The experimenters attempted to investigate the decomposition mechanism of hydrogen peroxide. When carbon monoxide is introduced into the system, the decomposition mechanism of hydrogen peroxide is considered to be as follows:

$\rm H_2O_2+M \to 2OH+M$

$\rm OH+H_2O_2 \to H_2O+HO_2$

$\rm 2HO_2 \to H_2O_2+O_2$

$\rm OH+CO \to CO_2+H$

$\rm H+H_2O_2 \to H_2O+OH$

$\rm H+H_2O_2 \to H_2+HO_2$

(In fact, there is also the reaction $\rm OH+H_2 \to H_2O+H$, but it is known that there is no $\rm H_2$ at the initial stage of the reaction, and the rate of $\rm{H}$ producing $\rm H_2$ is much lower than the rate of producing $\rm H_2O$ (about 350 times), so this elementary reaction is not considered in this question.)

The rate constants of these reactions are $\rm k_1$~$\rm k_6$ ( $\rm [M]$ has been included in $\rm k_1$, and $\rm [M]$ does not need to be involved in the calculation). Derive the rate equation for the decomposition of $\rm H_2O_2$ using the steady-state approximation method, and express it in terms of $\rm [H_2O_2], [CO], k_1, k_2, k_4, k_5, k_6$. The following statements are made:

1.  The rate equation for the decomposition of $\rm H_2O_2$ is $\rm r=2k_1[H_2O_2][ \frac{1+\frac{k_2[H_2O_2]}{k_4[CO]}}{\frac{k_6}{k_5+k_6}+k_2[H_2O_2]}]$

2.  When $\rm [CO]$ is very small or $\rm [CO]$ is very large, the order of the reaction with respect to $\rm H_2O_2$ is 1

3.  A mixed gas of $\rm CO、N_2$ is used to determine the rate constant. Let $\rm r_N$ be the reaction rate when $\rm [CO] = 0$, and $\rm r_{CO}$ be the reaction rate when $\rm [CO] \neq 0$. At a certain temperature, the measured values of $\rm r_N$/$\rm r_{CO}$ and $\rm [H_2O_2]/[CO]$ are shown below:

| $\rm r_N/r_{CO}$ | 0.153 | 0.217 | 0.318 | 0.474 | 0.516 |
| ------------------- | ------ | ------ | ------ | ------ | ------ |
| $\rm [H_2O_2]/[CO]$ | 0.00192 | 0.00681 | 0.0167 | 0.0384 | 0.0472 |

Then it can be calculated that $\rm k_2/k_4 = 17.2, k_5/k_6 = 7.02$

4.  After a series of experiments, researchers believe that the addition of the following elementary reaction can better describe the decomposition of hydrogen peroxide: $\rm HO_2+CO \to CO_2 + OH$

Let $\rm [HO_2]=a_1，r=b_1$ when this reaction is not considered, and $\rm [HO_2]=a_2，r=b_2$ when this reaction is considered. Keeping $\rm [H_2O_2]、[CO]$ constant, it can be concluded from the steady-state approximation method that $\rm a_1=a_2、b_1>b_2$

The following options contain all the correct statements:
实验人员尝试探究过氧化氢的分解机理。在体系中充入一氧化碳时，认为过氧化氢的分解机理如下：

$\rm H_2O_2+M \to 2OH+M$

$\rm OH+H_2O_2 \to H_2O+HO_2$

$\rm 2HO_2 \to H_2O_2+O_2$

$\rm OH+CO \to CO_2+H$

$\rm H+H_2O_2 \to H_2O+OH$

$\rm H+H_2O_2 \to H_2+HO_2$

（实际上还存在$\rm OH+H_2 \to H_2O+H$这个反应，但已知反应初始时不存在$\rm H_2$，且$\rm{H}$产生$\rm H_2$的速率远低于产生$\rm H_2O$的速率(约350倍)，因此本题不考虑此基元反应）

这几个反应的速率常数分别为$\rm k_1$~$\rm k_6$（$\rm [M]$已经算入$\rm k_1$，计算时不用涉及$\rm [M]$）。以稳态近似法推导$\rm H_2O_2$分解的速率方程，用$\rm [H_2O_2], [CO], k_1, k_2, k_4, k_5, k_6$表示。有下列说法：

