Ozone is an important air pollutant in the lower atmosphere, capable of triggering photochemical smog, among other effects. In alkaline aqueous solutions, ozone can decompose, while the small amount of \( \text{H}_2\text{O}_2 \) produced during decomposition also catalyzes ozone degradation. A possible mechanistic scheme is shown below. In alkaline solutions, \( \text{HO}_2^- \) can be considered fully ionized, and the ionization constant of water is \( K_w \).  

$$
\text{O}_3 + \text{OH}^- \rightarrow \text{HO}_2^- + \text{O}_2 \quad k_1
$$

$$
\text{H}_2\text{O}_2 = \text{HO}_2^- + \text{H}^+ \quad K_2
$$

$$
\text{O}_3 + \text{HO}_2^- \rightarrow \text{HO}\cdot + \text{O}_2^- + \text{O}_2 \quad k_3
$$

$$
\text{O}_3 + \text{HO}\cdot \rightarrow \text{HO}_2\cdot + \text{O}_2 \quad k_4
$$

$$
\text{H}_2\text{O}_2 + \text{HO}\cdot \rightarrow \text{HO}_2\cdot + \text{H}_2\text{O} \quad k_5
$$

$$
\text{HO}_2^- + \text{HO}\cdot \rightarrow \text{H}\text{O}_2\cdot + \text{OH}^- \quad k_6
$$

$$
\text{HO}_2\cdot = \text{H}^+ + \text{O}_2^- \quad K_7
$$

$$
\text{OH}^- + \text{OH}\cdot \rightarrow \text{H}_2\text{O} + \text{O}^- \quad k_8
$$

$$
\text{O}_3 + \text{O}_2^- \rightarrow \text{O}_3^- + \text{O}_2 \quad k_9
$$

$$
\text{O}_3^- + \text{H}^+ \rightarrow \text{HO}\cdot + \text{O}_2 \quad k_{10}
$$

$$
\text{O}^- + \text{H}_2\text{O} \rightarrow \text{HO}\cdot + \text{HO}^- \quad k_{11}
$$

Assuming that the chain termination step is solely the dimerization of hydroxyl radicals $\mathrm{OH\cdot}$ (with a rate constant of $ k_{12} $), and considering the chain initiation rate, derive the expression for $[\text{O}_2^-]$ in terms of the concentrations of $[\text{H}_2\text{O}_2]$, $[\text{OH}^-]$, and $[\text{O}_3]$, using reasonable steady-state approximations. Additionally, analyze $v_{\text{O}_3}$, the rate of ozone consumption, and select the most appropriate option from the choices below.
臭氧在底层大气中则是一种重要的空气污染物，可引发光化学烟雾等。在碱性水相中臭氧可以被分解，与此同时分解产生的少量 \( \text{H}_2\text{O}_2 \) 亦将催化臭氧的分解。其可能机理示意图如下。在碱性溶液中 \( \text{HO}_2^- \) 可视为完全电离，水的电离常数为 \( K_w \)。

$$
\text{O}_3 + \text{OH}^- \rightarrow \text{HO}_2^- + \text{O}_2 \quad k_1
$$

$$
\text{H}_2\text{O}_2 = \text{HO}_2^- + \text{H}^+ \quad K_2
$$

$$
\text{O}_3 + \text{HO}_2^- \rightarrow \text{HO}\cdot + \text{O}_2^- + \text{O}_2 \quad k_3
$$

$$
\text{O}_3 + \text{HO}\cdot \rightarrow \text{HO}_2\cdot + \text{O}_2 \quad k_4
$$

$$
\text{H}_2\text{O}_2 + \text{HO}\cdot \rightarrow \text{HO}_2\cdot + \text{H}_2\text{O} \quad k_5
$$

$$
\text{HO}_2^- + \text{HO}\cdot \rightarrow \text{H}\text{O}_2\cdot + \text{OH}^- \quad k_6
$$

$$
\text{HO}_2\cdot = \text{H}^+ + \text{O}_2^- \quad K_7
$$

$$
\text{OH}^- + \text{OH}\cdot \rightarrow \text{H}_2\text{O} + \text{O}^- \quad k_8
$$

$$
\text{O}_3 + \text{O}_2^- \rightarrow \text{O}_3^- + \text{O}_2 \quad k_9
$$

$$
\text{O}_3^- + \text{H}^+ \rightarrow \text{HO}\cdot + \text{O}_2 \quad k_{10}
$$

$$
\text{O}^- + \text{H}_2\text{O} \rightarrow \text{HO}\cdot + \text{HO}^- \quad k_{11}
$$

