Calculate the calorific value of octane a component of gasoline with the help of following reaction. `2C_(8)H_(18(l))+25O_(2(g)) rarr 16 CO_(2(g))+18H_(2)O, DeltaH=-10.920 J`

Options

A. `478.9 kJ`
B. `47.89 kJ`
C. `95.78 kJ`
D. `957.8 kJ`

Answer

To calculate the calorific value of octane (C₈H₁₈) from the given reaction, we will follow these steps:

### Step-by-Step Solution:

1. **Identify the Reaction and Given Data**:
The reaction provided is:
\[ 2C_{8}H_{18(l)} + 25O_{2(g)} \rightarrow 16CO_{2(g)} + 18H_{2}O \]
The enthalpy change (ΔH) for this reaction is given as:
\[ \Delta H = -10,920 \text{ J} \]
This value indicates the heat released during the combustion of 2 moles of octane.

2. **Calculate the Heat Released per Mole of Octane**:
Since the reaction involves 2 moles of octane, we need to find the heat released for 1 mole of octane:
\[ \text{Heat released per mole of octane} = \frac{-10,920 \text{ J}}{2} = -5,460 \text{ J/mol} \]

3. **Convert the Heat Released to Kilojoules**:
To express the calorific value in kilojoules, we convert joules to kilojoules:
\[ -5,460 \text{ J/mol} = -5.460 \text{ kJ/mol} \]

4. **Determine the Molar Mass of Octane**:
The molar mass of octane (C₈H₁₈) can be calculated as follows:
- Carbon (C): 12.01 g/mol × 8 = 96.08 g/mol
- Hydrogen (H): 1.008 g/mol × 18 = 18.144 g/mol
- Total molar mass = 96.08 g/mol + 18.144 g/mol = 114.224 g/mol

5. **Calculate the Calorific Value**:
The calorific value (CV) is defined as the heat released per unit mass of the fuel:
\[ \text{Calorific Value (CV)} = \frac{\text{Heat released (kJ/mol)}}{\text{Molar mass (kg/mol)}} \]
Convert the molar mass from grams to kilograms:
\[ \text{Molar mass in kg} = 114.224 \text{ g/mol} \times \frac{1 \text{ kg}}{1000 \text{ g}} = 0.114224 \text{ kg/mol} \]
Now, substitute the values into the formula:
\[ CV = \frac{-5.460 \text{ kJ/mol}}{0.114224 \text{ kg/mol}} \approx -47.89 \text{ kJ/kg} \]

6. **Final Result**:
The calorific value of octane is approximately:
\[ CV \approx 47.89 \text{ kJ/kg} \]