There is an empirical rule that can be used to estimate the activation energy of elementary reactions. Under the following conditions, the activation energy can be calculated using the corresponding methods in the table below:  
① Reactions without bond breaking;  
② Exothermic reactions between radicals and molecules;  
③ Reactions with incomplete bond breaking (e.g., concerted reactions);  
④ Reactions with complete bond breaking.  

| Activation Energy/BE | 0 | 0.05 | 0.3  | 1 |
|---|---|---|---|---|
| | ① | ② | ③ | ④ |

$\mathrm{BE}$ is the sum of the bond energies of the broken bonds in the reaction.  

Based on the above empirical rule and the provided bond energy information, estimate the activation energies of each elementary reaction in the reaction between hydrogen and chlorine, and compare them with the given experimental values.  

$(A)~Cl_2=2Cl\cdot$  $E_{a1}=242.8kJ\cdot mol^{-1}$  

$(B)~Cl\cdot+H_2=HCl+H\cdot$  
$E_{a2}=25.1kJ\cdot mol^{-1}$  

$(C)~H\cdot+Cl_2=HCl+Cl\cdot$  
$E_{a3}=12.6kJ\cdot mol^{-1}$  

$BE(Cl-Cl)=243kJ\cdot mol^{-1}, BE(H-H)=436kJ\cdot mol^{-1}, BE(H-Cl)=431kJ\cdot mol^{-1}$
有一个经验规则可以用于估算基元反应的活化能，遵循下列情况时，可分别通过对应下表中的计算方法计算活化能：
①反应中没有键的断裂的反应；②自由基与分子的放热反应；③反应中键不完全断裂(如协同反应)的反应；④反应中键完全断裂的反应。

| 活化能/BE | 0 | 0.05 | 0.3  | 1 |
|---|---|---|---|---|
| | ① | ② | ③ | ④ |

$\mathrm{BE}$ 是反应中断掉的键的键能之和。

根据上述经验规则及下面所提供的键能信息，估算氢气与氯气反应中各步基元反应的活化能，并和给出的活化能实验值作比较。

$(A)~Cl_2=2Cl\cdot$  $E_{a1}=242.8kJ\cdot mol^{-1}$

$(B)~Cl\cdot+H_2=HCl+H\cdot$  
$E_{a2}=25.1kJ\cdot mol^{-1}$

$(C)~H\cdot+Cl_2=HCl+Cl\cdot$  
$E_{a3}=12.6kJ\cdot mol^{-1}$

$BE(Cl-Cl)=243kJ\cdot mol^{-1},BE(H-H)=436kJ\cdot mol^{-1},BE(H-Cl)=431kJ\cdot mol^{-1}$
The estimated activation energy for reaction (C) is in complete agreement with the experimental value.
The activation energy of reaction (A) is 0
The estimated activation energy for reaction (B) is $3.3kJ\cdot mol^{-1}$ lower than the experimental value.
The estimated activation energy for reaction (B) is greater than the experimental value by $1.4kJ\cdot mol^{-1}$.
Using the method described in the problem, the relative error in the activation energy of reaction (C) is calculated to be 3.17%.
None of the above options are correct
反应(C)的活化能估算值和实验值完全一致
反应(A)的活化能为0
反应(B)的活化能估算值比实验值小$3.3kJ\cdot mol^{-1}$
反应(B)的活化能估算值比实验值大$1.4kJ\cdot mol^{-1}$
用题述方法算得反应(C)的活化能的相对误差为3.17%
以上选项均不对
Reaction (A): $E_{a1}=BE(Cl-Cl)=243kJ\cdot mol^{-1}$  

<Checkpoint>[$E_{a1}=243kJ\cdot mol^{-1}$](1)</Checkpoint>  

Reaction (B): $\Delta H=BE(H-H)-BE(H-Cl)=5kJ\cdot mol^{-1}>0$  

<Checkpoint>[$\Delta H>0$](1)</Checkpoint>  

The empirical formula $E_a=0.05\cdot BE(H-H)=21.8kJ\cdot mol^{-1}$ cannot be directly applied. Instead, calculate using the relationship: activation energy of the forward reaction - activation energy of the reverse reaction = $\Delta H$:  

$E_{a2}=E_{a2,\text{reverse}}+\Delta H=0.05\cdot BE(H-Cl)+5kJ\cdot mol^{-1}=26.5kJ\cdot mol^{-1}$  

<Checkpoint>[Calculate using activation energy of forward reaction - activation energy of reverse reaction = $\Delta H$](1)</Checkpoint>  

<Checkpoint>[$E_{a2}=26.5kJ\cdot mol^{-1}$](1)</Checkpoint>  

Reaction (C): $\Delta H=BE(Cl-Cl)-BE(H-Cl)=-188kJ\cdot mol^{-1}<0$. The empirical formula $E_{a3}=0.05·BE(Cl-Cl)=12.2 kJ\cdot mol^{-1}$ is directly applied.  

<Checkpoint>[$E_{a3}=12.2kJ\cdot mol^{-1}$](1)</Checkpoint>  

Note: The relative error mentioned in option E is $-3.17\%$.
反应(A)：$E_{a1}=BE(Cl-Cl)=243kJ\cdot mol^{-1}$

<Checkpoint>[$E_{a1}=243kJ\cdot mol^{-1}$](1)</Checkpoint>

反应(B)：$\Delta H=BE(H-H)-BE(H-Cl)=5kJ\cdot mol^{-1}>0$

<Checkpoint>[$\Delta H>0$](1)</Checkpoint>

不能直接使用经验公式$E_a=0.05\cdot BE(H-H)=21.8kJ\cdot mol^{-1}$，应用正反应活化能-逆反应活化能=$\Delta H$求算：

$E_{a2}=E_{a2逆}+\Delta H=0.05\cdot BE(H-Cl)+5kJ\cdot mol^{-1}=26.5kJ\cdot mol^{-1}$

<Checkpoint>[用正反应活化能-逆反应活化能=$\Delta H$求算](1)</Checkpoint>

<Checkpoint>[$E_{a2}=26.5kJ\cdot mol^{-1}$](1)</Checkpoint>

反应(C)：$\Delta H=BE(Cl-Cl)-BE(H-Cl)=-188kJ\cdot mol^{-1}<0$，直接使用经验公式$E_{a3}=0.05·BE(Cl-Cl)=12.2 kJ\cdot mol^{-1}$

<Checkpoint>[$E_{a3}=12.2kJ\cdot mol^{-1}$](1)</Checkpoint>

注：选项E提到的相对误差是$-3.17\%$