Calorific value of ethane, in k J/g if for the reaction `2C_(2)H_(6) + 7O_(2) rightarrow 4CO_(2) + 6 H_(2)O, Delta H = -745.6 kcal`

Options

A. `-12.4`
B. `-52`
C. `-24.8`
D. `-104`

Answer

To find the calorific value of ethane (C₂H₆) in kJ/g from the given reaction and ΔH, we can follow these steps:

### Step-by-step Solution:

1. **Identify the Reaction and ΔH**:
The reaction given is:
\[ 2C_{2}H_{6} + 7O_{2} \rightarrow 4CO_{2} + 6H_{2}O \]
The enthalpy change (ΔH) for this reaction is -745.6 kcal.

2. **Calculate ΔH for 1 Mole of Ethane**:
Since the reaction involves 2 moles of ethane, we need to find the heat of combustion per mole of ethane:
\[ \Delta H \text{ for 1 mole of } C_{2}H_{6} = \frac{-745.6 \text{ kcal}}{2} = -372.8 \text{ kcal} \]

3. **Convert kcal to kJ**:
We need to convert the heat of combustion from kcal to kJ. We know that:
\[ 1 \text{ kcal} = 4.184 \text{ kJ} \]
Therefore, we convert:
\[ -372.8 \text{ kcal} \times 4.184 \text{ kJ/kcal} = -1559.79 \text{ kJ} \]

4. **Calculate the Molecular Weight of Ethane**:
The molecular weight of ethane (C₂H₆) is calculated as follows:
- Carbon (C) has an atomic mass of approximately 12 g/mol, and there are 2 carbon atoms:
\[ 2 \times 12 = 24 \text{ g/mol} \]
- Hydrogen (H) has an atomic mass of approximately 1 g/mol, and there are 6 hydrogen atoms:
\[ 6 \times 1 = 6 \text{ g/mol} \]
- Therefore, the total molecular weight of ethane is:
\[ 24 + 6 = 30 \text{ g/mol} \]

5. **Calculate the Calorific Value**:
The calorific value (CV) in kJ/g is given by the heat of combustion per gram of ethane:
\[ \text{Calorific Value} = \frac{\text{Heat of combustion (kJ)}}{\text{Molecular weight (g/mol)}} \]
Substituting the values:
\[ \text{Calorific Value} = \frac{-1559.79 \text{ kJ}}{30 \text{ g/mol}} \approx -51.99 \text{ kJ/g} \]
Rounding this value gives approximately:
\[ \text{Calorific Value} \approx -52 \text{ kJ/g} \]

### Final Answer:
The calorific value of ethane is approximately **-52 kJ/g**.