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 of calculating the heat evolved when 39 g of benzene is burnt, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Reaction and Heat of Combustion**:
The combustion reaction of benzene is given by:
\[ C_6H_6(l) + \frac{7}{2} O_2(g) \rightarrow 6 CO_2(g) + 3 H_2O(l) \]
The heat evolved (ΔH) for this reaction is -781 kcal/mol. This means that when 1 mole of benzene is combusted, 781 kcal of heat is released.

2. **Calculate the Molar Mass of Benzene**:
The molar mass of benzene (C6H6) can be calculated as follows:
\[ \text{Molar mass of } C_6H_6 = (6 \times 12) + (6 \times 1) = 72 + 6 = 78 \text{ g/mol} \]

3. **Determine the Amount of Heat Released per Gram**:
Since 781 kcal is released for 78 g of benzene, we can find the heat released per gram:
\[ \text{Heat per gram} = \frac{781 \text{ kcal}}{78 \text{ g}} = 10.02 \text{ kcal/g} \]

4. **Calculate the Heat Released for 39 g of Benzene**:
Now, we can calculate the heat evolved when 39 g of benzene is burnt:
\[ \text{Heat evolved} = 39 \text{ g} \times 10.02 \text{ kcal/g} = 390.78 \text{ kcal} \]

5. **Round the Result**:
Rounding the value gives us approximately:
\[ \text{Heat evolved} \approx 390.5 \text{ kcal} \]

### Final Answer:
The heat evolved when 39 g of benzene is burnt is approximately **390.5 kcal**.

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