Heat of combustion of benzoic acid `(C_(6)H_(5)COOH)` at constant volume at `25^(@)C` is `-3233` kJ/mole. When 0.5 g of benzoic acid is burnt in bomb calorimeter, the temperature of calorimeter increased by `0.53^(@)C`. Now in the same bomb calorimeter 1g of `C_(2)H_(6)` burnt then temperature increased by `2.04 C`. `DeltaH` for combustion of `C_(2)H_(6)` is

Options

A. `-1530` kJ/mole
B. `-1536.2` kJ/mole
C. `-1522.8` kJ/mole
D. `+1536.2` kJ/mole

Explanation

To find the enthalpy change (ΔH) for the combustion of ethane (C2H6), we will use the data provided from the combustion of benzoic acid and the temperature change observed in the bomb calorimeter.

### Step-by-step Solution:

1. **Calculate the Heat Released from Benzoic Acid Combustion:**
The heat of combustion of benzoic acid is given as -3233 kJ/mole. We need to find out how much heat is released when 0.5 g of benzoic acid is burnt.

- Molar mass of benzoic acid (C6H5COOH) = 6*12 + 5*1 + 12 + 16 = 122 g/mol
- Moles of benzoic acid burnt = mass / molar mass = 0.5 g / 122 g/mol = 0.0041 moles
- Heat released (q) = moles * heat of combustion = 0.0041 moles * -3233 kJ/mole = -13.25 kJ

2. **Calculate the Heat Capacity of the Calorimeter:**
The temperature increase in the calorimeter when 0.5 g of benzoic acid is burnt is 0.53 °C. The heat absorbed by the calorimeter (q) can also be expressed as:

\[ q = C \cdot \Delta T \]
where C is the heat capacity of the calorimeter, and ΔT is the change in temperature.

- Rearranging gives us:
\[ C = \frac{q}{\Delta T} = \frac{-13.25 \text{ kJ}}{0.53 \text{ °C}} = -25.00 \text{ kJ/°C} \]

3. **Calculate the Heat Released from Ethane Combustion:**
Now, we will use the same calorimeter to find the ΔH for the combustion of ethane (C2H6). When 1 g of ethane is burnt, the temperature of the calorimeter increased by 2.04 °C.

- Moles of ethane burnt = mass / molar mass = 1 g / 30 g/mol = 0.0333 moles (molar mass of C2H6 = 30 g/mol)
- Heat absorbed by the calorimeter during the combustion of ethane:
\[ q = C \cdot \Delta T = -25.00 \text{ kJ/°C} \cdot 2.04 \text{ °C} = -51.00 \text{ kJ} \]

4. **Calculate the Enthalpy Change (ΔH) for Ethane:**
The heat released during the combustion of ethane can be calculated as follows:
\[ \Delta H = \frac{q}{\text{moles of ethane}} = \frac{-51.00 \text{ kJ}}{0.0333 \text{ moles}} = -1530.00 \text{ kJ/mole} \]

### Final Answer:
The enthalpy change (ΔH) for the combustion of ethane (C2H6) is approximately **-1530 kJ/mole**.