The difference between heats of reaction at constant pressure and constant volume for the reaction,
`2C_6H_6(l) + 15O_2(g) to 12CO_2(g) + 6H_2O (l)`
at `25^@C` in kJ is

Options

A. ` + 7.43`
B. ` +3.72`
C. `-7.43`
D. `-3.72`

Explanation

To solve the problem of finding the difference between the heats of reaction at constant pressure (ΔH) and constant volume (ΔU) for the reaction:

\[ 2C_6H_6(l) + 15O_2(g) \rightarrow 12CO_2(g) + 6H_2O(l) \]

at \( 25^\circ C \), we can follow these steps:

### Step 1: Understand the relationship between ΔH and ΔU
The relationship between the heat of reaction at constant pressure (ΔH) and the heat of reaction at constant volume (ΔU) is given by the equation:

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

Where:
- \( \Delta n_g \) = change in the number of moles of gas (moles of products - moles of reactants)
- \( R \) = universal gas constant (8.314 J/mol·K)
- \( T \) = temperature in Kelvin

### Step 2: Convert the temperature to Kelvin
Given that the temperature is \( 25^\circ C \):

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

### Step 3: Calculate Δng
Next, we need to calculate \( \Delta n_g \):

- Moles of gaseous products:
- From the products: \( 12 \, CO_2(g) \) = 12 moles of gas
- Moles of gaseous reactants:
- From the reactants: \( 15 \, O_2(g) \) = 15 moles of gas

Now, calculate \( \Delta n_g \):

\[ \Delta n_g = \text{(moles of products)} - \text{(moles of reactants)} = 12 - 15 = -3 \]

### Step 4: Substitute values into the equation
Now we can substitute \( \Delta n_g \) into the equation for the difference between ΔH and ΔU:

\[ \Delta H - \Delta U = \Delta n_g RT \]

Substituting the values:

\[ \Delta H - \Delta U = (-3) \times (8.314 \, \text{J/mol·K}) \times (298 \, K) \]

### Step 5: Calculate the result
Calculating the right side:

\[ \Delta H - \Delta U = -3 \times 8.314 \times 298 \]

Calculating:

\[ \Delta H - \Delta U = -7432.71 \, \text{J} \]

### Step 6: Convert to kJ
Since the question asks for the answer in kJ:

\[ \Delta H - \Delta U = \frac{-7432.71 \, \text{J}}{1000} = -7.43271 \, \text{kJ} \approx -7.43 \, \text{kJ} \]

### Final Answer
The difference between the heats of reaction at constant pressure and constant volume is:

\[ \Delta H - \Delta U = -7.43 \, \text{kJ} \]