`delta_(f) U^(Ó¨) "of formation of" CH_(4)(g)` at a certain temperature is `-393Jmol^(-1).` The value of `delta H^(Ó¨)is:`

Options

A. zero
B. `lt Delta_(f) U^(Θ)`
C. `gt Delta_(f) U^(Θ)`
D. equal to `Delta_(f) U^(Θ)`

Explanation

To find the value of \( \Delta H^{\circ} \) for the formation of \( CH_4(g) \) given that \( \Delta U^{\circ} \) is -393 J/mol, we can use the following thermodynamic relation:

### Step-by-Step Solution:

1. **Write the Relation**:
The relationship between the change in enthalpy (\( \Delta H^{\circ} \)) and the change in internal energy (\( \Delta U^{\circ} \)) is given by:
\[ \Delta H^{\circ} = \Delta U^{\circ} + \Delta N_g RT \]
where \( \Delta N_g \) is the change in the number of moles of gas, \( R \) is the universal gas constant, and \( T \) is the temperature in Kelvin.

2. **Identify the Reaction**:
The formation of methane (\( CH_4 \)) can be represented as:
\[ C(s) + 2H_2(g) \rightarrow CH_4(g) \]

3. **Calculate \( \Delta N_g \)**:
- In the reaction, the total number of moles of products is 1 (from \( CH_4 \)).
- The total number of moles of reactants is 2 (from \( 2H_2 \)).
- Therefore, the change in moles of gas (\( \Delta N_g \)) is:
\[ \Delta N_g = \text{moles of products} - \text{moles of reactants} = 1 - 2 = -1 \]

4. **Substitute Values**:
Now, substitute \( \Delta U^{\circ} = -393 \, \text{J/mol} \) and \( \Delta N_g = -1 \) into the equation:
\[ \Delta H^{\circ} = -393 + (-1)RT \]
This simplifies to:
\[ \Delta H^{\circ} = -393 - RT \]

5. **Analyze the Result**:
Since \( R \) (the gas constant) and \( T \) (temperature) are always positive, the term \( -RT \) will always be negative. Thus:
\[ \Delta H^{\circ} < -393 \, \text{J/mol} \]

### Conclusion:
From the above steps, we conclude that \( \Delta H^{\circ} \) is less than \( -393 \, \text{J/mol} \).

### Final Answer:
Thus, the value of \( \Delta H^{\circ} \) is less than \( -393 \, \text{J/mol} \).

---