At `27^(@) C`, the combustion of ethane takes place according to the reaction `C_(2) H_(6)(g) + 7/2O_(2)(g) rightarrow 2CO_(2)(g) + 3H_(2)O(l)`
`Delta E - Delta H ` for this reaction at `27^(@)C` will be

Options

A. `+1347.1 J`
B. `-1247.1 J`
C. -`6235.5 J`
D. `+6235.5 J`

Answer

To solve the problem, we need to find the value of \( \Delta E - \Delta H \) for the combustion of ethane at \( 27^\circ C \). We will use the relationship between \( \Delta E \) and \( \Delta H \) given by the equation:

\[ \Delta H = \Delta E + \Delta N_g RT \]

From this, we can rearrange it to find \( \Delta E - \Delta H \):

\[ \Delta E - \Delta H = -\Delta N_g RT \]

### Step 1: Calculate \( \Delta N_g \)

To find \( \Delta N_g \), we need to determine the change in the number of moles of gas during the reaction.

The reaction is:

\[ C_2H_6(g) + \frac{7}{2}O_2(g) \rightarrow 2CO_2(g) + 3H_2O(l) \]

- **Reactants:**
- Ethane (\( C_2H_6 \)): 1 mole (gas)
- Oxygen (\( O_2 \)): \( \frac{7}{2} \) moles (gas)

Total moles of gaseous reactants = \( 1 + \frac{7}{2} = \frac{9}{2} \) moles

- **Products:**
- Carbon Dioxide (\( CO_2 \)): 2 moles (gas)
- Water (\( H_2O \)): 3 moles (liquid, not counted)

Total moles of gaseous products = 2 moles

Now, we can calculate \( \Delta N_g \):

\[ \Delta N_g = \text{(moles of gaseous products)} - \text{(moles of gaseous reactants)} = 2 - \frac{9}{2} = 2 - 4.5 = -2.5 \]

### Step 2: Substitute \( \Delta N_g \) into the equation

Now we substitute \( \Delta N_g \) into the equation for \( \Delta E - \Delta H \):

\[ \Delta E - \Delta H = -(-2.5) RT = 2.5 RT \]

### Step 3: Calculate \( RT \)

We need to calculate \( RT \) at \( 27^\circ C \):

- Convert temperature to Kelvin:
\[ T = 27 + 273 = 300 \, K \]

- Use the gas constant \( R = 8.314 \, \text{J/(mol K)} \):

Now calculate \( RT \):

\[ RT = 8.314 \, \text{J/(mol K)} \times 300 \, K = 2494.2 \, \text{J/mol} \]

### Step 4: Calculate \( \Delta E - \Delta H \)

Now substitute \( RT \) back into the equation:

\[ \Delta E - \Delta H = 2.5 \times 2494.2 \, \text{J/mol} = 6235.5 \, \text{J} \]

### Final Answer

Thus, the value of \( \Delta E - \Delta H \) for the combustion of ethane at \( 27^\circ C \) is:

\[ \Delta E - \Delta H = 6235.5 \, \text{J} \]