The combustion of benzene(/) gives `CO_2(g) and H_2O(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. `4152.6`
B. `-452.46`
C. `3260`
D. `-3267.6`

Explanation

To find the heat of combustion of benzene at constant pressure (\( \Delta H \)), we will use the relationship between the change in internal energy (\( \Delta U \)) and the change in enthalpy (\( \Delta H \)) given by the equation:

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

Where:
- \( \Delta U \) is the heat of combustion at constant volume (given as \(-3263.9 \, \text{kJ/mol}\)),
- \( \Delta N_g \) is the change in the number of moles of gas,
- \( R \) is the universal gas constant (\(8.314 \, \text{J/K/mol}\)),
- \( T \) is the temperature in Kelvin.

### Step 1: Write the balanced chemical equation for the combustion of benzene.

The combustion of benzene (\( \text{C}_6\text{H}_6 \)) can be represented as:

\[ \text{C}_6\text{H}_6 (l) + \frac{15}{2} \text{O}_2 (g) \rightarrow 6 \text{CO}_2 (g) + 3 \text{H}_2\text{O} (l) \]

### Step 2: Calculate \( \Delta N_g \).

To find \( \Delta N_g \), we need to determine the number of moles of gaseous products and subtract the number of moles of gaseous reactants.

- Moles of gaseous products: \( 6 \, \text{CO}_2 \) (from the products) = 6
- Moles of gaseous reactants: \( \frac{15}{2} \, \text{O}_2 \) = 7.5

Now, calculate \( \Delta N_g \):

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

### Step 3: Convert temperature to Kelvin.

The temperature is given as \( 25^\circ C \). To convert to Kelvin:

\[ T = 25 + 273 = 298 \, K \]

### Step 4: Substitute values into the equation for \( \Delta H \).

Now we can substitute the values into the equation:

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

Substituting the known values:

\[ \Delta H = -3263.9 \, \text{kJ/mol} + (-1.5) \times (8.314 \, \text{J/K/mol}) \times (298 \, K) \]

### Step 5: Calculate \( \Delta N_g RT \).

First, calculate \( \Delta N_g RT \):

\[ \Delta N_g RT = -1.5 \times 8.314 \times 298 = -1.5 \times 2477.572 = -3716.36 \, \text{J/mol} \]

Convert \( -3716.36 \, \text{J/mol} \) to kJ/mol:

\[ -3716.36 \, \text{J/mol} = -3.71636 \, \text{kJ/mol} \]

### Step 6: Final calculation of \( \Delta H \).

Now substitute back into the equation for \( \Delta H \):

\[ \Delta H = -3263.9 \, \text{kJ/mol} - 3.71636 \, \text{kJ/mol} \]

\[ \Delta H = -3267.61636 \, \text{kJ/mol} \]

### Step 7: Round the answer.

Rounding to one decimal place, we get:

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

### Final Answer:

The heat of combustion of benzene at constant pressure is approximately \(-3267.6 \, \text{kJ/mol}\).

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