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 calculate the calorific value of ethane (C₂H₆) in kJ/g from the given reaction, we will follow these steps:

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

### Step 2: Calculate the Heat of Combustion per Mole of Ethane
The enthalpy change given is for the combustion of 2 moles of ethane. To find the heat of combustion per mole of ethane, we divide the total heat change by 2:
\[ \Delta H \text{ (per mole of ethane)} = \frac{-745.6 \text{ kcal}}{2} = -372.8 \text{ kcal} \]

### Step 3: Convert kcal to kJ
To convert kilocalories to kilojoules, we use the conversion factor:
\[ 1 \text{ kcal} = 4.184 \text{ kJ} \]
Thus,
\[ \Delta H \text{ (per mole of ethane in kJ)} = -372.8 \text{ kcal} \times 4.184 \text{ kJ/kcal} = -1555.8 \text{ kJ} \]

### Step 4: Calculate the Molar Mass of Ethane
The molar mass of ethane (C₂H₆) can be calculated as follows:
- Carbon (C): 12.01 g/mol, and there are 2 carbon atoms.
- Hydrogen (H): 1.008 g/mol, and there are 6 hydrogen atoms.

Calculating the molar mass:
\[ \text{Molar mass of } C_{2}H_{6} = (2 \times 12.01) + (6 \times 1.008) = 24.02 + 6.048 = 30.07 \text{ g/mol} \]

### Step 5: Calculate the Calorific Value in kJ/g
The calorific value (CV) is calculated by dividing the heat of combustion per mole by the molar mass:
\[ CV = \frac{\Delta H \text{ (in kJ)}}{\text{Molar mass (in g)}} = \frac{-1555.8 \text{ kJ}}{30.07 \text{ g}} \approx -51.7 \text{ kJ/g} \]

### Final Answer
The calorific value of ethane is approximately **-51.7 kJ/g**.

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