To calculate the enthalpy change, [tex]\( \Delta H_{\text{rxn}} \)[/tex], for the reaction:
[tex]\[ \text{CH}_4(g) + 2 \text{O}_2(g) \rightarrow \text{CO}_2(g) + 2 \text{H}_2\text{O}(l) \][/tex]
we need to combine the given chemical equations and their respective enthalpy changes. Here are the two given reactions:
1. [tex]\[ \text{CH}_4(g) + 2 \text{O}_2(g) \rightarrow \text{CO}_2(g) + 2 \text{H}_2\text{O}(g) \quad \Delta H_1 = -802 \text{ kJ} \][/tex]
2. [tex]\[ 2 \text{H}_2\text{O}(g) \rightarrow 2 \text{H}_2\text{O}(l) \quad \Delta H_2 = -88 \text{ kJ} \][/tex]
The target reaction is:
[tex]\[ \text{CH}_4(g) + 2 \text{O}_2(g) \rightarrow \text{CO}_2(g) + 2 \text{H}_2\text{O}(l) \][/tex]
This target reaction can be viewed as the combination of the two given reactions:
- The first reaction converts methane and oxygen into carbon dioxide and water vapor (gaseous).
- The second reaction converts the produced water vapor into liquid water.
By combining these steps, we transition from the reactants to the desired products in the target reaction.
Therefore, we need to sum the enthalpy changes from both reactions to determine the overall [tex]\( \Delta H_{\text{rxn}} \)[/tex] for the target reaction.
Adding the enthalpy changes:
[tex]\[ \Delta H_{\text{rxn}} = \Delta H_1 + \Delta H_2 \][/tex]
[tex]\[ \Delta H_{\text{rxn}} = -802 \text{ kJ} + (-88 \text{ kJ}) \][/tex]
[tex]\[ \Delta H_{\text{rxn}} = -890 \text{ kJ} \][/tex]
Thus, the enthalpy change for the given target reaction:
[tex]\[ \text{CH}_4(g) + 2 \text{O}_2(g) \rightarrow \text{CO}_2(g) + 2 \text{H}_2\text{O}(l) \][/tex]
is [tex]\( \Delta H_{\text{rxn}} = -890 \text{ kJ} \)[/tex].
