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The enthalpies of formation of the compounds in the combustion of methane are given:

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

[tex]\[
CH_4(g): \Delta H_f = -74.6 \, \text{kJ/mol}, \quad CO_2(g): \Delta H_f = -393.5 \, \text{kJ/mol}, \quad H_2O(g): \Delta H_f = -241.82 \, \text{kJ/mol}
\][/tex]

How much heat is released by the combustion of 2 mol of methane?

Use the equation:
[tex]\[
\Delta H_{\text{reaction}} = \sum (\Delta H_{\text{f, products}}) - \sum (\Delta H_{\text{f, reactants}})
\][/tex]

A. \(-80.3 \, \text{kJ}\)
B. \(802.5 \, \text{kJ}\)
C. \(-1,605.1 \, \text{kJ}\)
D. [tex]\(-6,420.3 \, \text{kJ}\)[/tex]



Answer :

GinnyAnswer
To calculate the heat released during the combustion of 2 moles of methane, we can follow a step-by-step approach using the given enthalpies of formation for each compound:

1. Write the balanced chemical equation:
[tex]\[ \text{CH}_4(g) + 2\text{O}_2(g) \rightarrow \text{CO}_2(g) + 2\text{H}_2\text{O}(g) \][/tex]

2. Identify the enthalpies of formation (\(\Delta H_f\)) for each compound:
- \(\Delta H_f (\text{CH}_4) = -74.6 \ \text{kJ/mol}\)
- \(\Delta H_f (\text{CO}_2) = -393.5 \ \text{kJ/mol}\)
- \(\Delta H_f (\text{H}_2\text{O}) = -241.82 \ \text{kJ/mol}\)
- \(\Delta H_f (\text{O}_2) = 0 \ \text{kJ/mol}\) (since \(\text{O}_2\) is in its standard state)

3. Calculate the total enthalpy of the reactants:
- Only \(\text{CH}_4\) and \(\text{O}_2\) are reactants, with their respective moles given:
[tex]\[ \text{Total } \Delta H_{\text{reactants}} = (1 \ \text{mol} \times -74.6 \ \text{kJ/mol}) + (2 \ \text{mol} \times 0 \ \text{kJ/mol}) = -74.6 \ \text{kJ} \][/tex]

4. Calculate the total enthalpy of the products:
- The products are \(\text{CO}_2\) and \(\text{H}_2\text{O}\):
[tex]\[ \text{Total } \Delta H_{\text{products}} = (1 \ \text{mol} \times -393.5 \ \text{kJ/mol}) + (2 \ \text{mol} \times -241.82 \ \text{kJ/mol}) = -393.5 + (2 \times -241.82) = -877.14 \ \text{kJ} \][/tex]

5. Calculate the enthalpy change for the reaction:
- Using the formula \(\Delta H_{\text{reaction}} = \sum \Delta H_{\text{products}} - \sum \Delta H_{\text{reactants}}\):
[tex]\[ \Delta H_{\text{reaction}} = -877.14 \ \text{kJ} - (-74.6 \ \text{kJ}) = -877.14 + 74.6 = -802.54 \ \text{kJ} \][/tex]

6. Determine the heat released for 2 moles of methane:
- Since the reaction enthalpy change is calculated per mole of methane, for 2 moles of methane:
[tex]\[ \text{Heat released} = \Delta H_{\text{reaction}} \times 2 = -802.54 \ \text{kJ/mol} \times 2 = -1605.08 \ \text{kJ} \][/tex]

Therefore, the heat released by the combustion of 2 moles of methane is \(-1605.08 \ \text{kJ}\). Given the options, the closest match is:

[tex]\[ \boxed{-1605.1 \ \text{kJ}} \][/tex]

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