The combustion of benzene (l) gives `CO_(2)(g)` and `H_(2)O(l)`. Given that heat of combustion of benzene at constant volume is `–3263.9 kJ mol^(–1)` at `25^(@)C`, heat of combustion (in kJ` mol^(–1)`) of benzene at constant pressure will be
(R = 8.314 JK–1 mol–1)

Options

A. `-3267.6`
B. `4152.6`
C. `-452.46`
D. 3260

Answer

To find the heat of combustion of benzene at constant pressure (ΔH), we can use the relationship between the change in internal energy (ΔU) and the change in enthalpy (ΔH) at constant pressure. The relevant equation is:

\[ \Delta H = \Delta U + nRT \]

Where:
- ΔH = change in enthalpy (heat of combustion at constant pressure)
- ΔU = change in internal energy (heat of combustion at constant volume)
- n = change in number of moles of gas
- R = universal gas constant (8.314 J K⁻¹ mol⁻¹)
- T = temperature in Kelvin

### Step 1: Write the balanced combustion reaction for benzene
The combustion of benzene (C₆H₆) in the presence of oxygen (O₂) produces carbon dioxide (CO₂) and water (H₂O). The balanced reaction is:

\[ C_6H_6 (l) + \frac{15}{2} O_2 (g) \rightarrow 6 CO_2 (g) + 3 H_2O (l) \]

### Step 2: Calculate the change in the number of moles of gas (Δn)
In the balanced equation:
- Moles of gaseous products = 6 (from 6 CO₂)
- Moles of gaseous reactants = \(\frac{15}{2} = 7.5\) (from \(\frac{15}{2} O₂\))

Thus, the change in the number of moles of gas (Δn) is:

\[ \Delta n = \text{moles of products} - \text{moles of reactants} = 6 - 7.5 = -1.5 \]

### Step 3: Substitute the values into the equation
Given:
- ΔU = -3263.9 kJ/mol
- R = 8.314 J K⁻¹ mol⁻¹ = 0.008314 kJ K⁻¹ mol⁻¹ (conversion from J to kJ)
- T = 25°C = 298 K

Now, substituting these values into the equation:

\[ \Delta H = \Delta U + nRT \]

\[ \Delta H = -3263.9 \text{ kJ/mol} + (-1.5) \times (0.008314 \text{ kJ K}^{-1} \text{ mol}^{-1}) \times (298 \text{ K}) \]

### Step 4: Calculate the second term
Calculating the second term:

\[ -1.5 \times 0.008314 \times 298 = -3.726 \text{ kJ} \]

### Step 5: Final calculation
Now, substituting this back into the ΔH equation:

\[ \Delta H = -3263.9 \text{ kJ/mol} - 3.726 \text{ kJ} = -3267.626 \text{ kJ/mol} \]

### Step 6: Round the final answer
Rounding to an appropriate number of significant figures, we find:

\[ \Delta H \approx -3267.6 \text{ kJ/mol} \]

### Conclusion
The heat of combustion of benzene at constant pressure is approximately **-3267.6 kJ/mol**.