One mole of solid Zn is placed in excess of dilute `H_(2)SO_(4)` at `27^(@)C` in a cylinder fitted with a piston . Find the work done for the process of the area of piston is `500 cm^(2)` and it moves out by 50 cm against a pressure of 1 atm during the reaction.
`Zn(s) + 2H^(+)(aq) hArr Zn^(2+) (aq) + H_(2)(g)`

Options

A. `-1.53 kJ`
B. `-2.53 kJ`
C. zero
D. `2.53 KJ`

Explanation

To find the work done during the reaction of zinc with dilute sulfuric acid, we can follow these steps:

### Step 1: Understand the reaction
The reaction of zinc with sulfuric acid can be represented as:
\[ \text{Zn(s)} + 2\text{H}^+(aq) \rightleftharpoons \text{Zn}^{2+}(aq) + \text{H}_2(g) \]

During this reaction, hydrogen gas is evolved, which pushes the piston upwards.

### Step 2: Identify the given data
- Area of the piston \( A = 500 \, \text{cm}^2 \)
- Displacement of the piston \( h = 50 \, \text{cm} \)
- External pressure \( P = 1 \, \text{atm} \)

### Step 3: Calculate the change in volume (\( \Delta V \))
The change in volume can be calculated using the formula:
\[ \Delta V = A \times h \]
Convert the area from cm² to m²:
\[ A = 500 \, \text{cm}^2 = 500 \times 10^{-4} \, \text{m}^2 = 0.05 \, \text{m}^2 \]
Convert the height from cm to m:
\[ h = 50 \, \text{cm} = 0.5 \, \text{m} \]
Now calculate \( \Delta V \):
\[ \Delta V = 0.05 \, \text{m}^2 \times 0.5 \, \text{m} = 0.025 \, \text{m}^3 \]

### Step 4: Convert volume to liters
Since \( 1 \, \text{m}^3 = 1000 \, \text{L} \):
\[ \Delta V = 0.025 \, \text{m}^3 = 25 \, \text{L} \]

### Step 5: Calculate the work done (\( W \))
The work done by the system can be calculated using the formula:
\[ W = -P \Delta V \]
Convert pressure from atm to the appropriate unit:
\[ 1 \, \text{atm} = 101.325 \, \text{J/L} \]
Now calculate the work done:
\[ W = -1 \, \text{atm} \times 25 \, \text{L} = -25 \, \text{atm} \cdot \text{L} \]
Convert this to joules:
\[ W = -25 \, \text{atm} \cdot \text{L} \times 101.325 \, \text{J/L} = -2533.125 \, \text{J} \]

### Step 6: Convert joules to kilojoules
\[ W = -2.533125 \, \text{kJ} \]

### Final Answer:
The work done during the reaction is approximately:
\[ W \approx -2.53 \, \text{kJ} \]