Answer :
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]
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]
