For combustion of 1 mole of benzene at `25^(@)C`, the heat of reaction at constant pressure is `-780.9" kcal."`
What will be the heat of reaction at constant volume ?
`C_(6)H_(6(l))+7(1)/(2)O_(2(g))rarr 6CO_(2(g))+3H_(2)O_((l))`

Options

A. `-781.8" kcal"`
B. `-"780.0 kcal"`
C. `+"781.8 kcal"`
D. `+"780.0 kcal"`

Explanation

To find the heat of reaction at constant volume (QV) for the combustion of 1 mole of benzene, we can use the relationship between the heat of reaction at constant pressure (QP) and the heat of reaction at constant volume (QV). The formula we will use is:

\[ Q_V = Q_P - \Delta N_G RT \]

Where:
- \( Q_P \) is the heat of reaction at constant pressure (given as -780.9 kcal).
- \( \Delta N_G \) is the change in the number of moles of gas during the reaction.
- \( R \) is the universal gas constant (approximately 1.987 cal/(mol·K)).
- \( T \) is the temperature in Kelvin (25°C = 298 K).

### Step 1: Calculate \( \Delta N_G \)

From the balanced reaction:

\[ C_6H_6(l) + \frac{7}{2} O_2(g) \rightarrow 6 CO_2(g) + 3 H_2O(l) \]

- Moles of gaseous products = 6 (from \( CO_2 \))
- Moles of gaseous reactants = 3.5 (from \( \frac{7}{2} O_2 \))

Now, calculate \( \Delta N_G \):

\[ \Delta N_G = \text{Moles of products} - \text{Moles of reactants} = 6 - 3.5 = 2.5 \]

### Step 2: Convert temperature to Kelvin

The temperature is given as 25°C. We convert this to Kelvin:

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

### Step 3: Calculate \( Q_V \)

Now we can substitute the values into the equation:

\[ Q_V = Q_P - \Delta N_G RT \]

Substituting the known values:

\[ Q_V = -780.9 \, \text{kcal} - (2.5) \times (1.987 \, \text{cal/(mol·K)}) \times (298 \, K) \]

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

\[ \Delta N_G RT = 2.5 \times 1.987 \times 298 \approx 1481.5 \, \text{cal} \]

Convert calories to kilocalories (1 kcal = 1000 cal):

\[ \Delta N_G RT \approx 1.4815 \, \text{kcal} \]

Now substitute this back into the equation for \( Q_V \):

\[ Q_V = -780.9 \, \text{kcal} - 1.4815 \, \text{kcal} \approx -782.3815 \, \text{kcal} \]

### Final Result

Thus, the heat of reaction at constant volume is approximately:

\[ Q_V \approx -782.4 \, \text{kcal} \]