][\mathrm{Aly}]\rbrace $](1)</Checkpoint>

代入反应速率表达式：
$$
r=d[\mathbf{P}]/dt=k_2[\mathbf{E}]=k_1k_2[\mathbf{D}][\mathrm{H}_2]/(k_{-1}+k_2)=k_1k_2K_{1} K_{2} K_{3} K_{4}[\mathbf{S}][\mathrm{CO}][\mathrm{Aly}][\mathrm{H}_2]/(k_{-1}+k_2)
$$

代入$[\mathbf{S}]$，得到反应速率表达式：
$$
r=k_1k_2K_{1} K_{2} K_{3} K_{4}[\mathrm{Rh}][\mathrm{CO}][\mathrm{Aly}][\mathrm{H}_2]/(k_{-1}+k_2)\lbrace1+K_6[\mathrm{PPh}_3]+[1+K_5[\mathrm{CO}]+k_1[\mathrm{H}_2]/(k_{-1}+k_2)]K_1K_2K_3K_4[\mathrm{CO}][\mathrm{Aly}]\rbrace
$$

<Checkpoint>[反应速率表达式$ r=k_1k_2K_{1} K_{2} K_{3} K_{4}[\mathrm{Rh}][\mathrm{CO}][\mathrm{Aly}][\mathrm{H}_2]/(k_{-1}+k_2)\lbrace1+K_6[\mathrm{PPh}_3]+[1+K_5[\mathrm{CO}]+k_1[\mathrm{H}_2]/(k_{-1}+k_2)]K_1K_2K_3K_4[\mathrm{CO}][\mathrm{Aly}]\rbrace $](1)</Checkpoint>

由反应速率表达式判断：

无论其他物种浓度取值如何，$[\mathrm{Rh}]$的指数项都为1，所以反应级数与其他物种无关，故选项A错误；

<Checkpoint>[$[\mathrm{Rh}]$的指数项都为1，所以反应级数与其他物种无关](1)</Checkpoint>

$[\mathrm{CO}]$足够大时，有：
$$
r=k_1k_2[\mathrm{Rh}][\mathrm{H}_2]/K_5(k_{-1}+k_2)[\mathrm{CO}]
$$

<Checkpoint>[$[\mathrm{CO}]$足够大时，有： $ r=k_1k_2[\mathrm{Rh}][\mathrm{H}_2]/K_5(k_{-1}+k_2)[\mathrm{CO}] $](1)</Checkpoint>

反应速率与$K_1$、$K_2$、$K_3$、$K_4$的值无关，故选项B错误；

$[\mathrm{H_2}]$足够大时，有：
$$
r=k_2[\mathrm{Rh}]
$$

<Checkpoint>[$[\mathrm{H_2}]$足够大时，有： $ r=k_2[\mathrm{Rh}] $](1)</Checkpoint>

故选项C正确；

$[\mathrm{H_2}]$足够小时，有：
$$
r=k_1k_2K_{1} K_{2} K_{3} K_{4}[\mathrm{Rh}][\mathrm{CO}][\mathrm{Aly}][\mathrm{H}_2]/(k_{-1}+k_2)\lbrace1+K_6[\mathrm{PPh}_3]+[1+K_5[\mathrm{CO}]]K_1K_2K_3K_4[\mathrm{CO}][\mathrm{Aly}]\rbrace
$$

<Checkpoint>[$[\mathrm{H_2}]$足够小时，有： $$ r=k_1k_2K_{1} K_{2} K_{3} K_{4}[\mathrm{Rh}][\mathrm{CO}][\mathrm{Aly}][\mathrm{H}_2]/(k_{-1}+k_2)\lbrace1+K_6[\mathrm{PPh}_3]+[1+K_5[\mathrm{CO}]]K_1K_2K_3K_4[\mathrm{CO}][\mathrm{Aly}]\rbrace $$](1)</Checkpoint>

题目中所有速率常数和平衡常数均与反应速率有关，故选项D错误。

$[\mathrm{PPh}_3]$足够大时，有：
$$
r=k_{1}k_{2}K_{1}K_{2}K_{3}K_{4}[\mathrm{Rh}][\mathrm{CO}][\mathrm{Aly}][\mathrm{H}_2]/(k_{-1}+k_{2})K_{6}[\mathrm{PPh}_3]
$$

<Checkpoint>[$[\mathrm{PPh}_3]$足够大时，有： $$ r=k_{1}k_{2}K_{1}K_{2}K_{3}K_{4}[\mathrm{Rh}][\mathrm{CO}][\mathrm{Aly}][\mathrm{H}_2]/(k_{-1}+k_{2})K_{6}[\mathrm{PPh}_3] $$](1)</Checkpoint>

速率与$[\mathrm{H}_2]$和$[\mathrm{CO}]$有关，故E选项错误。

$[\mathrm{Aly}]$足够大时，有：
$$
r=k_1k_2[\mathrm{Rh}][\mathrm{H}_2]/(k_{-1}+k_2)[1+K_5[\mathrm{CO}]+k_1[\mathrm{H}_2]/(k_{-1}+k_2)]
$$

<Checkpoint>[$[\mathrm{Aly}]$足够大时，有： $$ r=k_1k_2[\mathrm{Rh}][\mathrm{H}_2]/(k_{-1}+k_2)[1+K_5[\mathrm{CO}]+k_1[\mathrm{H}_2]/(k_{-1}+k_2)] $$](1)</Checkpoint>

速率与 $[\mathrm{PPh}_3]$ 无关，F选项错误。

综上所述，应当选择C选项。