1. $\rm H_2O_2$分解的速率方程为$\rm r=2k_1[H_2O_2][ \frac{1+\frac{k_2[H_2O_2]}{k_4[CO]}}{\frac{k_6}{k_5+k_6}+k_2[H_2O_2]}]$

2. 当$\rm [CO]$很小或$\rm [CO]$很大的时候该反应对于$\rm H_2O_2$的级数为1

3. 使用$\rm CO、N_2$的混合气体进行速率常数测定，记当$\rm [CO] = 0$时的反应速率为$\rm r_N$，$\rm [CO] \neq 0$时的反应速率为$\rm r_{CO}$。在一定的温度下，测得$\rm r_N$/$\rm r_{CO}$和$\rm [H_2O_2]/[CO]$的值如下所示：

| $\rm r_N/r_{CO}$ | 0.153 | 0.217 | 0.318 | 0.474 | 0.516 |
| ------------------- | ------ | ------ | ------ | ------ | ------ |
| $\rm [H_2O_2]/[CO]$ | 0.00192 | 0.00681 | 0.0167 | 0.0384 | 0.0472 |

则可以计算得到$\rm k_2/k_4 = 17.2, k_5/k_6 = 7.02$

4. 经过一系列的实验，研究人员认为下面的基元反应的加入可以更好地描述过氧化氢的分解：$\rm HO_2+CO \to CO_2 + OH$

记不考虑该反应时$\rm [HO_2]=a_1，r=b_1$，考虑该反应时$\rm [HO_2]=a_2，r=b_2$。维持$\rm [H_2O_2]、[CO]$不变，利用稳态近似法可以得出$\rm a_1=a_2、b_1>b_2$

下列选项中包含全部正确说法的是：
All other options are incorrect
1,2
1
2,3
2,4
1,3,4
1,2,4
3
2,3,4
1,2,3,4
1,4
3,4
其他选项均不正确
1,2
1
2,3
2,4
1,3,4
1,2,4
3
2,3,4
1,2,3,4
1,4
3,4
Perform steady-state approximation for the three intermediates $\rm OH、HO_2、H$:

$\rm d[OH]/dt = 2k_1[H_2O_2] - k_2[OH][H_2O_2] + k_5[H][H_2O_2] - k_4[OH][CO] = 0$  (1)

<Checkpoint>[$\rm d[OH]/dt = 2k_1[H_2O_2] - k_2[OH][H_2O_2] + k_5[H][H_2O_2] - k_4[OH][CO] = 0$](1)</Checkpoint>

$\rm d[HO_2]/dt = k_2[OH][H_2O_2] + k_6[H][H_2O_2] - 2k_3[HO_2]^2 = 0$  (2)

<Checkpoint>[$\rm d[HO_2]/dt = k_2[OH][H_2O_2] + k_6[H][H_2O_2] - 2k_3[HO_2]^2 = 0$](1)</Checkpoint>

$\rm d[H]/dt = k_4[OH][CO] - k_5[H][H_2O_2] - k_6[H][H_2O_2] = 0$  (3)

<Checkpoint>[$\rm d[H]/dt = k_4[OH][CO] - k_5[H][H_2O_2] - k_6[H][H_2O_2] = 0$](1)</Checkpoint>

(1) + (3)：$\rm k_2[OH] + k_6[H] = 2k_1，[OH] = (2k_1 - k_6[H])/k_2$  (4)

Substituting (4) into (3)：$\rm [H]= \frac{2k_1k_4[CO]}{k_4k_6[CO]+k_2(k_5+k_6)[H_2O_2]}$

And $\rm r=k_1[H_2O_2] + k_2[OH][H_2O_2] - k_3[HO_2]^2 + k_5[H][H_2O_2] + k_6[H][H_2O_2] = 2k_1[H_2O_2] + 
k_5[H][H_2O_2]$ (5)

Thus $\rm r=2k_1[H_2O_2][ \frac{1+\frac{k_2[H_2O_2]}{k_4[CO]}}{\frac{k_6}{k_5+k_6}+\frac{k_2[H_2O_2]}{k_4[CO]}}]$

<Checkpoint>[$\rm r=2k_1[H_2O_2][ \frac{1+\frac{k_2[H_2O_2]}{k_4[CO]}}{\frac{k_6}{k_5+k_6}+\frac{k_2[H_2O_2]}{k_4[CO]}}]$](3)</Checkpoint>

Statement 1: The denominator omits $\rm k_4[CO]$, incorrect.

