For the combustion of n-octane
`C_(8)H_(18)+O_(2)rarrCO_(2)+H_(2)O` at `25^(@)C` (ingnoring resonance in `CO_(2)`)

Options

A. `DeltaH = DeltaE -5.5xx8.31xx0.298` in kJ/mol
B. `DeltaH = DeltaE +4.5xx8.31xx0.298` in kJ/mol
C. `DeltaH = DeltaE -4.5xx8.31xx298` in kJ/mol
D. `DeltaH = DeltaE -4.5xx8.31xx0.298` in kJ/mol

Answer

To solve the problem regarding the combustion of n-octane (\(C_8H_{18}\)) and the relationship between \(\Delta H\) (enthalpy change) and \(\Delta E\) (internal energy change), we will follow these steps:

### Step 1: Write the balanced chemical equation for the combustion of n-octane
The combustion of n-octane can be represented as:
\[ C_8H_{18} + O_2 \rightarrow CO_2 + H_2O \]

### Step 2: Balance the chemical equation
1. **Carbon atoms**: There are 8 carbon atoms in n-octane, so we need 8 \(CO_2\):
\[ C_8H_{18} + O_2 \rightarrow 8 CO_2 + H_2O \]

2. **Hydrogen atoms**: There are 18 hydrogen atoms in n-octane, so we need 9 \(H_2O\):
\[ C_8H_{18} + O_2 \rightarrow 8 CO_2 + 9 H_2O \]

3. **Oxygen atoms**: On the product side, we have:
- From \(8 CO_2\): \(8 \times 2 = 16\) oxygen atoms
- From \(9 H_2O\): \(9 \times 1 = 9\) oxygen atoms
- Total oxygen needed = \(16 + 9 = 25\) oxygen atoms

Therefore, we need \( \frac{25}{2} O_2\) on the reactant side:
\[ C_8H_{18} + \frac{25}{2} O_2 \rightarrow 8 CO_2 + 9 H_2O \]

### Step 3: Calculate \(\Delta N_G\)
\(\Delta N_G\) is defined as the change in the number of moles of gaseous products minus the change in the number of moles of gaseous reactants.
- **Gaseous products**: \(8 CO_2 + 9 H_2O = 17\) moles
- **Gaseous reactants**: \(\frac{25}{2} O_2 = 12.5\) moles

Thus,
\[ \Delta N_G = 17 - 12.5 = 4.5 \]

### Step 4: Use the relation between \(\Delta H\) and \(\Delta E\)
The relationship is given by:
\[ \Delta H = \Delta E + \Delta N_G RT \]
Where:
- \(R\) is the gas constant, \(R = 8.314 \, \text{J/(mol K)} = 0.008314 \, \text{kJ/(mol K)}\)
- \(T\) is the temperature in Kelvin. At \(25^\circ C\), \(T = 25 + 273 = 298 \, K\)

### Step 5: Substitute the values into the equation
Substituting the values into the equation:
\[ \Delta H = \Delta E + (4.5)(0.008314)(298) \]

Calculating the term:
\[ 4.5 \times 0.008314 \times 298 = 11.77 \, \text{kJ} \]

Thus, the equation becomes:
\[ \Delta H = \Delta E + 11.77 \]

### Final Answer
The relationship between \(\Delta H\) and \(\Delta E\) for the combustion of n-octane is:
\[ \Delta H = \Delta E + 11.77 \, \text{kJ} \]