To calculate the enthalpy change ([tex]\(\Delta H_{rxn}\)[/tex]) for the given reaction:
[tex]\[ \text{CH}_3\text{Cl} + O_2 \rightarrow CO + \text{HCl} + \text{H}_2\text{O} \][/tex]
we will use Hess's Law, which states that the total enthalpy change for a given reaction is the sum of the enthalpy changes of the intermediate steps that lead to that overall reaction.
We will use the following given thermochemical equations and their enthalpy changes ([tex]\(\Delta H\)[/tex]):
1. [tex]\(2 \text{H}_2 + O_2 \rightarrow 2 \text{H}_2\text{O}, \quad \Delta H_1 = -571 \text{ kJ}\)[/tex]
2. [tex]\(\text{CO} + 2 \text{H}_2 \rightarrow \text{CH}_3\text{OH}, \quad \Delta H_2 = -139 \text{ kJ}\)[/tex]
3. [tex]\(\text{CH}_3\text{OH} + \text{HCl} \rightarrow \text{CH}_3\text{Cl} + \text{H}_2\text{O}, \quad \Delta H_3 = -28 \text{ kJ}\)[/tex]
First, we will reverse reaction 3, since we need [tex]\(\text{CH}_3\text{Cl}\)[/tex] as a reactant and not as a product. Reversing the reaction also reverses the sign of [tex]\(\Delta H_3\)[/tex].
[tex]\[ \text{CH}_3\text{Cl} + \text{H}_2\text{O} \rightarrow \text{CH}_3\text{OH} + \text{HCl}, \quad \Delta H = +28 \text{ kJ} \][/tex]
Now, we add this reversed [tex]\( \Delta H \)[/tex] value to the [tex]\(\Delta H\)[/tex] values of reactions 1 and 2:
[tex]\[
\begin{align*}
1. & \quad 2 \text{H}_2 + O_2 \rightarrow 2 \text{H}_2\text{O}, \quad \Delta H_1 = -571 \text{ kJ} \\
2. & \quad \text{CO} + 2 \text{H}_2 \rightarrow \text{CH}_3\text{OH}, \quad \Delta H_2 = -139 \text{ kJ} \\
3. & \quad \text{CH}_3\text{Cl} + \text{H}_2\text{O} \rightarrow \text{CH}_3\text{OH} + \text{HCl}, \quad \Delta H = +28 \text{ kJ} \\
\end{align*}
\][/tex]
Summing these up:
[tex]\[
\Delta H_{\text{reaction}} = \Delta H_1 + \Delta H_2 + \Delta H\text{ (reversed 3)}
\][/tex]
[tex]\[
\Delta H_{\text{reaction}} = -571 \text{ kJ} + (-139 \text{ kJ}) + (28 \text{ kJ})
\][/tex]
[tex]\[
\Delta H_{\text{reaction}} = -571 - 139 + 28 \text{ kJ}
\][/tex]
[tex]\[
\Delta H_{\text{reaction}} = -682 \text{ kJ}
\][/tex]
Therefore, the enthalpy change ([tex]\(\Delta H_{\text{r\times n}}\)[/tex]) for the given reaction is:
[tex]\[
\boxed{-682 \text{ kJ}}
\][/tex]