若链终止步骤仅为羟基自由基$\mathrm{OH\cdot}$二聚（速率常数为 $ k_{12} $），考虑链引发速率，请通过合理的稳态近似，求出以 $[\text{H}_2\text{O}_2]$、$[\text{OH}^-]$ 及 $[\text{O}_3]$ 三者浓度所表示的 $[\text{O}_2^-]$ 的表达式。并考虑$v_{\text{O}_3}$亦即臭氧消耗的速率，从下面选项中选择最符合题意的一项。
Expand the original expression completely into a sum of single terms with addition and subtraction operations (excluding the operations inside the square root), where each single term is the product of the coefficient, the power of the concentration term, and the operation inside the square root (if any). The expression for $[\text{O}_2^-]$ includes 1 term, and the expression for $v_{\text{O}_3}$ includes 4 terms.
Expand the original expression completely into a sum of monomials involving addition and subtraction operations (excluding the content inside the square root), where each monomial is the product of a coefficient and the power of a concentration term multiplied by the operation inside the square root (if any). The expression for $[\text{O}_2^-]$ includes 1 term, and the expression for $v_{\text{O}_3}$ includes 3 terms.
Expand the original expression completely into the sum of monomials involving addition and subtraction operations (excluding the content inside the square root), where each monomial is the product of a coefficient, a power term of concentration, and the operation inside the square root (if any). The expression for $[\text{O}_2^-]$ includes 2 terms, and the expression for $v_{\text{O}_3}$ includes 3 terms.
Expand the original expression completely into a sum of individual terms with addition and subtraction operations (excluding the contents inside the square root). Each individual term is the product of a coefficient, a power of the concentration term, and the operation inside the square root (if any). The $[\text{O}_2^-]$ expression includes 2 terms, and the $v_{\text{O}_3}$ expression includes 2 terms.
Expand the original expression completely into the sum of monomials involving addition and subtraction operations (excluding the content inside the square root), where each monomial is the product of a coefficient, the power of the concentration term, and the operation inside the square root (if any). The expression for $[\text{O}_2^-]$ includes 3 terms, and the expression for $v_{\text{O}_3}$ includes 2 terms.
Expand the original expression completely into a sum of monomials involving addition and subtraction operations (excluding the operations inside the square root), where each monomial is the product of a coefficient, the power of a concentration term, and the operation inside the square root (if applicable). The expression for $[\text{O}_2^-]$ includes 3 terms, and the expression for $v_{\text{O}_3}$ includes 1 term.
Fully expand the original expression into the sum of single terms with addition and subtraction operations (excluding the operations inside the square root), where each single term is the product of the coefficient, the power of the concentration term, and the operation inside the square root (if any). The $[\text{O}_2^-]$ expression includes 4 terms, and the $v_{\text{O}_3}$ expression includes 1 term.
Expand the original expression completely into a sum of single terms with addition and subtraction operations (excluding the operations inside the square root), where each single term is the product of a coefficient and the power of the concentration term multiplied by the operation inside the square root (if any). The expression for $[\text{O}_2^-]$ includes 4 terms, and the expression for $v_{\text{O}_3}$ includes 4 terms.
Expand the original expression fully into the sum of monomials involving addition and subtraction (excluding operations inside the square root), where each monomial is the product of a coefficient, the power of concentration terms, and the operation inside the square root (if any). The sum of the integer coefficients in the $[\text{O}_2^-]$ expression that do not include rate constants or equilibrium constants is 1, and the sum of the integer coefficients in the $v_{\text{O}_3}$ expression that do not include rate constants or equilibrium constants is 10.
