If 100 mole of `H_(2)O_(2)` decompose at 1 bar and 300 K, the work done (kJ) by one mole of `O_(2)(g)` as it expands against 1 bar pressure is
`2H_(2)O_(2) (l) overset(rarr)(larr) 2H_(2)O(l)+O_(2) (g)`
`(R=8.3 JK^(-1) mol^(-1))`

Options

A. `498.00`
B. `62.25`
C. `124.50`
D. `249.00`

Explanation

To solve the problem of calculating the work done by one mole of \( O_2(g) \) as it expands against a pressure of 1 bar during the decomposition of hydrogen peroxide, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Reaction**: The decomposition of hydrogen peroxide is given by the equation:
\[ 2H_2O_2 (l) \rightleftharpoons 2H_2O (l) + O_2 (g) \]

2. **Determine the Change in Moles of Gas (\( \Delta n \))**:
- From the reaction, we see that 2 moles of \( H_2O_2 \) produce 1 mole of \( O_2 \).
- Therefore, when 100 moles of \( H_2O_2 \) decompose, the number of moles of \( O_2 \) produced is:
\[ \Delta n = \frac{100 \text{ moles of } H_2O_2}{2} = 50 \text{ moles of } O_2 \]

3. **Use the Work Done Formula**:
- The work done (\( W \)) in an isobaric process (constant pressure) is given by:
\[ W = -P \Delta V \]
- For gases, we can relate the change in volume to the change in moles using the ideal gas law:
\[ \Delta V = \Delta n \cdot \frac{RT}{P} \]
- Therefore, substituting into the work done formula:
\[ W = -P \left( \Delta n \cdot \frac{RT}{P} \right) = -\Delta n RT \]

4. **Substitute Known Values**:
- Here, \( \Delta n = 50 \) moles (from step 2), \( R = 8.314 \, J \, K^{-1} \, mol^{-1} \), and \( T = 300 \, K \).
- Thus, the work done becomes:
\[ W = -50 \cdot 8.314 \cdot 300 \]

5. **Calculate Work Done**:
- Performing the calculation:
\[ W = -50 \cdot 8.314 \cdot 300 = -124,710 \, J \]
- Converting to kilojoules:
\[ W = -124.71 \, kJ \]

6. **Final Answer**: The work done by one mole of \( O_2(g) \) as it expands against a pressure of 1 bar is approximately:
\[ W \approx 124.71 \, kJ \]