The heat of combustion of solid benzoic acid at constant volume is `-321.3 kJ at `27` degree Celsius. 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 from the heat of combustion at constant volume, we can use the relationship between the two. The equation we will use is:

\[ \Delta H = \Delta U + \Delta n_g R T \]

Where:
- \(\Delta H\) = change in enthalpy (heat of combustion at constant pressure)
- \(\Delta U\) = change in internal energy (heat of combustion at constant volume)
- \(\Delta n_g\) = change in the number of moles of gas
- \(R\) = universal gas constant (approximately \(8.314 \, \text{J/mol·K}\))
- \(T\) = temperature in Kelvin

### Step 1: Identify the given values
- The heat of combustion at constant volume (\(\Delta U\)) is given as \(-321.3 \, \text{kJ}\).
- The temperature is given as \(27 \, \text{°C}\), which we need to convert to Kelvin.

**Hint:** Remember to convert Celsius to Kelvin by adding 273.

### Step 2: Convert temperature to Kelvin
\[ T = 27 + 273 = 300 \, \text{K} \]

### Step 3: Write the balanced combustion reaction for benzoic acid
The combustion reaction of benzoic acid (\(C_6H_5COOH\)) is:
\[ C_6H_5COOH + O_2 \rightarrow CO_2 + H_2O \]

Balancing the reaction gives:
\[ C_6H_5COOH + 7.5 O_2 \rightarrow 6 CO_2 + 3 H_2O \]

### Step 4: Calculate \(\Delta n_g\)
From the balanced equation, we can see:
- Moles of gaseous products = \(6 \, (CO_2) + 3 \, (H_2O) = 9\)
- Moles of gaseous reactants = \(7.5 \, (O_2)\)

Thus, the change in the number of moles of gas (\(\Delta n_g\)) is:
\[ \Delta n_g = 9 - 7.5 = 1.5 \]

**Hint:** Count the moles of gaseous products and subtract the moles of gaseous reactants.

### Step 5: Substitute values into the equation
Now we can substitute the values into the equation:
\[ \Delta H = \Delta U + \Delta n_g R T \]
\[ \Delta H = -321.3 \, \text{kJ} + (1.5) (8.314 \, \text{J/mol·K}) (300 \, \text{K}) \]

### Step 6: Convert \(R\) to kJ
Since \(\Delta H\) is in kJ, convert \(R\) to kJ:
\[ R = 8.314 \, \text{J/mol·K} = 0.008314 \, \text{kJ/mol·K} \]

### Step 7: Calculate the term \(\Delta n_g R T\)
\[ \Delta n_g R T = 1.5 \times 0.008314 \, \text{kJ/mol·K} \times 300 \, \text{K} \]
\[ = 1.5 \times 0.008314 \times 300 = 3.7401 \, \text{kJ} \]

### Step 8: Substitute back to find \(\Delta H\)
\[ \Delta H = -321.3 \, \text{kJ} + 3.7401 \, \text{kJ} \]
\[ \Delta H = -317.5599 \, \text{kJ} \]

### Step 9: Round the answer
Rounding to three significant figures, we get:
\[ \Delta H \approx -317.6 \, \text{kJ} \]

### Final Answer
The heat of combustion of solid benzoic acid at constant pressure is approximately \(-317.6 \, \text{kJ}\).

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