Answer :
Alright, let's break down the problem step-by-step:
We are given two chemical reactions with their respective enthalpy changes (ΔH):
1. [tex]\( \text{CH}_4(g) + 2 \text{O}_2(g) \rightarrow \text{CO}_2(g) + 2 \text{H}_2\text{O}(g) \)[/tex]
[tex]\( \Delta H_1 = -802 \text{ kJ} \)[/tex]
2. [tex]\( 2 \text{H}_2\text{O}(g) \rightarrow 2 \text{H}_2\text{O}(l) \)[/tex]
[tex]\( \Delta H_2 = -88 \text{ kJ} \)[/tex]
We want to find the enthalpy change (ΔH) for the following 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]
To achieve this, notice that the target reaction can essentially be seen as the sum of the two given reactions. Specifically, we start with the reaction of methane combusting in air to form carbon dioxide and water vapor, then we transform the water vapor into liquid water.
Thus, the total enthalpy change for the target reaction is simply the sum of the enthalpy changes for the two individual reactions:
[tex]\[ \Delta H_{\text{target}} = \Delta H_1 + \Delta H_2 \][/tex]
Substituting the given values:
[tex]\[ \Delta H_{\text{target}} = -802 \text{ kJ} + (-88 \text{ kJ}) \][/tex]
[tex]\[ \Delta H_{\text{target}} = -802 \text{ kJ} - 88 \text{ kJ} \][/tex]
[tex]\[ \Delta H_{\text{target}} = -890 \text{ kJ} \][/tex]
Therefore, the equation to calculate [tex]\(\Delta H_{\text{ixn}}\)[/tex] for the target reaction is:
[tex]\[ \Delta H_{\text{ixn}} = \Delta H_1 + \Delta H_2 \][/tex]
And plugging in the values gives us:
[tex]\[ \Delta H_{\text{ixn}} = -890 \text{ kJ} \][/tex]
We are given two chemical reactions with their respective enthalpy changes (ΔH):
1. [tex]\( \text{CH}_4(g) + 2 \text{O}_2(g) \rightarrow \text{CO}_2(g) + 2 \text{H}_2\text{O}(g) \)[/tex]
[tex]\( \Delta H_1 = -802 \text{ kJ} \)[/tex]
2. [tex]\( 2 \text{H}_2\text{O}(g) \rightarrow 2 \text{H}_2\text{O}(l) \)[/tex]
[tex]\( \Delta H_2 = -88 \text{ kJ} \)[/tex]
We want to find the enthalpy change (ΔH) for the following 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]
To achieve this, notice that the target reaction can essentially be seen as the sum of the two given reactions. Specifically, we start with the reaction of methane combusting in air to form carbon dioxide and water vapor, then we transform the water vapor into liquid water.
Thus, the total enthalpy change for the target reaction is simply the sum of the enthalpy changes for the two individual reactions:
[tex]\[ \Delta H_{\text{target}} = \Delta H_1 + \Delta H_2 \][/tex]
Substituting the given values:
[tex]\[ \Delta H_{\text{target}} = -802 \text{ kJ} + (-88 \text{ kJ}) \][/tex]
[tex]\[ \Delta H_{\text{target}} = -802 \text{ kJ} - 88 \text{ kJ} \][/tex]
[tex]\[ \Delta H_{\text{target}} = -890 \text{ kJ} \][/tex]
Therefore, the equation to calculate [tex]\(\Delta H_{\text{ixn}}\)[/tex] for the target reaction is:
[tex]\[ \Delta H_{\text{ixn}} = \Delta H_1 + \Delta H_2 \][/tex]
And plugging in the values gives us:
[tex]\[ \Delta H_{\text{ixn}} = -890 \text{ kJ} \][/tex]