<Checkpoint>[Statement 1 is incorrect](1)</Checkpoint>

Statement 2: When $\rm [CO]$ concentration is very small, $\frac{k_2[H_2O_2]}{k_4[CO]}>>1$, at this time $\rm r \approx 2k_1[H_2O_2]$; when $\rm [CO]$ concentration is very large, $\frac{k_2[H_2O_2]}{k_4[CO]}<<1$, at this time $\rm r \approx 2k_1[H_2O_2] \frac{k_5 + k_6}{k_6}$. For $\rm [H_2O_2]$ all are first order, Statement 2 is correct.

<Checkpoint>[When $\rm [CO]$ is very large or very small, for $\rm [H_2O_2]$ all are first order, Statement (2) is correct](2)</Checkpoint>

Statement 3: Transformation yields:
$\rm \frac{r_N}{r_{CO}}=\frac{k_6}{k_5+k_6}+\frac{k_2[H_2O_2]}{k_4[CO]}(1-\frac{r_N}{r_{CO}})$

<Checkpoint>[$\rm \frac{r_N}{r_{CO}}=\frac{k_6}{k_5+k_6}+\frac{k_2[H_2O_2]}{k_4[CO]}(1-\frac{r_N}{r_{CO}})$](2)</Checkpoint>

Substituting numerical values, for linear regression of $\rm \frac{[H_2O_2]}{[CO]}(1-\frac{r_N}{r_{CO}})- \frac{r_N}{r_{CO}}$: $\rm k_2/k_4 = 17.2，k_5/k_6 = 7.02$, Statement 3 is correct.

<Checkpoint>[$\rm k_2/k_4 = 17.2，k_5/k_6 = 7.02$, Statement 3 is correct](2)</Checkpoint>

Statement 4: From equation (1)+(2)+(3), it can be known that $\rm 2k_1[H_2O_2]=k_3[HO_2]^2$, considering that the seventh reaction does not affect this result, thus $\rm a_1 = a_2$

<Checkpoint>[$\rm a_1 = a_2$](1)</Checkpoint>

Similarly, rewriting the expression for $\rm r$ can reveal that $\rm r=[OH](k_2[H_2O_2]+k_4[CO])$, adding a reaction that generates $\rm OH$ leads to an increase in its concentration, thus $\rm r_2>r_1$

<Checkpoint>[$\rm r_2>r_1$](1)</Checkpoint>

Statement 4 is incorrect.

Therefore, choose D.

In fact, $\rm r=2k_1[H_2O_2][1+ \frac{k_7[CO]}{ \sqrt {k_1k_3[H_2O_2]}}][ \frac{1+\frac{k_2[H_2O_2]}{k_4[CO]}}{\frac{k_6}{k_5+k_6}+\frac{k_2[H_2O_2]}{k_4[CO]}}]$
对三个中间体$\rm OH、HO_2、H$进行稳态近似：

$\rm d[OH]/dt = 2k_1[H_2O_2] - k_2[OH][H_2O_2] + k_5[H][H_2O_2] - k_4[OH][CO] = 0$  (1)

<Checkpoint>[$\rm d[OH]/dt = 2k_1[H_2O_2] - k_2[OH][H_2O_2] + k_5[H][H_2O_2] - k_4[OH][CO] = 0$](1)</Checkpoint>

$\rm d[HO_2]/dt = k_2[OH][H_2O_2] + k_6[H][H_2O_2] - 2k_3[HO_2]^2 = 0$  (2)

<Checkpoint>[$\rm d[HO_2]/dt = k_2[OH][H_2O_2] + k_6[H][H_2O_2] - 2k_3[HO_2]^2 = 0$](1)</Checkpoint>

