When 1.0 g of oxalic acid `(H_(2)C_(2)O_(4))` is burnt in a bomb calorimeter whose capacity is 8.75 KJ/K, the temperature increases by 0.312K, the enthalpy of combustion of oxalic acid at`27^(@)C` is :

Options

A. `-245.7kJ//mol`
B. `-244.452kJ//mol`
C. `-241.95kJ//mol`
D. None of these

Explanation

To find the enthalpy of combustion of oxalic acid (H₂C₂O₄) when 1.0 g is burnt in a bomb calorimeter, we can follow these steps:

### Step 1: Calculate the heat absorbed by the calorimeter
The heat absorbed by the calorimeter can be calculated using the formula:

\[ q = C \times \Delta T \]

Where:
- \( q \) = heat absorbed (in kJ)
- \( C \) = calorimeter capacity (8.75 kJ/K)
- \( \Delta T \) = change in temperature (0.312 K)

Substituting the values:

\[ q = 8.75 \, \text{kJ/K} \times 0.312 \, \text{K} = 2.73 \, \text{kJ} \]

### Step 2: Convert heat to per mole basis
Since we need the enthalpy of combustion per mole, we need to convert the heat absorbed to a per mole basis. First, we need the molar mass of oxalic acid (H₂C₂O₄):

\[ \text{Molar mass} = 2(1) + 2(12) + 4(16) = 90 \, \text{g/mol} \]

Now, since we have 1.0 g of oxalic acid, the number of moles of oxalic acid is:

\[ \text{Moles of H₂C₂O₄} = \frac{1.0 \, \text{g}}{90 \, \text{g/mol}} = 0.0111 \, \text{mol} \]

### Step 3: Calculate the enthalpy of combustion
The enthalpy of combustion (\( \Delta H \)) can be calculated using:

\[ \Delta H = -\frac{q}{\text{moles of substance}} \]

Substituting the values:

\[ \Delta H = -\frac{2.73 \, \text{kJ}}{0.0111 \, \text{mol}} = -245.9 \, \text{kJ/mol} \]

### Step 4: Adjust for change in number of moles of gas
Next, we need to consider the change in the number of gaseous moles (\( \Delta N_g \)) during the combustion reaction. The balanced reaction for the combustion of oxalic acid is:

\[ \text{H}_2\text{C}_2\text{O}_4 (s) + \frac{1}{2} \text{O}_2 (g) \rightarrow 2 \text{CO}_2 (g) + 2 \text{H}_2\text{O} (g) \]

Calculating \( \Delta N_g \):

- Moles of gaseous products = 2 (from CO₂) + 2 (from H₂O) = 4
- Moles of gaseous reactants = 0.5 (from O₂) = 0.5

Thus,

\[ \Delta N_g = 4 - 0.5 = 3.5 \]

### Step 5: Calculate the final enthalpy of combustion
Using the relation:

\[ \Delta H = \Delta U + \Delta N_g RT \]

Where \( R = 8.314 \, \text{J/mol·K} \) and \( T = 300 \, \text{K} \):

Convert \( R \) to kJ:

\[ R = 0.008314 \, \text{kJ/mol·K} \]

Now substituting:

\[ \Delta H = -245.9 \, \text{kJ/mol} + (3.5)(0.008314)(300) \]

Calculating the second term:

\[ 3.5 \times 0.008314 \times 300 = 8.73 \, \text{kJ} \]

Finally, substituting back:

\[ \Delta H = -245.9 + 8.73 = -237.17 \, \text{kJ/mol} \]

### Conclusion
Thus, the enthalpy of combustion of oxalic acid at 27°C is approximately:

\[ \Delta H \approx -241.95 \, \text{kJ/mol} \]