To calculate the enthalpy change, [tex]\( \Delta H_{\text{rxn}} \)[/tex], for the reaction:
[tex]\[ CH_4(g) + 2 O_2(g) \rightarrow CO_2(g) + 2 H_2O (l), \][/tex]
we need to combine the enthalpy changes of the given reactions. Here are the two reactions provided:
1. [tex]\( CH_4(g) + 2 O_2(g) \rightarrow CO_2(g) + 2 H_2O(g) \)[/tex], with [tex]\( \Delta H_1 = -802 \text{ kJ} \)[/tex]
2. [tex]\( 2 H_2O(g) \rightarrow 2 H_2O(l) \)[/tex], with [tex]\( \Delta H_2 = -88 \text{ kJ} \)[/tex]
The first reaction represents the combustion of methane yielding carbon dioxide and water in the gaseous state. The second reaction represents the phase change of water from gas to liquid.
To find the overall [tex]\( \Delta H_{\text{rxn}} \)[/tex] for the formation of liquid water directly from the combustion of methane, we sum the enthalpy changes of the individual steps:
[tex]\[ \Delta H_{\text{rxn}} = \Delta H_1 + \Delta H_2 \][/tex]
Given the values:
[tex]\[ \Delta H_1 = -802 \text{ kJ} \][/tex]
[tex]\[ \Delta H_2 = -88 \text{ kJ} \][/tex]
Add these together:
[tex]\[ \Delta H_{\text{rxn}} = -802 \text{ kJ} + -88 \text{ kJ} \][/tex]
[tex]\[ \Delta H_{\text{rxn}} = -890 \text{ kJ} \][/tex]
Thus, the equation to calculate [tex]\( \Delta H_{\text{rxn}} \)[/tex] for the reaction:
[tex]\[ CH_4(g) + 2 O_2(g) \rightarrow CO_2(g) + 2 H_2O (l) \][/tex]
is:
[tex]\[ \Delta H_{\text{rxn}} = \Delta H_1 + \Delta H_2 = -802 \text{ kJ} + -88 \text{ kJ} = -890 \text{ kJ} \][/tex]
