`DeltaU^@` of combustion of methane is `-X kJ mol^(-1)`. The value of `DeltaH^@` is

Options

A. ` = DeltaU^@`
B. `gt DeltaU^@`
C. `lt DeltaU^@`
D. `=0`

Explanation

To find the value of ΔH° (enthalpy change) for the combustion of methane given that ΔU° (internal energy change) is -X kJ/mol, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Reaction**:
The combustion of methane (CH₄) can be represented by the following balanced chemical equation:
\[ \text{CH}_4(g) + 2\text{O}_2(g) \rightarrow \text{CO}_2(g) + 2\text{H}_2\text{O}(l) \]

2. **Determine the Change in Moles of Gas (ΔN₍g₎)**:
- **Products**: 1 mole of CO₂ (g) + 2 moles of H₂O (l) = 1 mole of gas
- **Reactants**: 1 mole of CH₄ (g) + 2 moles of O₂ (g) = 3 moles of gas
- Therefore, the change in moles of gas (ΔN₍g₎) is:
\[ ΔN₍g₎ = \text{moles of products} - \text{moles of reactants} = 1 - 3 = -2 \]

3. **Use the Relation Between ΔH° and ΔU°**:
The relationship between the change in enthalpy (ΔH°) and the change in internal energy (ΔU°) is given by:
\[ ΔH° = ΔU° + ΔN₍g₎RT \]
Where R is the universal gas constant (approximately 8.314 J/(mol·K)) and T is the temperature in Kelvin.

4. **Substitute the Values**:
Given that ΔU° = -X kJ/mol, we can rewrite the equation:
\[ ΔH° = -X + (-2)RT \]
This indicates that ΔH° is equal to ΔU° plus a term that is negative (since ΔN₍g₎ is -2).

5. **Analyze the Result**:
Since we are adding a negative value (-2RT) to ΔU°, we can conclude that:
\[ ΔH° < ΔU° \]
Hence, ΔH° is less than ΔU°.

### Conclusion:
The value of ΔH° for the combustion of methane is less than ΔU°, which means:
\[ \Delta H° < -X \text{ kJ/mol} \]