Explore Westonci.ca, the leading Q&A site where experts provide accurate and helpful answers to all your questions. Get quick and reliable solutions to your questions from knowledgeable professionals on our comprehensive Q&A platform. Join our Q&A platform to connect with experts dedicated to providing accurate answers to your questions in various fields.
Sagot :
To solve the problem of calculating the amount of heat released by the combustion of 2 moles of methane (\( CH_4 \)), we need to use the given enthalpies of formation (\( \Delta H_f \)). The combustion reaction is:
[tex]\[ CH_4(g) + 2 O_2(g) \rightarrow CO_2(g) + 2 H_2O(g). \][/tex]
The given enthalpies of formation are:
- \( \Delta H_f \) for \( CH_4(g) \) = -74.6 kJ/mol
- \( \Delta H_f \) for \( CO_2(g) \) = -393.5 kJ/mol
- \( \Delta H_f \) for \( H_2O(g) \) = -241.82 kJ/mol
We use the formula for the enthalpy change of the reaction:
[tex]\[ \Delta H_{reaction} = \sum\left(\Delta H_{f,\text{products}}\right) - \sum\left(\Delta H_{f,\text{reactants}}\right) \][/tex]
Step-by-Step Solution:
1. Calculate the total enthalpy of formation for the products:
In the balanced chemical equation, the products are 1 mole of \( CO_2(g) \) and 2 moles of \( H_2O(g) \).
[tex]\[ \Delta H_{f,\text{products}} = \Delta H_f \text{ (for } CO_2) + 2 \times \Delta H_f \text{ (for } H_2O) \][/tex]
Substituting the given values:
[tex]\[ \Delta H_{f,\text{products}} = (-393.5 \, \text{kJ/mol}) + 2 \times (-241.82 \, \text{kJ/mol}) \][/tex]
[tex]\[ \Delta H_{f,\text{products}} = -393.5 \, \text{kJ/mol} - 483.64 \, \text{kJ/mol} \][/tex]
[tex]\[ \Delta H_{f,\text{products}} = -877.14 \, \text{kJ} \][/tex]
2. Calculate the total enthalpy of formation for the reactants:
In the balanced chemical equation, the reactants are 1 mole of \( CH_4(g) \) and 2 moles of \( O_2(g) \). Since \( O_2(g) \) is in its elemental form, its enthalpy of formation is zero.
[tex]\[ \Delta H_{f,\text{reactants}} = \Delta H_f \text{ (for } CH_4) \][/tex]
Substituting the given values:
[tex]\[ \Delta H_{f,\text{reactants}} = (-74.6 \, \text{kJ/mol}) \][/tex]
3. Calculate the enthalpy change (\( \Delta H \)) for the combustion of 1 mole of methane:
[tex]\[ \Delta H_{reaction} = \Delta H_{f,\text{products}} - \Delta H_{f,\text{reactants}} \][/tex]
Substituting the values:
[tex]\[ \Delta H_{reaction} = -877.14 \, \text{kJ} - (-74.6 \, \text{kJ}) \][/tex]
[tex]\[ \Delta H_{reaction} = -877.14 \, \text{kJ} + 74.6 \, \text{kJ} \][/tex]
[tex]\[ \Delta H_{reaction} = -802.54 \, \text{kJ/mol} \][/tex]
4. Calculate the heat released for the combustion of 2 moles of methane:
For 2 moles of \( CH_4 \):
[tex]\[ \text{Total heat released} = 2 \times \Delta H_{reaction} \][/tex]
Substituting the values:
[tex]\[ \text{Total heat released} = 2 \times (-802.54 \, \text{kJ/mol}) \][/tex]
[tex]\[ \text{Total heat released} = -1605.08 \, \text{kJ} \][/tex]
Thus, the correct answer is [tex]\( \boxed{-1605.1 \, \text{kJ}} \)[/tex].
[tex]\[ CH_4(g) + 2 O_2(g) \rightarrow CO_2(g) + 2 H_2O(g). \][/tex]
The given enthalpies of formation are:
- \( \Delta H_f \) for \( CH_4(g) \) = -74.6 kJ/mol
- \( \Delta H_f \) for \( CO_2(g) \) = -393.5 kJ/mol
- \( \Delta H_f \) for \( H_2O(g) \) = -241.82 kJ/mol
We use the formula for the enthalpy change of the reaction:
[tex]\[ \Delta H_{reaction} = \sum\left(\Delta H_{f,\text{products}}\right) - \sum\left(\Delta H_{f,\text{reactants}}\right) \][/tex]
Step-by-Step Solution:
1. Calculate the total enthalpy of formation for the products:
In the balanced chemical equation, the products are 1 mole of \( CO_2(g) \) and 2 moles of \( H_2O(g) \).
[tex]\[ \Delta H_{f,\text{products}} = \Delta H_f \text{ (for } CO_2) + 2 \times \Delta H_f \text{ (for } H_2O) \][/tex]
Substituting the given values:
[tex]\[ \Delta H_{f,\text{products}} = (-393.5 \, \text{kJ/mol}) + 2 \times (-241.82 \, \text{kJ/mol}) \][/tex]
[tex]\[ \Delta H_{f,\text{products}} = -393.5 \, \text{kJ/mol} - 483.64 \, \text{kJ/mol} \][/tex]
[tex]\[ \Delta H_{f,\text{products}} = -877.14 \, \text{kJ} \][/tex]
2. Calculate the total enthalpy of formation for the reactants:
In the balanced chemical equation, the reactants are 1 mole of \( CH_4(g) \) and 2 moles of \( O_2(g) \). Since \( O_2(g) \) is in its elemental form, its enthalpy of formation is zero.
[tex]\[ \Delta H_{f,\text{reactants}} = \Delta H_f \text{ (for } CH_4) \][/tex]
Substituting the given values:
[tex]\[ \Delta H_{f,\text{reactants}} = (-74.6 \, \text{kJ/mol}) \][/tex]
3. Calculate the enthalpy change (\( \Delta H \)) for the combustion of 1 mole of methane:
[tex]\[ \Delta H_{reaction} = \Delta H_{f,\text{products}} - \Delta H_{f,\text{reactants}} \][/tex]
Substituting the values:
[tex]\[ \Delta H_{reaction} = -877.14 \, \text{kJ} - (-74.6 \, \text{kJ}) \][/tex]
[tex]\[ \Delta H_{reaction} = -877.14 \, \text{kJ} + 74.6 \, \text{kJ} \][/tex]
[tex]\[ \Delta H_{reaction} = -802.54 \, \text{kJ/mol} \][/tex]
4. Calculate the heat released for the combustion of 2 moles of methane:
For 2 moles of \( CH_4 \):
[tex]\[ \text{Total heat released} = 2 \times \Delta H_{reaction} \][/tex]
Substituting the values:
[tex]\[ \text{Total heat released} = 2 \times (-802.54 \, \text{kJ/mol}) \][/tex]
[tex]\[ \text{Total heat released} = -1605.08 \, \text{kJ} \][/tex]
Thus, the correct answer is [tex]\( \boxed{-1605.1 \, \text{kJ}} \)[/tex].
Thanks for using our service. We aim to provide the most accurate answers for all your queries. Visit us again for more insights. We hope our answers were useful. Return anytime for more information and answers to any other questions you have. Thank you for visiting Westonci.ca. Stay informed by coming back for more detailed answers.