Fully expand the original expression into the sum of individual terms involving addition and subtraction operations (excluding operations inside the square root), where each term is the product of a coefficient, the power of the concentration term, and the operation inside the square root (if present). The sum of the integer coefficients in the $[\text{O}_2^-]$ expression that do not include rate constants or equilibrium constants is 2, and the sum of the integer coefficients in the $v_{\text{O}_3}$ expression that do not include rate constants or equilibrium constants is 9.
Expand the original expression completely into a sum of monomials with addition and subtraction operations (excluding operations inside the square root), where each monomial is the product of a coefficient, power terms of concentration terms, and operations inside the square root (if any). The sum of the integer coefficients in the $[\text{O}_2^-]$ expression that do not include rate constants or equilibrium constants is 3, and the sum of the integer coefficients in the $v_{\text{O}_3}$ expression that do not include rate constants or equilibrium constants is 8.
Fully expand the original expression into a sum of monomials with addition and subtraction operations (excluding operations inside square roots), where each monomial is the product of a coefficient, power terms of concentration terms, and operations inside square roots (if any). The sum of integer coefficients in the $[\text{O}_2^-]$ expression that do not include rate constants or equilibrium constants is 4, and the sum of integer coefficients in the $v_{\text{O}_3}$ expression that do not include rate constants or equilibrium constants is 7.
Fully expand the original expression into the sum of monomials involving addition and subtraction operations (excluding operations inside the square root), where each monomial is the product of a coefficient and the power of concentration terms multiplied by the operations inside the square root (if any). The sum of integer coefficients in the $[\text{O}_2^-]$ expression that do not include rate constants or equilibrium constants is 5, and the sum of integer coefficients in the $v_{\text{O}_3}$ expression that do not include rate constants or equilibrium constants is 6.
Expand the original expression completely into the sum of monomials with addition and subtraction operations (excluding the operations inside the square root), where each monomial is the product of a coefficient and the power of the concentration term and the operation inside the square root (if any). The sum of the integer coefficients in the $[\text{O}_2^-]$ expression that do not include rate constants or equilibrium constants is 6, and the sum of the integer coefficients in the $v_{\text{O}_3}$ expression that do not include rate constants or equilibrium constants is 5.
Expand the original expression completely into the sum of monomials involving addition and subtraction operations (excluding the operations inside the square root), where each monomial is the product of a coefficient, the power of a concentration term, and the operation inside the square root (if any). The sum of the integer coefficients in the $[\text{O}_2^-]$ expression that do not include rate constants or equilibrium constants is 7, and the sum of the integer coefficients in the $v_{\text{O}_3}$ expression that do not include rate constants or equilibrium constants is 4.
Expand the original expression completely into the sum of single terms with addition and subtraction operations (excluding the operations inside the square root), where each single term is the product of a coefficient, the power of the concentration term, and the operation inside the square root (if any). The sum of the integer coefficients in the $[\text{O}_2^-]$ expression that do not include rate constants or equilibrium constants is 8, and the sum of the integer coefficients in the $v_{\text{O}_3}$ expression that do not include rate constants or equilibrium constants is 3.
