The difference between `Delta H` and `Delta E` (on a molar basis) for the combustion of n-octane `(l)` at `25^(@)C` would be:

Options

A. `13.6 kJ`
B. `-1.14 kJ`
C. `-11.15 kJ`
D. `+11.15 kJ`

Explanation

To find the difference between \( \Delta H \) and \( \Delta E \) for the combustion of n-octane at \( 25^\circ C \), we will follow these steps:

### Step 1: Write the balanced chemical equation for the combustion of n-octane.
The combustion of n-octane \((C_8H_{18})\) can be represented as:
\[ C_8H_{18}(l) + \frac{25}{2} O_2(g) \rightarrow 8 CO_2(g) + 9 H_2O(l) \]

### Step 2: Calculate the change in the number of moles of gas (\( \Delta N_G \)).
To find \( \Delta N_G \):
\[ \Delta N_G = \text{moles of gaseous products} - \text{moles of gaseous reactants} \]
From the balanced equation:
- Gaseous products: \( 8 \, CO_2 \) (8 moles)
- Gaseous reactants: \( \frac{25}{2} \, O_2 \) (12.5 moles)

Thus,
\[ \Delta N_G = 8 - 12.5 = -4.5 \]

### Step 3: Use the relationship between \( \Delta H \) and \( \Delta E \).
According to the thermodynamic relation:
\[ \Delta H = \Delta E + \Delta N_G RT \]
Rearranging gives:
\[ \Delta H - \Delta E = \Delta N_G RT \]

### Step 4: Substitute the values into the equation.
- \( \Delta N_G = -4.5 \)
- \( R = 8.314 \, \text{J/mol·K} \)
- \( T = 25^\circ C = 298 \, \text{K} \)

Substituting these values:
\[ \Delta H - \Delta E = (-4.5) \times (8.314) \times (298) \]

### Step 5: Calculate the value.
Calculating the right side:
\[ \Delta H - \Delta E = -4.5 \times 8.314 \times 298 \]
\[ = -4.5 \times 2477.572 \approx -11148.66 \, \text{J} \]
Converting to kilojoules:
\[ \Delta H - \Delta E \approx -11.15 \, \text{kJ} \]

### Final Answer:
The difference between \( \Delta H \) and \( \Delta E \) for the combustion of n-octane at \( 25^\circ C \) is approximately:
\[ \Delta H - \Delta E \approx -11.15 \, \text{kJ} \]

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