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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].
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