Expand the original expression completely into the sum of individual terms involving addition and subtraction (excluding operations inside the square root), where each term is the product of a coefficient and the power of a concentration term multiplied by the operation inside the square root (if any). The sum of the integer coefficients in the $[\text{O}_2^-]$ expression that do not include rate constants or equilibrium constants is 9, and the sum of the integer coefficients in the $v_{\text{O}_3}$ expression that do not include rate constants or equilibrium constants is 2.
Expand the original expression completely into the sum of monomials involving addition and subtraction operations (excluding the operations inside the square root), where each monomial is the product of a coefficient and the power of concentration terms multiplied by the operations inside the square root (if any). The sum of the integer coefficients in the $[\text{O}_2^-]$ expression that do not include rate constants or equilibrium constants is 10, and the sum of the integer coefficients in the $v_{\text{O}_3}$ expression that do not include rate constants or equilibrium constants is 1.
Due to certain reasons, options A-R are all correct
For certain reasons, options A-R are all incorrect.
Options A-R have more than 3 correct answers
Option A-R has exactly 2 correct.
将原式完全展开为加减操作的单项之和（不包括根号内部），其中每个单项是系数与浓度项的幂次与根号内部运算（如有）相乘后，$[\text{O}_2^-]$表达式包括1项，$v_{\text{O}_3}$表达式包括4项
将原式完全展开为加减操作的单项之和（不包括根号内部），其中每个单项是系数与浓度项的幂次与根号内部运算（如有）相乘后，$[\text{O}_2^-]$表达式包括1项，$v_{\text{O}_3}$表达式包括3项
将原式完全展开为加减操作的单项之和（不包括根号内部），其中每个单项是系数与浓度项的幂次与根号内部运算（如有）相乘后，$[\text{O}_2^-]$表达式包括2项，$v_{\text{O}_3}$表达式包括3项
将原式完全展开为加减操作的单项之和（不包括根号内部），其中每个单项是系数与浓度项的幂次与根号内部运算（如有）相乘后，$[\text{O}_2^-]$表达式包括2项，$v_{\text{O}_3}$表达式包括2项
将原式完全展开为加减操作的单项之和（不包括根号内部），其中每个单项是系数与浓度项的幂次与根号内部运算（如有）相乘后，$[\text{O}_2^-]$表达式包括3项，$v_{\text{O}_3}$表达式包括2项
将原式完全展开为加减操作的单项之和（不包括根号内部），其中每个单项是系数与浓度项的幂次与根号内部运算（如有）相乘后，$[\text{O}_2^-]$表达式包括3项，$v_{\text{O}_3}$表达式包括1项
将原式完全展开为加减操作的单项之和（不包括根号内部），其中每个单项是系数与浓度项的幂次与根号内部运算（如有）相乘后，$[\text{O}_2^-]$表达式包括4项，$v_{\text{O}_3}$表达式包括1项
将原式完全展开为加减操作的单项之和（不包括根号内部），其中每个单项是系数与浓度项的幂次与根号内部运算（如有）相乘后，$[\text{O}_2^-]$表达式包括4项，$v_{\text{O}_3}$表达式包括4项
将原式完全展开为加减操作的单项之和（不包括根号内部），其中每个单项是系数与浓度项的幂次与根号内部运算（如有）相乘后，$[\text{O}_2^-]$表达式系数中不含速率常数、平衡常数的整数系数之和为1，$v_{\text{O}_3}$表达式系数中不含速率常数、平衡常数的整数系数之和为10
将原式完全展开为加减操作的单项之和（不包括根号内部），其中每个单项是系数与浓度项的幂次与根号内部运算（如有）相乘后，$[\text{O}_2^-]$表达式系数中不含速率常数、平衡常数的整数系数之和为2，$v_{\text{O}_3}$表达式系数不含速率常数、平衡常数的整数系数之和为9
将原式完全展开为加减操作的单项之和（不包括根号内部），其中每个单项是系数与浓度项的幂次与根号内部运算（如有）相乘后，$[\text{O}_2^-]$表达式系数中不含速率常数、平衡常数的整数系数之和为3，$v_{\text{O}_3}$表达式系数中不含速率常数、平衡常数的整数系数之和为8
将原式完全展开为加减操作的单项之和（不包括根号内部），其中每个单项是系数与浓度项的幂次与根号内部运算（如有）相乘后，$[\text{O}_2^-]$表达式系数中不含速率常数、平衡常数的整数系数之和为4，$v_{\text{O}_3}$表达式系数中不含速率常数、平衡常数的整数系数之和为7
将原式完全展开为加减操作的单项之和（不包括根号内部），其中每个单项是系数与浓度项的幂次与根号内部运算（如有）相乘后，$[\text{O}_2^-]$表达式系数中不含速率常数、平衡常数的整数系数之和为5，$v_{\text{O}_3}$表达式系数中不含速率常数、平衡常数的整数系数之和为6
将原式完全展开为加减操作的单项之和（不包括根号内部），其中每个单项是系数与浓度项的幂次与根号内部运算（如有）相乘后，$[\text{O}_2^-]$表达式系数中不含速率常数、平衡常数的整数系数之和为6，$v_{\text{O}_3}$表达式系数中不含速率常数、平衡常数的整数系数之和为5
将原式完全展开为加减操作的单项之和（不包括根号内部），其中每个单项是系数与浓度项的幂次与根号内部运算（如有）相乘后，$[\text{O}_2^-]$表达式系数中不含速率常数、平衡常数的整数系数之和为7，$v_{\text{O}_3}$表达式系数中不含速率常数、平衡常数的整数系数之和为4
将原式完全展开为加减操作的单项之和（不包括根号内部），其中每个单项是系数与浓度项的幂次与根号内部运算（如有）相乘后，$[\text{O}_2^-]$表达式系数中不含速率常数、平衡常数的整数系数之和为8，$v_{\text{O}_3}$表达式系数中不含速率常数、平衡常数的整数系数之和为3
将原式完全展开为加减操作的单项之和（不包括根号内部），其中每个单项是系数与浓度项的幂次与根号内部运算（如有）相乘后，$[\text{O}_2^-]$表达式系数中不含速率常数、平衡常数的整数系数之和为9，$v_{\text{O}_3}$表达式系数中不含速率常数、平衡常数的整数系数之和为2
将原式完全展开为加减操作的单项之和（不包括根号内部），其中每个单项是系数与浓度项的幂次与根号内部运算（如有）相乘后，$[\text{O}_2^-]$表达式系数中不含速率常数、平衡常数的整数系数之和为10，$v_{\text{O}_3}$表达式系数中不含速率常数、平衡常数的整数系数之和为1
由于某些原因，选项A-R均正确
由于某些原因，选项A-R均不正确
选项A-R有3个以上正确
选项A-R恰有2个正确
According to the problem, the chain termination step is solely the dimerization of hydroxyl radicals, i.e.:

