`DeltaU^(@)` of combustoin of methane is `-XKJ mol ^(-1)` The value
of `DeltaH^(@)` is :-

Options

A. `-X-596R`
B. `-X+596R`
C. `X+596R`
D. `X-596R`

Explanation

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

### Step 1: Understand the relationship between ΔH° and ΔU°
The relationship between the standard enthalpy change (ΔH°) and the standard internal energy change (ΔU°) is given by the equation:
\[ \Delta H^\circ = \Delta U^\circ + \Delta N_g R T \]
where:
- ΔN_g = change in the number of moles of gas
- R = universal gas constant (8.314 J/(mol·K))
- T = temperature in Kelvin (298 K for standard conditions)

### Step 2: Write the combustion reaction of methane
The balanced chemical equation for the combustion of methane (CH₄) is:
\[ \text{CH}_4(g) + 2\text{O}_2(g) \rightarrow \text{CO}_2(g) + 2\text{H}_2\text{O}(l) \]

### Step 3: Calculate ΔN_g
ΔN_g is calculated as:
\[ \Delta N_g = \text{moles of gaseous products} - \text{moles of gaseous reactants} \]
From the reaction:
- Moles of gaseous products = 1 (from CO₂)
- Moles of gaseous reactants = 1 (from CH₄) + 2 (from O₂) = 3

Thus,
\[ \Delta N_g = 1 - 3 = -2 \]

### Step 4: Substitute values into the equation
Now we can substitute ΔU°, ΔN_g, R, and T into the equation:
\[ \Delta H^\circ = \Delta U^\circ + \Delta N_g R T \]
Substituting the known values:
\[ \Delta H^\circ = -X + (-2)(8.314 \, \text{J/(mol·K)})(298 \, \text{K}) \]

### Step 5: Convert R to kJ
Since ΔU° is given in kJ, we need to convert R from J to kJ:
\[ R = 8.314 \, \text{J/(mol·K)} = 0.008314 \, \text{kJ/(mol·K)} \]

### Step 6: Calculate ΔH°
Now substituting R in kJ:
\[ \Delta H^\circ = -X - 2(0.008314)(298) \]
Calculating the second term:
\[ \Delta H^\circ = -X - 4.964 \, \text{kJ} \]

### Final Answer
Thus, the final expression for ΔH° is:
\[ \Delta H^\circ = -X - 4.964 \, \text{kJ/mol} \]