The difference between heats of reaction at constant pressure and constant volume for reaction:
`2C_(6)H_(6)(l)+15O_(2)(g)to12CO_(2)(g)+6H_(2)O(l)` at `25^(@)C` in kJ is

Options

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

Explanation

To find the difference between the heats of reaction at constant pressure and constant volume for the given reaction, we can follow these steps:

### Step 1: Identify the Reaction
The reaction given is:
\[ 2C_{6}H_{6}(l) + 15O_{2}(g) \rightarrow 12CO_{2}(g) + 6H_{2}O(l) \]

### Step 2: Determine the Change in Moles of Gaseous Reactants and Products
Next, we need to calculate the change in the number of moles of gaseous products and reactants.

- **Gaseous Products**:
- \(12CO_{2}(g)\) contributes 12 moles.

- **Gaseous Reactants**:
- \(15O_{2}(g)\) contributes 15 moles.

Now, calculate the change in moles of gas (\(\Delta n_g\)):
\[ \Delta n_g = \text{(moles of gaseous products)} - \text{(moles of gaseous reactants)} = 12 - 15 = -3 \]

### Step 3: Use the Relation Between \(\Delta H\) and \(\Delta U\)
The relationship between the change in enthalpy (\(\Delta H\)) and the change in internal energy (\(\Delta U\)) at constant pressure and volume is given by:
\[ \Delta H = \Delta U + \Delta n_g \cdot R \cdot T \]
Where:
- \(R\) is the universal gas constant (approximately \(8.314 \, \text{J/mol K}\))
- \(T\) is the temperature in Kelvin (which is \(298 \, \text{K}\) at \(25^\circ C\))

### Step 4: Substitute the Values
Now, substituting the values into the equation:
\[ \Delta H - \Delta U = \Delta n_g \cdot R \cdot T \]
\[ \Delta H - \Delta U = (-3) \cdot (8.314 \, \text{J/mol K}) \cdot (298 \, \text{K}) \]
Calculating this gives:
\[ \Delta H - \Delta U = -3 \cdot 8.314 \cdot 298 = -7435.74 \, \text{J} \approx -7.44 \, \text{kJ} \]

### Step 5: Final Answer
Thus, the difference between the heats of reaction at constant pressure and constant volume is:
\[ \Delta H - \Delta U \approx -7.44 \, \text{kJ} \]