$$
2\text{HO}\cdot \rightarrow \text{H}_2\text{O}_2\quad k_{12}
$$

For other intermediate products (radicals and reactive intermediates), we can apply the steady-state approximation, meaning their generation rate equals their consumption rate.

The steady-state approximation yields:

$$
d[\text{HO}\cdot]/dt = k_3 [\text{O}_3][\text{HO}_2^-] + k_{10} [\text{O}_3^-][\text{H}^+] + k_{11} [\text{O}^-][\text{H}_2\text{O}] - k_4 [\text{O}_3][\text{HO}\cdot] - k_5 [\text{H}_2\text{O}_2][\text{HO}\cdot] - k_6 [\text{HO}_2^-][\text{HO}\cdot] - k_8 [\text{OH}^-][\text{HO}\cdot] - 2 k_{12} [\text{HO}\cdot]^2 = 0
$$

$$
d[\text{HO}_2\cdot]/dt = k_4 [\text{O}_3][\text{HO}\cdot] + k_5 [\text{H}_2\text{O}_2][\text{HO}\cdot] + k_6 [\text{HO}_2^-][\text{HO}\cdot] - k_7 [\text{HO}_2\cdot] + k_{-7} [\text{H}^+][\text{O}_2^-] = 0
$$

$$
d[\text{O}_2^-]/dt = k_3 [\text{O}_3][\text{HO}_2^-] + k_7 [\text{HO}_2\cdot] - k_9 [\text{O}_3][\text{O}_2^-] - k_{-7} [\text{H}^+][\text{O}_2^-] = 0
$$

$$
d[\text{O}_3^-]/dt = k_9 [\text{O}_3][\text{O}_2^-] - k_{10} [\text{O}_3^-][\text{H}^+] = 0
$$

$$
d[\text{O}^-]/dt = k_8 [\text{OH}^-][\text{HO}\cdot] - k_{11} [\text{O}^-][\text{H}_2\text{O}] = 0
$$


Adding the five radical equations, we easily obtain:

$$
2 k_3 [\text{O}_3][\text{HO}_2^-] - 2k_{12}[\text{HO}\cdot]^2=0
$$

Thus, we can solve for:

$$
[\text{HO}\cdot] = \sqrt{\frac{k_{3}[\text{O}_3][\text{HO}_2^-]}{k_{12}}}
$$

<Checkpoint>[Solved:  $ [\text{HO}\cdot] = \sqrt{\frac{k_{3}[\text{O}_3][\text{HO}_2^-]}{k_{12}}} $](1.0)</Checkpoint>

We know that in alkaline solutions, $\text{H}_2\text{O}_2$ can be considered almost completely ionized, and from the reaction:

$$
\text{H}_2\text{O}_2 = \text{HO}_2^- + \text{H}^+ \quad K_2
$$

we can write:

$$
\text{H}_2\text{O}_2 + \text{OH}^- = \text{HO}_2^- + \text{H}_2\text{O} \quad \frac{K_2}{K_w}
$$

Thus:

$$
[\text{HO}_2^-] = \frac{K_2}{K_w}[\text{H}_2\text{O}_2][\text{OH}^-]
$$

That is:

$$
[\text{HO}\cdot] = \sqrt{\frac{k_{3}K_2}{k_{12}K_w}[\text{O}_3][\text{H}_2\text{O}_2][\text{OH}^-]}
$$

<Checkpoint>[Solved: $ [\text{HO}\cdot] = \sqrt{\frac{k_{3}K_2}{k_{12}K_w}[\text{O}_3][\text{H}_2\text{O}_2][\text{OH}^-]} $](1.0)</Checkpoint>

