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The enthalpies of formation of the compounds in the combustion of methane are given as follows:

[tex]\[ CH_4(g) + 2 O_2(g) \rightarrow CO_2(g) + 2 H_2O(g) \][/tex]

[tex]\[ \Delta H_f \text{ of } CH_4(g) = -74.6 \, \text{kJ/mol} \][/tex]
[tex]\[ \Delta H_f \text{ of } CO_2(g) = -393.5 \, \text{kJ/mol} \][/tex]
[tex]\[ \Delta H_f \text{ of } H_2O(g) = -241.82 \, \text{kJ/mol} \][/tex]

How much heat is released by the combustion of 2 mol of methane?

Use the formula:

[tex]\[ \Delta H_{combustion} = \sum (\Delta H_{f,\text{products}}) - \sum (\Delta H_{f,\text{reactants}}) \][/tex]

Possible answers:

A. [tex]\(-80.3 \, \text{kJ}\)[/tex]
B. [tex]\(-802.5 \, \text{kJ}\)[/tex]
C. [tex]\(-1,605.1 \, \text{kJ}\)[/tex]
D. [tex]\(-6,420.3 \, \text{kJ}\)[/tex]



Answer :

GinnyAnswer
To determine how much heat is released by the combustion of 2 mol of methane, we need to use the enthalpies of formation (ΔH_f) for each compound involved in the reaction. The reaction is given by:

[tex]\[ CH_4(g) + 2 O_2(g) \rightarrow CO_2(g) + 2 H_2O(g) \][/tex]

The enthalpies of formation are given as:
- [tex]\(\Delta H_f \text{ of } CH_4(g) = -74.6 \text{ kJ/mol} \)[/tex]
- [tex]\(\Delta H_f \text{ of } CO_2(g) = -393.5 \text{ kJ/mol} \)[/tex]
- [tex]\(\Delta H_f \text{ of } H_2O(g) = -241.82 \text{ kJ/mol} \)[/tex]

### Step-by-Step Solution

1. Determine the moles of each compound:
- For methane ([tex]\(CH_4\)[/tex]), moles = 2 (as we are considering the combustion of 2 moles of methane).
- For carbon dioxide ([tex]\(CO_2\)[/tex]), moles = 2 (1 mole of [tex]\(CO_2\)[/tex] per 1 mole of [tex]\(CH_4\)[/tex]).
- For water ([tex]\(H_2O\)[/tex]), moles = 4 (2 moles of [tex]\(H_2O\)[/tex] per 1 mole of [tex]\(CH_4\)[/tex]).

2. Calculate the total enthalpy change for the reactants:
- Total enthalpy of the reactants [tex]\( \Delta H_{\text{reactants}} \)[/tex]:
[tex]\[ \Delta H_{\text{reactants}} = n_{\text{CH}_4} \times \Delta H_f \text{ of } CH_4 = 2 \times (-74.6) = -149.2 \text{ kJ} \][/tex]

3. Calculate the total enthalpy change for the products:
- Total enthalpy of the products [tex]\( \Delta H_{\text{products}} \)[/tex]:
[tex]\[ \Delta H_{\text{products}} = (n_{\text{CO}_2} \times \Delta H_f \text{ of } CO_2) + (n_{\text{H}_2O} \times \Delta H_f \text{ of } H_2O) \][/tex]
Substituting the values:
[tex]\[ \Delta H_{\text{products}} = (2 \times -393.5) + (4 \times -241.82) \][/tex]
[tex]\[ = -787 + (-967.28) \][/tex]
[tex]\[ = -1754.28 \text{ kJ} \][/tex]

4. Calculate the enthalpy change (ΔH_combustion) for the combustion reaction:
[tex]\[ \Delta H_{\text{combustion}} = \Delta H_{\text{products}} - \Delta H_{\text{reactants}} \][/tex]
Substituting the values:
[tex]\[ \Delta H_{\text{combustion}} = -1754.28 - (-149.2) \][/tex]
[tex]\[ = -1754.28 + 149.2 \][/tex]
[tex]\[ = -1605.08 \text{ kJ} \][/tex]

Therefore, the heat released by the combustion of 2 moles of methane is [tex]\(-1605.08\)[/tex] kJ. The correct answer is:

[tex]\[ -1,605.1 \text{ kJ} \][/tex]

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