`DeltaU^(@)` of combustion of `CH_(4(g))` at certain temperature is `-"393 kJ mol"^(-1)`. The value of `DeltaH^(@)` is

Options

A. zero
B. `lt DeltaH^(@)`
C. `gt DeltaU^(@)`
D. equal to `DeltaU^(@)`

Explanation

To find the value of \(\Delta H^\circ\) for the combustion of methane (\(CH_4(g)\)), we can use the relationship between \(\Delta H\) and \(\Delta U\). The equation we will use is:

\[ \Delta H^\circ = \Delta U^\circ + \Delta N_g RT \]

where:
- \(\Delta H^\circ\) is the change in enthalpy,
- \(\Delta U^\circ\) is the change in internal energy,
- \(\Delta N_g\) is the change in the number of moles of gas,
- \(R\) is the universal gas constant (approximately \(8.314 \, \text{J/mol·K}\)),
- \(T\) is the temperature in Kelvin.

### Step 1: Write the balanced equation for the combustion of methane
The balanced equation for the combustion of methane is:

\[ CH_4(g) + 2 O_2(g) \rightarrow CO_2(g) + 2 H_2O(l) \]

### Step 2: Calculate \(\Delta N_g\)
To find \(\Delta N_g\), we need to count the number of moles of gaseous reactants and products.

- **Reactants**:
- \(CH_4(g)\): 1 mole
- \(O_2(g)\): 2 moles
- Total = \(1 + 2 = 3\) moles of gas

- **Products**:
- \(CO_2(g)\): 1 mole
- \(H_2O(l)\): 0 moles (liquid is not counted)
- Total = \(1\) mole of gas

Now, we can calculate \(\Delta N_g\):

\[ \Delta N_g = \text{(moles of products)} - \text{(moles of reactants)} = 1 - 3 = -2 \]

### Step 3: Substitute values into the equation
Now we can substitute the values into the equation for \(\Delta H^\circ\):

Given:
- \(\Delta U^\circ = -393 \, \text{kJ/mol}\) (which is \(-393,000 \, \text{J/mol}\) for consistency in units),
- \(\Delta N_g = -2\),
- \(R = 8.314 \, \text{J/mol·K}\),
- Assume \(T\) is at standard temperature, \(T = 298 \, \text{K}\).

Now substituting into the equation:

\[ \Delta H^\circ = \Delta U^\circ + \Delta N_g RT \]
\[ \Delta H^\circ = -393,000 \, \text{J/mol} + (-2)(8.314 \, \text{J/mol·K})(298 \, \text{K}) \]

Calculating the second term:

\[ \Delta H^\circ = -393,000 \, \text{J/mol} - 2 \times 8.314 \times 298 \]
\[ = -393,000 \, \text{J/mol} - 4965.68 \, \text{J/mol} \]
\[ = -393,000 \, \text{J/mol} - 4,965.68 \, \text{J/mol} = -397,965.68 \, \text{J/mol} \]

Converting back to kJ:

\[ \Delta H^\circ = -397.97 \, \text{kJ/mol} \]

### Final Answer
Thus, the value of \(\Delta H^\circ\) for the combustion of methane is approximately:

\[ \Delta H^\circ \approx -398 \, \text{kJ/mol} \]