The heat evolved in the combustion of benzene is given by the equation
`C_(6)H_(6)(l)+7(1)/(2)O_(2)(g)to 6CO_(2)(g)+3H_(2)O(l) , Delta H =- 781 Kcal mol^(-1)`
Which of the following quantities of heat will be evolved when 39 g of benzene is burnt in an open container

Options

A. 122.25 kcal
B. 390.5 kcal
C. 1562.0 kcal
D. 586.75 kcal

Answer

To solve the problem, we need to determine how much heat is evolved when 39 grams of benzene (C₆H₆) is burned, given that the heat evolved in the combustion of one mole of benzene is -781 kcal.

### Step-by-Step Solution:

1. **Determine the Molar Mass of Benzene (C₆H₆)**:
- The molar mass of benzene can be calculated as follows:
\[ \text{Molar mass of C₆H₆} = (6 \times 12) + (6 \times 1) = 72 + 6 = 78 \text{ g/mol} \]

2. **Calculate the Number of Moles of Benzene in 39 g**:
- The number of moles (n) can be calculated using the formula:
\[ n = \frac{\text{mass}}{\text{molar mass}} = \frac{39 \text{ g}}{78 \text{ g/mol}} = 0.5 \text{ moles} \]

3. **Calculate the Heat Evolved for 0.5 Moles of Benzene**:
- Since the heat evolved for 1 mole of benzene is -781 kcal, the heat evolved for 0.5 moles can be calculated as:
\[ \text{Heat evolved} = -781 \text{ kcal/mol} \times 0.5 \text{ moles} = -390.5 \text{ kcal} \]

4. **Final Result**:
- The heat evolved when 39 g of benzene is burned is -390.5 kcal.

### Summary of the Solution:
When 39 g of benzene is combusted, the heat evolved is -390.5 kcal.