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, we can use the relationship between the change in enthalpy (ΔH) and the change in internal energy (ΔU). The formula we will use is:

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

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

### Step 1: Write the combustion reaction of benzene

The combustion of benzene (C₆H₆) can be represented as follows:

\[ C_6H_6 (l) + O_2 (g) \rightarrow 6 CO_2 (g) + 3 H_2O (l) \]

### Step 2: Calculate ΔN_g

ΔN_g is calculated as the difference between the moles of gaseous products and the moles of gaseous reactants.

- Moles of gaseous products (6 CO₂) = 6
- Moles of gaseous reactants (O₂) = 15/2 = 7.5

Thus,

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

### Step 3: Convert ΔU to Joules

Given that ΔU (heat of combustion at constant volume) is -3263.9 kJ/mol, we convert this to Joules:

\[ \Delta U = -3263.9 \text{ kJ/mol} \times 1000 \text{ J/kJ} = -3263900 \text{ J/mol} \]

### Step 4: Calculate ΔH using the formula

Now we can substitute the values into the equation:

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

Where:
- R = 8.314 J K⁻¹ mol⁻¹
- T = 25°C = 298 K

Substituting the values:

\[ \Delta H = -3263900 \text{ J/mol} + (-1.5) \times (8.314 \text{ J K}^{-1} \text{ mol}^{-1}) \times (298 \text{ K}) \]

Calculating the second term:

\[ \Delta N_g RT = -1.5 \times 8.314 \times 298 = -1.5 \times 2477.572 = -3716.358 \text{ J/mol} \]

Now substituting this back into the ΔH equation:

\[ \Delta H = -3263900 \text{ J/mol} - 3716.358 \text{ J/mol} = -3267616.358 \text{ J/mol} \]

### Step 5: Convert ΔH back to kJ

Finally, convert ΔH back to kJ:

\[ \Delta H = -3267616.358 \text{ J/mol} \div 1000 = -3267.616 \text{ kJ/mol} \]

Thus, the heat of combustion of benzene at constant pressure is approximately:

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

### Final Answer

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

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