From:

$$
d[\text{O}_2^-]/dt = k_3 [\text{O}_3][\text{HO}_2^-] + k_7 [\text{HO}_2\cdot] - k_9 [\text{O}_3][\text{O}_2^-] - k_{-7} [\text{H}^+][\text{O}_2^-] = 0
$$

and

$$
d[\text{HO}_2\cdot]/dt = k_4 [\text{O}_3][\text{HO}\cdot] + k_5 [\text{H}_2\text{O}_2][\text{HO}\cdot] + k_6 [\text{HO}_2^-][\text{HO}\cdot] - k_7 [\text{HO}_2\cdot] + k_{-7} [\text{H}^+][\text{O}_2^-] = 0
$$

we can eliminate the equilibrium terms, yielding:

$$
k_3 [\text{O}_3][\text{HO}_2^-] + k_4 [\text{O}_3][\text{HO}\cdot] + k_5 [\text{H}_2\text{O}_2][\text{HO}\cdot] + k_6 [\text{HO}_2^-][\text{HO}\cdot] - k_9 [\text{O}_3][\text{O}_2^-] = 0
$$

Thus, we obtain:

$$
[\text{O}_2^-] = \frac{k_3 [\text{O}_3][\text{HO}_2^-] + k_4 [\text{O}_3][\text{HO}\cdot] + k_5 [\text{H}_2\text{O}_2][\text{HO}\cdot] + k_6 [\text{HO}_2^-][\text{HO}\cdot]}{k_9[\text{O}_3]}
$$

<Checkpoint>[Eliminated terms to obtain: $ [\text{O}_2^-] = \frac{k_3 [\text{O}_3][\text{HO}_2^-] + k_4 [\text{O}_3][\text{HO}\cdot] + k_5 [\text{H}_2\text{O}_2][\text{HO}\cdot] + k_6 [\text{HO}_2^-][\text{HO}\cdot]}{k_9[\text{O}_3]} $](1.0)</Checkpoint>

Substituting our earlier results, we get:

$$
[\text{O}_2^-] = \frac{\frac{k_3K_2}{K_w} [\text{O}_3][\text{H}_2\text{O}_2][\text{OH}^-] + (k_4 [\text{O}_3] + k_5 [\text{H}_2\text{O}_2] +\frac{k_6K_2}{K_w}[\text{H}_2\text{O}_2][\text{OH}^-])\sqrt{\frac{k_{3}K_2}{k_{12}K_w}[\text{O}_3][\text{H}_2\text{O}_2][\text{OH}^-]}}{k_9[\text{O}_3]}
$$

<Checkpoint>[Calculated: $ [\text{O}_2^-] = \frac{\frac{k_3K_2}{K_w} [\text{O}_3][\text{H}_2\text{O}_2][\text{OH}^-] + (k_4 [\text{O}_3] + k_5 [\text{H}_2\text{O}_2] +\frac{k_6K_2}{K_w}[\text{H}_2\text{O}_2][\text{OH}^-])\sqrt{\frac{k_{3}K_2}{k_{12}K_w}[\text{O}_3][\text{H}_2\text{O}_2][\text{OH}^-]}}{k_9[\text{O}_3]} $](1.0)</Checkpoint>

Next, we calculate the ozone consumption rate:

$$
v_{\text{O}_3} = k_1[\text{O}_3][\text{OH}^-] + k_3[\text{O}_3][\text{HO}_2^-] + k_4[\text{O}_3][\text{HO}\cdot] + k_9[\text{O}_3][\text{O}_2^-]
$$

Substituting the previous results, we find:

$$
v_{\text{O}_3} = k_1[\text{O}_3][\text{OH}^-] + 2\frac{K_2k_3}{K_w}[\text{OH}^-][\text{O}_3][\text{H}_2\text{O}_2] + (2k_4[\text{O}_3] + k_5[\text{H}_2\text{O}_2] + \frac{K_2k_6}{K_w}[\text{OH}^-][\text{H}_2\text{O}_2])\sqrt{\frac{k_{3}K_2}{k_{12}K_w}[\text{O}_3][\text{H}_2\text{O}_2][\text{OH}^-]}
$$

<Checkpoint>[Calculated ozone consumption rate: $ v_{\text{O}_3} = k_1[\text{O}_3][\text{OH}^-] + 2\frac{K_2k_3}{K_w}[\text{OH}^-][\text{O}_3][\text{H}_2\text{O}_2] + (2k_4[\text{O}_3] + k_5[\text{H}_2\text{O}_2] + \frac{K_2k_6}{K_w}[\text{OH}^-][\text{H}_2\text{O}_2])\sqrt{\frac{k_{3}K_2}{k_{12}K_w}[\text{O}_3][\text{H}_2\text{O}_2][\text{OH}^-]} $](1.0)</Checkpoint>

