To calculate the enthalpy change [tex]\(\Delta H_{rxn}\)[/tex] for the reaction:
[tex]\[ \text{CH}_3\text{Cl} + \text{O}_2 \rightarrow \text{CO} + \text{HCl} + \text{H}_2\text{O} \][/tex]
we need to manipulate the given thermochemical equations:
1. [tex]\(2 \text{H}_2 + \text{O}_2 \rightarrow 2 \text{H}_2\text{O}\)[/tex], [tex]\(\Delta H_1 = -571 \, \text{kJ}\)[/tex]
2. [tex]\(\text{CO} + 2 \text{H}_2 \rightarrow \text{CH}_3\text{OH}\)[/tex], [tex]\(\Delta H_2 = -139 \, \text{kJ}\)[/tex]
3. [tex]\(\text{CH}_3\text{OH} + \text{HCl} \rightarrow \text{CH}_3\text{Cl} + \text{H}_2\text{O}\)[/tex], [tex]\(\Delta H_3 = -28 \, \text{kJ}\)[/tex]
First, we reverse equation 3 to match our desired reactants and products:
[tex]\[
\text{CH}_3\text{Cl} + \text{H}_2\text{O} \rightarrow \text{CH}_3\text{OH} + \text{HCl}, \Delta H_{\text{reverse 3}} = +28 \, \text{kJ}
\][/tex]
Next, we combine equations to achieve the target reaction. We add equation 1 and the reversed equation 3:
[tex]\[
\begin{align*}
& \text{CH}_3\text{Cl} + \text{H}_2\text{O} + 2 \text{H}_2 + \text{O}_2 \rightarrow \text{CH}_3\text{OH} + \text{HCl} + 2 \text{H}_2\text{O} \\
& \Delta H = -571 \, \text{kJ} + 28 \, \text{kJ} = -543 \, \text{kJ}
\end{align*}
\][/tex]
Then, add equation 2 to this resulting equation:
[tex]\[
\begin{align*}
& \text{CH}_3\text{Cl} + \text{H}_2\text{O} + 2 \text{H}_2 + \text{O}_2 + \text{CO} + 2\text{H}_2 \rightarrow \text{CH}_3\text{OH} + \text{HCl} + 2\text{H}_2\text{O} + \text{CH}_3\text{OH} \\
& \Delta H = -543 \, \text{kJ} - 139 \, \text{kJ} = -682 \, \text{kJ}
\end{align*}
\][/tex]
Thus, the enthalpy change for the reaction:
[tex]\[ \text{CH}_3\text{Cl} + \text{O}_2 \rightarrow \text{CO} + \text{HCl} + \text{H}_2\text{O} \][/tex]
is [tex]\(\Delta H_{rxn} = -682 \, \text{kJ}\)[/tex]
