The standard molar heat of formation of ethane `CO_2` and water (l) are respectively -21.1 ,-94.1 and -68.3 kcal .the standard molar heat of combustion of ethane will be

Options

A. (-372kcal)
B. 162 kcal
C. (-240 kcal)
D. 183.5 kcal

Answer

To find the standard molar heat of combustion of ethane (C2H6), we can use the standard heats of formation of the products and reactants involved in the combustion reaction. The combustion of ethane can be represented by the following balanced chemical equation:

\[ \text{C}_2\text{H}_6(g) + 7\text{O}_2(g) \rightarrow 2\text{CO}_2(g) + 3\text{H}_2\text{O}(l) \]

### Step 1: Write the equation for the heat of combustion
The heat of combustion (\( \Delta H_{comb} \)) can be calculated using the standard heats of formation (\( \Delta H_f \)) of the products and reactants:

\[ \Delta H_{comb} = \Delta H_f (\text{products}) - \Delta H_f (\text{reactants}) \]

### Step 2: Identify the standard heats of formation
From the problem statement, we have the following values:
- \( \Delta H_f \) of ethane (C2H6) = -21.1 kcal
- \( \Delta H_f \) of carbon dioxide (CO2) = -94.1 kcal
- \( \Delta H_f \) of water (H2O) = -68.3 kcal

### Step 3: Calculate the heat of formation for the products
For the products, we have:
- 2 moles of CO2: \( 2 \times (-94.1 \text{ kcal}) = -188.2 \text{ kcal} \)
- 3 moles of H2O: \( 3 \times (-68.3 \text{ kcal}) = -204.9 \text{ kcal} \)

Adding these together gives:
\[ \Delta H_f (\text{products}) = -188.2 \text{ kcal} + (-204.9 \text{ kcal}) = -393.1 \text{ kcal} \]

### Step 4: Calculate the heat of formation for the reactants
For the reactants, we have:
- 1 mole of C2H6: \( -21.1 \text{ kcal} \)
- 7 moles of O2: The standard heat of formation for O2 is 0 kcal (as it is in its elemental form).

Thus, the total for the reactants is:
\[ \Delta H_f (\text{reactants}) = -21.1 \text{ kcal} + 0 \text{ kcal} = -21.1 \text{ kcal} \]

### Step 5: Substitute into the heat of combustion equation
Now we can substitute the values into the equation:
\[ \Delta H_{comb} = \Delta H_f (\text{products}) - \Delta H_f (\text{reactants}) \]
\[ \Delta H_{comb} = (-393.1 \text{ kcal}) - (-21.1 \text{ kcal}) = -393.1 \text{ kcal} + 21.1 \text{ kcal} = -372.0 \text{ kcal} \]

### Final Answer
The standard molar heat of combustion of ethane is:
\[ \Delta H_{comb} = -372.0 \text{ kcal} \]