To find the enthalpy change [tex]\(\Delta H_{rxn}\)[/tex] for the reaction:
[tex]\[CH_4(g) + 2O_2(g) \rightarrow CO_2(g) + 2H_2O(l)\][/tex]
we need to combine the enthalpy changes of the given reactions step-by-step. Let's go through the steps methodically:
1. Identify the given reactions and their enthalpy changes:
[tex]\[
\begin{array}{l}
CH_4(g) + 2O_2(g) \rightarrow CO_2(g) + 2H_2O(g) \quad \Delta H_1 = -802 \, \text{kJ} \\
2H_2O(g) \rightarrow 2H_2O(l) \quad \Delta H_2 = -88 \, \text{kJ}
\end{array}
\][/tex]
2. Combine these reactions to form the desired reaction:
We start with the initial reaction:
[tex]\[
CH_4(g) + 2O_2(g) \rightarrow CO_2(g) + 2H_2O(g)
\][/tex]
Next, we convert the water from the gas phase to the liquid phase:
[tex]\[
2H_2O(g) \rightarrow 2H_2O(l)
\][/tex]
3. Add the enthalpy changes of the reactions to find [tex]\(\Delta H_{rxn}\)[/tex] for the overall reaction:
When combining reactions, the enthalpy changes are added together. Hence, we sum [tex]\(\Delta H_1\)[/tex] and [tex]\(\Delta H_2\)[/tex]:
[tex]\[
\Delta H_{rxn} = \Delta H_1 + \Delta H_2
\][/tex]
Substituting the given values:
[tex]\[
\Delta H_{rxn} = -802 \, \text{kJ} + (-88 \, \text{kJ})
\][/tex]
4. Perform the addition:
[tex]\[
\Delta H_{rxn} = -802 \, \text{kJ} - 88 \, \text{kJ} = -890 \, \text{kJ}
\][/tex]
Thus, the enthalpy change [tex]\(\Delta H_{rxn}\)[/tex] for the reaction [tex]\(CH_4(g) + 2O_2(g) \rightarrow CO_2(g) + 2H_2O(l)\)[/tex] is [tex]\(-890 \, \text{kJ}\)[/tex].