From the rate expression, it is evident that only option L is correct.
根据题意，链终止步骤仅为羟基自由基二聚，即：

$$
2\text{HO}\cdot \rightarrow \text{H}_2\text{O}_2\quad k_{12}
$$

对于其他中间产物（自由基和活性中间体），我们可以应用稳态近似，即它们生成速率等于消耗速率。

稳态近似给出：

$$
d[\text{HO}\cdot]/dt = k_3 [\text{O}_3][\text{HO}_2^-] + k_{10} [\text{O}_3^-][\text{H}^+] + k_{11} [\text{O}^-][\text{H}_2\text{O}] - k_4 [\text{O}_3][\text{HO}\cdot] - k_5 [\text{H}_2\text{O}_2][\text{HO}\cdot] - k_6 [\text{HO}_2^-][\text{HO}\cdot] - k_8 [\text{OH}^-][\text{HO}\cdot] - 2 k_{12} [\text{HO}\cdot]^2 = 0
$$

$$
d[\text{HO}_2\cdot]/dt = k_4 [\text{O}_3][\text{HO}\cdot] + k_5 [\text{H}_2\text{O}_2][\text{HO}\cdot] + k_6 [\text{HO}_2^-][\text{HO}\cdot] - k_7 [\text{HO}_2\cdot] + k_{-7} [\text{H}^+][\text{O}_2^-] = 0
$$

$$
d[\text{O}_2^-]/dt = k_3 [\text{O}_3][\text{HO}_2^-] + k_7 [\text{HO}_2\cdot] - k_9 [\text{O}_3][\text{O}_2^-] - k_{-7} [\text{H}^+][\text{O}_2^-] = 0
$$

$$
d[\text{O}_3^-]/dt = k_9 [\text{O}_3][\text{O}_2^-] - k_{10} [\text{O}_3^-][\text{H}^+] = 0
$$

$$
d[\text{O}^-]/dt = k_8 [\text{OH}^-][\text{HO}\cdot] - k_{11} [\text{O}^-][\text{H}_2\text{O}] = 0
$$


五个自由基的式子相加，我们容易得到：

$$
2 k_3 [\text{O}_3][\text{HO}_2^-] - 2k_{12}[\text{HO}\cdot]^2=0
$$

那么可以解出：

$$
[\text{HO}\cdot] = \sqrt{\frac{k_{3}[\text{O}_3][\text{HO}_2^-]}{k_{12}}}
$$

<Checkpoint>[解出：  $ [\text{HO}\cdot] = \sqrt{\frac{k_{3}[\text{O}_3][\text{HO}_2^-]}{k_{12}}} $](1.0)</Checkpoint>

我们知道在碱性溶液中可以视为$\text{H}_2\text{O}_2$几乎完全电离，并且由反应式：

$$
\text{H}_2\text{O}_2 = \text{HO}_2^- + \text{H}^+ \quad K_2
$$

可以写出：

$$
\text{H}_2\text{O}_2 + \text{OH}^- = \text{HO}_2^- + \text{H}_2\text{O} \quad \frac{K_2}{K_w}
$$

那么：

$$
[\text{HO}_2^-] = \frac{K_2}{K_w}[\text{H}_2\text{O}_2][\text{OH}^-]
$$

也就是：

$$
[\text{HO}\cdot] = \sqrt{\frac{k_{3}K_2}{k_{12}K_w}[\text{O}_3][\text{H}_2\text{O}_2][\text{OH}^-]}
$$

<Checkpoint>[解出：$ [\text{HO}\cdot] = \sqrt{\frac{k_{3}K_2}{k_{12}K_w}[\text{O}_3][\text{H}_2\text{O}_2][\text{OH}^-]} $](1.0)</Checkpoint>

根据：

$$
d[\text{O}_2^-]/dt = k_3 [\text{O}_3][\text{HO}_2^-] + k_7 [\text{HO}_2\cdot] - k_9 [\text{O}_3][\text{O}_2^-] - k_{-7} [\text{H}^+][\text{O}_2^-] = 0
$$

