The difference between heat of reaction and change in internal energy at constant volume for the reaction given below at `25^@C` in kJ is
`2C_6H_6(I)+15O_2(g) rarr12CO_2(g)+6H_2O(l)`

Options

A. `-7.43`
B. `+3.72`
C. `-3.72`
D. `+7.43`

Answer

To solve the problem, we need to find the difference between the heat of reaction (ΔH) and the change in internal energy (ΔU) at constant volume for the given reaction:

\[ 2C_6H_6(l) + 15O_2(g) \rightarrow 12CO_2(g) + 6H_2O(l) \]

### Step-by-Step Solution:

1. **Identify the Reaction Components**:
- Reactants: 2 moles of liquid benzene (C₆H₆) and 15 moles of gaseous oxygen (O₂).
- Products: 12 moles of gaseous carbon dioxide (CO₂) and 6 moles of liquid water (H₂O).

2. **Determine the Change in Moles of Gas (Δn)**:
- Δn = (moles of gaseous products) - (moles of gaseous reactants)
- Moles of gaseous products = 12 (from CO₂)
- Moles of gaseous reactants = 15 (from O₂)
- Therefore, Δn = 12 - 15 = -3.

3. **Use the Relationship Between ΔH and ΔU**:
- The relationship is given by the equation:
\[ ΔH = ΔU + PΔV \]
- At constant pressure, we can express \( PΔV \) in terms of Δn:
\[ PΔV = Δn_gRT \]
- Here, \( R \) is the universal gas constant (8.314 J/(mol·K)), and \( T \) is the temperature in Kelvin.

4. **Convert Temperature to Kelvin**:
- Given temperature = 25°C
- Convert to Kelvin: \( T = 25 + 273 = 298 \, K \).

5. **Calculate \( PΔV \)**:
- Substitute the values into the equation:
\[ PΔV = Δn_gRT = (-3)(8.314 \, \text{J/(mol·K)})(298 \, K) \]
- Calculate:
\[ PΔV = -3 \times 8.314 \times 298 = -7432.716 \, J \]

6. **Convert Joules to Kilojoules**:
- Since \( 1 \, \text{kJ} = 1000 \, \text{J} \):
\[ PΔV = -7432.716 \, J \div 1000 = -7.432716 \, kJ \approx -7.43 \, kJ \]

7. **Find the Difference \( ΔH - ΔU \)**:
- From the relationship \( ΔH - ΔU = PΔV \):
\[ ΔH - ΔU = -7.43 \, kJ \]

### Final Answer:
The difference between the heat of reaction and the change in internal energy at constant volume for the reaction is approximately **-7.43 kJ**.

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