The heat of combustion of `CH_(4)(g)` at constant volume is measured in a bomb calorimeter at 298.2 K and found to be -885389 `J // mol` . Find the value of enthalpy change.

Answer

To find the enthalpy change (ΔH) for the combustion of methane (CH₄), we start with the heat of combustion measured at constant volume, which is given as -885389 J/mol. We will follow these steps:

### Step-by-Step Solution:

1. **Convert Heat of Combustion to kJ/mol**:
\[ \Delta U = -885389 \, \text{J/mol} = -885.389 \, \text{kJ/mol} \]

2. **Identify the Reaction**:
The balanced chemical equation for the combustion of methane is:
\[ \text{CH}_4(g) + 2\text{O}_2(g) \rightarrow \text{CO}_2(g) + 2\text{H}_2\text{O}(l) \]

3. **Determine Δn (Change in Moles of Gas)**:
- Count the moles of gaseous products and reactants:
- Gaseous products: 1 (from CO₂)
- Gaseous reactants: 3 (1 from CH₄ and 2 from O₂)
\[ \Delta n = \text{(moles of gaseous products)} - \text{(moles of gaseous reactants)} = 1 - 3 = -2 \]

4. **Use the Relationship Between ΔH and ΔU**:
The relationship is given by:
\[ \Delta H = \Delta U + R T \Delta n \]
Where:
- \( R = 8.314 \, \text{J/mol·K} = 0.008314 \, \text{kJ/mol·K} \)
- \( T = 298.2 \, \text{K} \)

5. **Calculate \( R T \Delta n \)**:
\[ R T \Delta n = 0.008314 \, \text{kJ/mol·K} \times 298.2 \, \text{K} \times (-2) \]
\[ R T \Delta n = -4.96 \, \text{kJ/mol} \]

6. **Calculate ΔH**:
\[ \Delta H = -885.389 \, \text{kJ/mol} + (-4.96 \, \text{kJ/mol}) = -890.349 \, \text{kJ/mol} \]

7. **Final Result**:
\[ \Delta H \approx -890.34 \, \text{kJ/mol} \]

### Summary:
The enthalpy change (ΔH) for the combustion of methane (CH₄) is approximately -890.34 kJ/mol.