Answer :
To determine the amount of heat released by the combustion of 2 mol of methane (\(CH_4\)), we will use the given enthalpies of formation (\(\Delta H_f\)) and the balanced chemical equation:
[tex]\[ CH_4(g) + 2 O_2(g) \rightarrow CO_2(g) + 2 H_2O(g) \][/tex]
### Step-by-Step Solution:
1. Identify the enthalpies of formation for each compound:
- \(\Delta H_f \text{ (CH}_4(g)\text{)} = -74.6 \text{ kJ/mol}\)
- \(\Delta H_f \text{ (CO}_2(g)\text{)} = -393.5 \text{ kJ/mol}\)
- \(\Delta H_f \text{ (H}_2O(g)\text{)} = -241.82 \text{ kJ/mol}\)
2. Determine the coefficients from the balanced chemical equation:
- \(1 \text{ mol of CH}_4\) reacts with \(2 \text{ mols of O}_2\) to produce \(1 \text{ mol of CO}_2\) and \(2 \text{ mols of H}_2O\).
3. Apply the formula for the enthalpy change of the reaction \(\Delta H_{\text{rxn}}\):
[tex]\[ \Delta H_{\text{rxn}} = \sum (\Delta H_{\text{f,products}}) - \sum (\Delta H_{\text{f,reactants}}) \][/tex]
4. Calculate the sum of the enthalpies of the products and reactants:
- Products:
- \(\Delta H_f \text{ (CO}_2(g)\text{)} = -393.5 \text{ kJ/mol}\)
- \(2 \times \Delta H_f \text{ (H}_2O(g)\text{)} = 2 \times (-241.82 \text{ kJ/mol}) = -483.64 \text{ kJ}\)
- Total \(\Delta H_{\text{f,products}}\) = \(-393.5 + (-483.64) = -877.14 \text{ kJ}\)
- Reactant (\(CH_4\)):
- \(\Delta H_f \text{ (CH}_4(g)\text{)} = -74.6 \text{ kJ/mol}\)
- Reactant (\(O_2\)):
- Note: \(\Delta H_f \text{ (O}_2(g)\text{)}\) is zero for elemental forms in their standard states.
5. Calculate \(\Delta H_{\text{rxn}} \text{ per mol}\):
[tex]\[ \Delta H_{\text{rxn per mol}} = -877.14 \text{ kJ} - (-74.6 \text{ kJ}) = -802.54 \text{ kJ/mol} \][/tex]
6. Calculate the total enthalpy change for the combustion of 2 mol of \(CH_4\):
[tex]\[ \Delta H_{\text{total}} = 2 \text{ mols} \times (-802.54 \text{ kJ/mol}) = -1605.08 \text{ kJ} \][/tex]
### Answer
The amount of heat released by the combustion of 2 mol of methane is \(-1605.1 \text{ kJ}\). This corresponds to the third option provided:
[tex]\[ \boxed{-1605.1 \text{ kJ}} \][/tex]
[tex]\[ CH_4(g) + 2 O_2(g) \rightarrow CO_2(g) + 2 H_2O(g) \][/tex]
### Step-by-Step Solution:
1. Identify the enthalpies of formation for each compound:
- \(\Delta H_f \text{ (CH}_4(g)\text{)} = -74.6 \text{ kJ/mol}\)
- \(\Delta H_f \text{ (CO}_2(g)\text{)} = -393.5 \text{ kJ/mol}\)
- \(\Delta H_f \text{ (H}_2O(g)\text{)} = -241.82 \text{ kJ/mol}\)
2. Determine the coefficients from the balanced chemical equation:
- \(1 \text{ mol of CH}_4\) reacts with \(2 \text{ mols of O}_2\) to produce \(1 \text{ mol of CO}_2\) and \(2 \text{ mols of H}_2O\).
3. Apply the formula for the enthalpy change of the reaction \(\Delta H_{\text{rxn}}\):
[tex]\[ \Delta H_{\text{rxn}} = \sum (\Delta H_{\text{f,products}}) - \sum (\Delta H_{\text{f,reactants}}) \][/tex]
4. Calculate the sum of the enthalpies of the products and reactants:
- Products:
- \(\Delta H_f \text{ (CO}_2(g)\text{)} = -393.5 \text{ kJ/mol}\)
- \(2 \times \Delta H_f \text{ (H}_2O(g)\text{)} = 2 \times (-241.82 \text{ kJ/mol}) = -483.64 \text{ kJ}\)
- Total \(\Delta H_{\text{f,products}}\) = \(-393.5 + (-483.64) = -877.14 \text{ kJ}\)
- Reactant (\(CH_4\)):
- \(\Delta H_f \text{ (CH}_4(g)\text{)} = -74.6 \text{ kJ/mol}\)
- Reactant (\(O_2\)):
- Note: \(\Delta H_f \text{ (O}_2(g)\text{)}\) is zero for elemental forms in their standard states.
5. Calculate \(\Delta H_{\text{rxn}} \text{ per mol}\):
[tex]\[ \Delta H_{\text{rxn per mol}} = -877.14 \text{ kJ} - (-74.6 \text{ kJ}) = -802.54 \text{ kJ/mol} \][/tex]
6. Calculate the total enthalpy change for the combustion of 2 mol of \(CH_4\):
[tex]\[ \Delta H_{\text{total}} = 2 \text{ mols} \times (-802.54 \text{ kJ/mol}) = -1605.08 \text{ kJ} \][/tex]
### Answer
The amount of heat released by the combustion of 2 mol of methane is \(-1605.1 \text{ kJ}\). This corresponds to the third option provided:
[tex]\[ \boxed{-1605.1 \text{ kJ}} \][/tex]