和

$$
d[\text{HO}_2\cdot]/dt = k_4 [\text{O}_3][\text{HO}\cdot] + k_5 [\text{H}_2\text{O}_2][\text{HO}\cdot] + k_6 [\text{HO}_2^-][\text{HO}\cdot] - k_7 [\text{HO}_2\cdot] + k_{-7} [\text{H}^+][\text{O}_2^-] = 0
$$

可以消除平衡项，也就是得到：

$$
k_3 [\text{O}_3][\text{HO}_2^-] + k_4 [\text{O}_3][\text{HO}\cdot] + k_5 [\text{H}_2\text{O}_2][\text{HO}\cdot] + k_6 [\text{HO}_2^-][\text{HO}\cdot] - k_9 [\text{O}_3][\text{O}_2^-] = 0
$$



那么可以得到：

$$
[\text{O}_2^-] = \frac{k_3 [\text{O}_3][\text{HO}_2^-] + k_4 [\text{O}_3][\text{HO}\cdot] + k_5 [\text{H}_2\text{O}_2][\text{HO}\cdot] + k_6 [\text{HO}_2^-][\text{HO}\cdot]}{k_9[\text{O}_3]}
$$

<Checkpoint>[消除部分项得到： $ [\text{O}_2^-] = \frac{k_3 [\text{O}_3][\text{HO}_2^-] + k_4 [\text{O}_3][\text{HO}\cdot] + k_5 [\text{H}_2\text{O}_2][\text{HO}\cdot] + k_6 [\text{HO}_2^-][\text{HO}\cdot]}{k_9[\text{O}_3]} $](1.0)</Checkpoint>

把我们之前计算出的结果带入，可以得到：

$$
[\text{O}_2^-] = \frac{\frac{k_3K_2}{K_w} [\text{O}_3][\text{H}_2\text{O}_2][\text{OH}^-] + (k_4 [\text{O}_3] + k_5 [\text{H}_2\text{O}_2] +\frac{k_6K_2}{K_w}[\text{H}_2\text{O}_2][\text{OH}^-])\sqrt{\frac{k_{3}K_2}{k_{12}K_w}[\text{O}_3][\text{H}_2\text{O}_2][\text{OH}^-]}}{k_9[\text{O}_3]}
$$

<Checkpoint>[计算出：$ [\text{O}_2^-] = \frac{\frac{k_3K_2}{K_w} [\text{O}_3][\text{H}_2\text{O}_2][\text{OH}^-] + (k_4 [\text{O}_3] + k_5 [\text{H}_2\text{O}_2] +\frac{k_6K_2}{K_w}[\text{H}_2\text{O}_2][\text{OH}^-])\sqrt{\frac{k_{3}K_2}{k_{12}K_w}[\text{O}_3][\text{H}_2\text{O}_2][\text{OH}^-]}}{k_9[\text{O}_3]} $](1.0)</Checkpoint>

下面我们计算臭氧消耗的速率：

$$
v_{\text{O}_3} = k_1[\text{O}_3][\text{OH}^-] + k_3[\text{O}_3][\text{HO}_2^-] + k_4[\text{O}_3][\text{HO}\cdot] + k_9[\text{O}_3][\text{O}_2^-]
$$

带入前面的结果知道：

$$
v_{\text{O}_3} = k_1[\text{O}_3][\text{OH}^-] + 2\frac{K_2k_3}{K_w}[\text{OH}^-][\text{O}_3][\text{H}_2\text{O}_2] + (2k_4[\text{O}_3] + k_5[\text{H}_2\text{O}_2] + \frac{K_2k_6}{K_w}[\text{OH}^-][\text{H}_2\text{O}_2])\sqrt{\frac{k_{3}K_2}{k_{12}K_w}[\text{O}_3][\text{H}_2\text{O}_2][\text{OH}^-]}
$$

<Checkpoint>[计算出臭氧消耗速率：$ v_{\text{O}_3} = k_1[\text{O}_3][\text{OH}^-] + 2\frac{K_2k_3}{K_w}[\text{OH}^-][\text{O}_3][\text{H}_2\text{O}_2] + (2k_4[\text{O}_3] + k_5[\text{H}_2\text{O}_2] + \frac{K_2k_6}{K_w}[\text{OH}^-][\text{H}_2\text{O}_2])\sqrt{\frac{k_{3}K_2}{k_{12}K_w}[\text{O}_3][\text{H}_2\text{O}_2][\text{OH}^-]} $](1.0)</Checkpoint>

根据速率表达式可知，只有选项L是正确的。