The difference between `DeltaH` and `DeltaE` on a molar basis for the combustion of methane gas at T K (kelvin) would be:

Options

A. Zero
B. `-RT`
C. `-2RT`
D. `-3RT`

Explanation

To find the difference between ΔH (enthalpy change) and ΔE (internal energy change) for the combustion of methane gas, we can follow these steps:

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

### Step 2: Identify the number of moles of gaseous reactants and products.
From the balanced equation:
- Reactants: 1 mole of CH₄ and 2 moles of O₂ = 3 moles of gas (reactants)
- Products: 1 mole of CO₂ and 2 moles of H₂O (liquid) = 1 mole of gas (products)

### Step 3: Calculate the change in the number of moles of gas (ΔN).
\[ \Delta N = \text{Moles of gaseous products} - \text{Moles of gaseous reactants} = 1 - 3 = -2 \]

### Step 4: Use the relationship between ΔH and ΔE.
The relationship between the change in enthalpy (ΔH) and the change in internal energy (ΔE) is given by:
\[ \Delta H = \Delta E + \Delta N \cdot R \cdot T \]
Where:
- R = universal gas constant (8.314 J/(mol·K))
- T = temperature in Kelvin

### Step 5: Substitute ΔN into the equation.
Substituting ΔN = -2 into the equation:
\[ \Delta H = \Delta E - 2RT \]

### Step 6: Rearrange to find the difference ΔH - ΔE.
Rearranging gives us:
\[ \Delta H - \Delta E = -2RT \]

### Conclusion:
Thus, the difference between ΔH and ΔE on a molar basis for the combustion of methane gas at T K is:
\[ \Delta H - \Delta E = -2RT \]