Calculate the work done (in J) when 4.5 g of `H_(2)O_(2)` reacts against a pressure of 1.0 atm at `25^(@)C` `2H_(2)O_(2)(l)rarr O_(2)(g) +2H_(2)O(l)`

Options

A. `-1.63 xx 10^(2)`
B. `4.5 xx 10^(2)`
C. `3.2 xx 10^(2)`
D. `-6.1 xx 10^(2)`

Explanation

To calculate the work done when 4.5 g of \( H_2O_2 \) reacts against a pressure of 1.0 atm at \( 25^\circ C \), we can follow these steps:

### Step 1: Write the balanced chemical equation
The reaction is given as:
\[ 2 H_2O_2 (l) \rightarrow O_2 (g) + 2 H_2O (l) \]

### Step 2: Calculate the number of moles of \( H_2O_2 \)
To find the number of moles of \( H_2O_2 \), we use the formula:
\[ \text{Number of moles} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}} \]

The molar mass of \( H_2O_2 \) (Hydrogen Peroxide) is approximately 34 g/mol. Thus:
\[ \text{Number of moles of } H_2O_2 = \frac{4.5 \, \text{g}}{34 \, \text{g/mol}} \]

### Step 3: Determine the change in moles of gas (\( \Delta N_g \))
From the balanced equation, we can see that:
- 2 moles of \( H_2O_2 \) produce 1 mole of \( O_2 \) (g) and 2 moles of \( H_2O \) (l).
- Therefore, the change in moles of gas (\( \Delta N_g \)) is:
\[ \Delta N_g = \text{moles of products} - \text{moles of reactants} = 1 - 2 = -1 \]

### Step 4: Calculate the work done (\( W \))
The work done against a constant pressure is given by the formula:
\[ W = -P \Delta V \]
Using the ideal gas law, we can express this as:
\[ W = -\Delta N_g \cdot R \cdot T \]
Where:
- \( R = 8.314 \, \text{J/(mol K)} \)
- \( T = 298 \, \text{K} \) (25°C)

Substituting the values:
\[ W = -(-1) \cdot \Delta N_g \cdot R \cdot T \]

### Step 5: Substitute the values into the equation
Now we need to calculate the number of moles of \( H_2O_2 \):
\[ \text{Number of moles of } H_2O_2 = \frac{4.5}{34} \approx 0.1324 \, \text{mol} \]

Now substituting into the work equation:
\[ W = -(-1) \cdot (0.1324) \cdot (8.314) \cdot (298) \]

### Step 6: Calculate the work done
Calculating this gives:
\[ W \approx 0.1324 \cdot 8.314 \cdot 298 \approx 329.4 \, \text{J} \]

Thus, the work done is:
\[ W \approx -1.63 \times 10^2 \, \text{J} \]

### Final Answer
The work done when 4.5 g of \( H_2O_2 \) reacts against a pressure of 1.0 atm at \( 25^\circ C \) is approximately:
\[ W \approx -163 \, \text{J} \]

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