`DeltaU^(Θ)` of combustion of methane is `-XkJ mol^(-1)`. The value of `DeltaH^(Θ)` is

Options

A. `=DeltaU^(Θ)`
B. `gt DeltaU^(Θ)`
C. `lt DeltaU^(Θ)`
D. zero

Explanation

To find the value of ΔH for the combustion of methane when ΔU is given as -X kJ/mol, we can follow these steps:

### Step-by-Step Solution:

1. **Write the Combustion Reaction of Methane:**
The balanced chemical equation for the combustion of methane (CH₄) is:
\[ \text{CH}_4(g) + 2\text{O}_2(g) \rightarrow \text{CO}_2(g) + 2\text{H}_2\text{O}(l) \]

2. **Identify the States of Matter:**
- Reactants: CH₄ (g), O₂ (g)
- Products: CO₂ (g), H₂O (l)

3. **Calculate ΔNG (Change in Moles of Gas):**
- **Products:** 1 mole of CO₂ (g) = 1 mole
- **Reactants:** 1 mole of CH₄ (g) + 2 moles of O₂ (g) = 3 moles
- Therefore, ΔNG = Moles of gaseous products - Moles of gaseous reactants:
\[ \Delta N_G = 1 - 3 = -2 \]

4. **Use the Relationship Between ΔH and ΔU:**
The relationship is given by the equation:
\[ \Delta H = \Delta U + \Delta N_G RT \]
Substituting ΔNG and ΔU:
\[ \Delta H = -X + (-2)RT \]

5. **Analyze the Sign of ΔH:**
Since ΔNG is negative (-2), the term \(-2RT\) will also be negative, which means:
\[ \Delta H = -X - 2RT \]
This indicates that ΔH is less than ΔU because we are subtracting a positive quantity (2RT) from ΔU.

6. **Conclusion:**
Therefore, the value of ΔH is:
\[ \Delta H < \Delta U \]
Since ΔU is given as -X, we conclude that ΔH is less than -X.

### Final Answer:
ΔH is less than ΔU.