The difference between heat of reaction at constant pressure and at constant volume for the reaction `C_6H_12(l) + 90_2(g) rightarrow 6CO_2(g)+6H_2O(l), at 27^oC` is

Options

A. `- 14.96 kJ`
B. `- 7.48 kJ`
C. `+ 7.43 kJ`
D. `- 10.23 kJ`

Answer

To solve the problem of finding the difference between the heat of reaction at constant pressure and at constant volume for the given reaction, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Reaction**:
The reaction given is:
\[ C_6H_{12}(l) + 9O_2(g) \rightarrow 6CO_2(g) + 6H_2O(l) \]

2. **Understand the Terms**:
- \( \Delta H \) is the change in enthalpy (heat of reaction at constant pressure).
- \( \Delta U \) is the change in internal energy (heat of reaction at constant volume).
- The relationship between \( \Delta H \) and \( \Delta U \) is given by:
\[ \Delta H = \Delta U + \Delta N_g RT \]
where \( \Delta N_g \) is the change in the number of moles of gaseous products minus gaseous reactants, \( R \) is the gas constant (8.314 J/mol·K), and \( T \) is the temperature in Kelvin.

3. **Calculate \( \Delta N_g \)**:
- Count the moles of gaseous products and reactants:
- Products: \( 6 \, \text{moles of } CO_2 \)
- Reactants: \( 9 \, \text{moles of } O_2 \)
- Calculate \( \Delta N_g \):
\[ \Delta N_g = \text{(moles of gaseous products)} - \text{(moles of gaseous reactants)} = 6 - 9 = -3 \]

4. **Convert Temperature to Kelvin**:
- The given temperature is \( 27^\circ C \).
- Convert to Kelvin:
\[ T = 27 + 273 = 300 \, K \]

5. **Substitute Values into the Equation**:
- Now substitute \( \Delta N_g \), \( R \), and \( T \) into the equation:
\[ \Delta H - \Delta U = \Delta N_g RT \]
\[ \Delta H - \Delta U = (-3)(8.314)(300) \]

6. **Calculate the Right Side**:
- Calculate \( (-3)(8.314)(300) \):
\[ \Delta H - \Delta U = -3 \times 8.314 \times 300 = -7474.2 \, J \]
- Convert to kilojoules:
\[ -7474.2 \, J = -7.4742 \, kJ \]

7. **Final Result**:
- The difference between the heat of reaction at constant pressure and at constant volume is approximately:
\[ \Delta H - \Delta U \approx -7.47 \, kJ \]

### Conclusion:
The difference between the heat of reaction at constant pressure and at constant volume for the reaction is approximately **-7.47 kJ**.