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Consider the chemical equations shown here.

[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]

Which equation shows how to calculate [tex]\(\Delta H_{\text{rxn}}\)[/tex] for the equation below?

[tex]\[
CH_4(g) + 2O_2(g) \rightarrow CO_2(g) + 2H_2O(l)
\][/tex]


Sagot :

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].