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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]

What is [tex]\(\Delta H_{\text{rxn}}\)[/tex] for the overall reaction?
[tex]\[ \boxed{\text{kJ}} \][/tex]

Sagot :

GinnyAnswer
To determine the enthalpy change ([tex]\(\Delta H_{\text{rxn}}\)[/tex]) for the overall 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 consider the enthalpy changes ([tex]\(\Delta H\)[/tex]) of the two given steps:

1. For the reaction:
[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. For the phase change:
[tex]\[ 2 \text{H}_2\text{O}(g) \rightarrow 2 \text{H}_2\text{O}(l), \quad \Delta H_2 = -88 \, \text{kJ} \][/tex]

We can combine these two reactions to find the overall enthalpy change for the conversion of methane and oxygen into carbon dioxide and liquid water.

The overall reaction is formed by summing these two reactions:

[tex]\[ \left( \text{CH}_4(g) + 2 \text{O}_2(g) \rightarrow \text{CO}_2(g) + 2 \text{H}_2\text{O}(g), \quad \Delta H_1 = -802 \, \text{kJ} \right) \][/tex]

[tex]\[ + \][/tex]

[tex]\[ \left( 2 \text{H}_2\text{O}(g) \rightarrow 2 \text{H}_2\text{O}(l), \quad \Delta H_2 = -88 \, \text{kJ} \right) \][/tex]

Thus, the overall 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]

To find the overall enthalpy change ([tex]\(\Delta H_{\text{rxn}}\)[/tex]), we add the enthalpy changes of the two steps:

[tex]\[ \Delta H_{\text{rxn}} = \Delta H_1 + \Delta H_2 \][/tex]

Given that:

[tex]\[ \Delta H_1 = -802 \, \text{kJ} \][/tex]
[tex]\[ \Delta H_2 = -88 \, \text{kJ} \][/tex]

Combining these, we get:

[tex]\[ \Delta H_{\text{rxn}} = -802 \, \text{kJ} + -88 \, \text{kJ} \][/tex]
[tex]\[ \Delta H_{\text{rxn}} = -890 \, \text{kJ} \][/tex]

Therefore, the enthalpy change ([tex]\(\Delta H_{\text{rxn}}\)[/tex]) for the overall reaction is:

[tex]\[ -890 \, \text{kJ} \][/tex]
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