IDNLearn.com: Your trusted platform for finding reliable answers. Join our interactive Q&A platform to receive prompt and accurate responses from experienced professionals in various fields.
Sagot :
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]
Thank you for being part of this discussion. Keep exploring, asking questions, and sharing your insights with the community. Together, we can find the best solutions. For clear and precise answers, choose IDNLearn.com. Thanks for stopping by, and come back soon for more valuable insights.