The heat of combustion of solid benzoic acid at constant volume is `-321.3 kJ at 27 degree Celcius . The heat of combustion at constant pressure is

Options

A. `-321.3 - 300R`
B. `-321.30 + 300R`
C. `-321.3 - 150R`
D. `-321.3 + 900R`

Answer

To find the heat of combustion of solid benzoic acid at constant pressure, we can use the relationship between the heat of combustion at constant volume (ΔU) and at constant pressure (ΔH). The relationship is given by the equation:

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

Where:
- ΔH = heat of combustion at constant pressure
- ΔU = heat of combustion at constant volume (given as -321.3 kJ)
- Δn = change in the number of moles of gas (products - reactants)
- R = universal gas constant (8.314 J/mol·K)
- T = temperature in Kelvin

### Step 1: Calculate Δn
The balanced combustion reaction for solid benzoic acid (C₆H₅COOH) is:

\[ C_6H_5COOH (s) + \frac{15}{2} O_2 (g) \rightarrow 7 CO_2 (g) + 3 H_2O (l) \]

Now, we need to calculate Δn, which is the change in the number of moles of gaseous products minus the number of moles of gaseous reactants.

- Moles of gaseous products (CO₂) = 7
- Moles of gaseous reactants (O₂) = 15/2 = 7.5

Thus,

\[ \Delta n = \text{Moles of products} - \text{Moles of reactants} = 7 - 7.5 = -0.5 \]

### Step 2: Convert the temperature to Kelvin
The temperature given is 27°C. To convert this to Kelvin:

\[ T(K) = 27 + 273 = 300 \, K \]

### Step 3: Substitute values into the equation
Now we can substitute the values into the equation for ΔH:

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

Substituting the known values:

\[ \Delta H = -321.3 \, \text{kJ} + (-0.5) \times (8.314 \, \text{J/mol·K}) \times (300 \, K) \]

First, convert -321.3 kJ to J:

\[ -321.3 \, \text{kJ} = -321300 \, \text{J} \]

Now calculate ΔnRT:

\[ \Delta nRT = -0.5 \times 8.314 \times 300 = -1249.1 \, \text{J} \]

### Step 4: Calculate ΔH
Now we can calculate ΔH:

\[ \Delta H = -321300 \, \text{J} - 1249.1 \, \text{J} = -322549.1 \, \text{J} \]

Convert back to kJ:

\[ \Delta H = -322.55 \, \text{kJ} \]

### Final Answer
The heat of combustion at constant pressure is approximately:

\[ \Delta H \approx -322.55 \, \text{kJ} \]