The difference in `Delta H` and `Delta E` for the combustion of methane at `25^(@)C` would be :-

Options

A. Zero
B. `2xx298xx -2` Cals
C. `2xx298xx -3` Cals
D. `2xx25xx -3` Cals

Answer

To find the difference between ΔH (change in enthalpy) and ΔE (change in internal energy) for the combustion of methane at 25°C, we will follow these steps:

### Step 1: Write the combustion reaction
The balanced equation for the combustion of methane (CH₄) is:
\[ \text{CH}_4(g) + 2 \text{O}_2(g) \rightarrow \text{CO}_2(g) + 2 \text{H}_2O(l) \]

### Step 2: Identify the change in number of gaseous moles (ΔN₍g₎)
ΔN₍g₎ is calculated as the difference between the number of moles of gaseous products and the number of moles of gaseous reactants.

- **Reactants:** 1 mole of CH₄ + 2 moles of O₂ = 3 moles of gas
- **Products:** 1 mole of CO₂ (g) + 2 moles of H₂O (l) = 1 mole of gas

Thus,
\[ \Delta N₍g₎ = \text{Moles of products} - \text{Moles of reactants} = 1 - 3 = -2 \]

### Step 3: Use the first law of thermodynamics
According to the first law of thermodynamics:
\[ \Delta H = \Delta E + \Delta N₍g₎ \cdot R \cdot T \]
Where:
- R = gas constant = 2 cal/(K·mol) (in calories)
- T = temperature in Kelvin = 25°C + 273 = 298 K

### Step 4: Substitute the values into the equation
Substituting the values into the equation:
\[ \Delta H = \Delta E + (-2) \cdot R \cdot T \]
\[ \Delta H = \Delta E - 2 \cdot (2 \, \text{cal/(K·mol)}) \cdot (298 \, \text{K}) \]

### Step 5: Calculate the difference
Calculating the term:
\[ -2 \cdot 2 \cdot 298 = -4 \cdot 298 = -1192 \, \text{cal} \]

Thus, the difference between ΔH and ΔE is:
\[ \Delta H - \Delta E = -1192 \, \text{cal} \]

### Final Answer
The difference in ΔH and ΔE for the combustion of methane at 25°C is **-1192 cal**.

---