The enthalpies of formation of methane,carbon dioxide and water ( liquid) are - 74.8, -393.5 and-286.2 kJ respectively . Calculate the enthalpy of combustion of methane at ordinary temperature.

Explanation

To calculate the enthalpy of combustion of methane (CH₄) at ordinary temperature, we can use the enthalpies of formation of the reactants and products involved in the combustion reaction. The combustion of methane can be represented by the following balanced chemical equation:

\[ \text{CH}_4(g) + 2 \text{O}_2(g) \rightarrow \text{CO}_2(g) + 2 \text{H}_2\text{O}(l) \]

### Step-by-Step Solution:

1. **Identify the Enthalpies of Formation**:
- Enthalpy of formation of methane (CH₄): \( \Delta H_f^\circ (\text{CH}_4) = -74.8 \, \text{kJ/mol} \)
- Enthalpy of formation of carbon dioxide (CO₂): \( \Delta H_f^\circ (\text{CO}_2) = -393.5 \, \text{kJ/mol} \)
- Enthalpy of formation of water (H₂O): \( \Delta H_f^\circ (\text{H}_2\text{O}) = -286.2 \, \text{kJ/mol} \)

2. **Write the Enthalpy of Combustion Equation**:
The enthalpy of combustion (\( \Delta H_c \)) can be calculated using the formula:
\[ \Delta H_c = \sum (\Delta H_f^\circ \text{ of products}) - \sum (\Delta H_f^\circ \text{ of reactants}) \]

3. **Substitute the Values**:
- Products: 1 mole of CO₂ and 2 moles of H₂O
- Reactants: 1 mole of CH₄ and 2 moles of O₂ (O₂ is in its elemental form, so its enthalpy of formation is 0)

Therefore, we have:
\[ \Delta H_c = [\Delta H_f^\circ (\text{CO}_2) + 2 \times \Delta H_f^\circ (\text{H}_2\text{O})] - [\Delta H_f^\circ (\text{CH}_4) + 2 \times \Delta H_f^\circ (\text{O}_2)] \]
\[ \Delta H_c = [-393.5 \, \text{kJ} + 2 \times (-286.2 \, \text{kJ})] - [-74.8 \, \text{kJ} + 0] \]

4. **Calculate the Enthalpy of Combustion**:
- Calculate the total enthalpy of products:
\[ \Delta H_c = [-393.5 \, \text{kJ} - 572.4 \, \text{kJ}] - [-74.8 \, \text{kJ}] \]
\[ = -965.9 \, \text{kJ} + 74.8 \, \text{kJ} \]
\[ = -891.1 \, \text{kJ} \]

5. **Final Result**:
The enthalpy of combustion of methane at ordinary temperature is:
\[ \Delta H_c = -891.1 \, \text{kJ/mol} \]