$\rm d[H]/dt = k_4[OH][CO] - k_5[H][H_2O_2] - k_6[H][H_2O_2] = 0$  (3)

<Checkpoint>[$\rm d[H]/dt = k_4[OH][CO] - k_5[H][H_2O_2] - k_6[H][H_2O_2] = 0$](1)</Checkpoint>

(1) + (3)：$\rm k_2[OH] + k_6[H] = 2k_1，[OH] = (2k_1 - k_6[H])/k_2$  (4)

(4)代入(3)：$\rm [H]= \frac{2k_1k_4[CO]}{k_4k_6[CO]+k_2(k_5+k_6)[H_2O_2]}$

而 $\rm r=k_1[H_2O_2] + k_2[OH][H_2O_2] - k_3[HO_2]^2 + k_5[H][H_2O_2] + k_6[H][H_2O_2] = 2k_1[H_2O_2] + 
k_5[H][H_2O_2]$ (5)

故 $\rm r=2k_1[H_2O_2][ \frac{1+\frac{k_2[H_2O_2]}{k_4[CO]}}{\frac{k_6}{k_5+k_6}+\frac{k_2[H_2O_2]}{k_4[CO]}}]$

<Checkpoint>[$\rm r=2k_1[H_2O_2][ \frac{1+\frac{k_2[H_2O_2]}{k_4[CO]}}{\frac{k_6}{k_5+k_6}+\frac{k_2[H_2O_2]}{k_4[CO]}}]$](3)</Checkpoint>

说法1：分母漏了$\rm k_4[CO]$，错误

<Checkpoint>[说法1错误](1)</Checkpoint>

说法2：$\rm [CO]$浓度很小时$\frac{k_2[H_2O_2]}{k_4[CO]}>>1$，此时$\rm r \approx 2k_1[H_2O_2]$；$\rm [CO]$浓度很大时$\frac{k_2[H_2O_2]}{k_4[CO]}<<1$，此时$\rm r \approx 2k_1[H_2O_2] \frac{k_5 + k_6}{k_6}$。对于$\rm [H_2O_2]$均为一级，说法2正确

<Checkpoint>[$\rm [CO]$很大或很小时，对于$\rm [H_2O_2]$均为一级，说法（2）正确](2)</Checkpoint>

说法3：变形可得：
$\rm \frac{r_N}{r_{CO}}=\frac{k_6}{k_5+k_6}+\frac{k_2[H_2O_2]}{k_4[CO]}(1-\frac{r_N}{r_{CO}})$

<Checkpoint>[$\rm \frac{r_N}{r_{CO}}=\frac{k_6}{k_5+k_6}+\frac{k_2[H_2O_2]}{k_4[CO]}(1-\frac{r_N}{r_{CO}})$](2)</Checkpoint>

代入数值，对于$\rm \frac{[H_2O_2]}{[CO]}(1-\frac{r_N}{r_{CO}})- \frac{r_N}{r_{CO}}$线性回归：$\rm k_2/k_4 = 17.2，k_5/k_6 = 7.02$，说法3正确

<Checkpoint>[$\rm k_2/k_4 = 17.2，k_5/k_6 = 7.02$，说法3正确](2)</Checkpoint>

说法4：由式(1)+(2)+(3)可知$\rm 2k_1[H_2O_2]=k_3[HO_2]^2$，考虑第七个反应不影响此结果，故$\rm a_1 = a_2$

<Checkpoint>[$\rm a_1 = a_2$](1)</Checkpoint>

类似的，改写式$\rm r$的表达式可以发现$\rm r=[OH](k_2[H_2O_2]+k_4[CO])$，加入一个生成$\rm OH$的反应导致其浓度升高，故$\rm r_2>r_1$

<Checkpoint>[$\rm r_2>r_1$](1)</Checkpoint>

说法4错误

故选D

事实上$\rm r=2k_1[H_2O_2][1+ \frac{k_7[CO]}{ \sqrt {k_1k_3[H_2O_2]}}][ \frac{1+\frac{k_2[H_2O_2]}{k_4[CO]}}{\frac{k_6}{k_5+k_6}+\frac{k_2[H_2O_2]}{k_4[CO]}}]$