The degree of dissociation of `I_(2)` molecule at `1000^(@)C` and under `1.0 atm` is `40%` by volume. If the dissociation is reduced to `20%` at the same temperature, the total equilibrium pressure on the gas will be:

Options

A. `1.57 atm`
B. `2.57 atm`
C. `3.57 atm`
D. `4.57 atm`

Explanation

To solve the problem, we need to find the total equilibrium pressure when the degree of dissociation of \( I_2 \) is reduced to 20% at 1000°C. We will use the concept of equilibrium and the degree of dissociation.

### Step-by-Step Solution:

1. **Understanding the Reaction**:
The dissociation of iodine can be represented as:
\[ I_2 \rightleftharpoons 2I \]

2. **Initial Moles**:
Let's assume we start with 1 mole of \( I_2 \).

3. **Degree of Dissociation**:
The degree of dissociation (\( \alpha \)) is the fraction of \( I_2 \) that dissociates.
- For 40% dissociation: \( \alpha = 0.4 \)
- For 20% dissociation: \( \alpha = 0.2 \)

4. **Equilibrium Moles Calculation**:
- At 40% dissociation:
- Moles of \( I_2 \) at equilibrium = \( 1 - \alpha = 1 - 0.4 = 0.6 \)
- Moles of \( I \) formed = \( 2\alpha = 2 \times 0.4 = 0.8 \)
- Total moles at equilibrium = \( 0.6 + 0.8 = 1.4 \)

- At 20% dissociation:
- Moles of \( I_2 \) at equilibrium = \( 1 - \alpha = 1 - 0.2 = 0.8 \)
- Moles of \( I \) formed = \( 2\alpha = 2 \times 0.2 = 0.4 \)
- Total moles at equilibrium = \( 0.8 + 0.4 = 1.2 \)

5. **Calculating \( K_p \)**:
The equilibrium constant \( K_p \) can be expressed in terms of partial pressures:
\[ K_p = \frac{(P_I)^2}{P_{I_2}} \]
Where:
- \( P_I \) is the partial pressure of iodine atoms.
- \( P_{I_2} \) is the partial pressure of iodine molecules.

At 40% dissociation:
- Total pressure \( P = 1.0 \, \text{atm} \)
- \( P_{I_2} = \frac{0.6}{1.4} \times P \)
- \( P_I = \frac{0.8}{1.4} \times P \)

Substituting into \( K_p \):
\[ K_p = \frac{\left(\frac{0.8}{1.4} P\right)^2}{\frac{0.6}{1.4} P} = \frac{0.64 P^2}{0.6 P} = \frac{0.64 P}{0.6} = \frac{32}{30} P \]

6. **Finding New Pressure**:
Now, we calculate \( K_p \) for 20% dissociation:
\[ K_p = \frac{(P_I)^2}{P_{I_2}} = \frac{\left(\frac{0.4}{1.2} P'\right)^2}{\frac{0.8}{1.2} P'} = \frac{0.16 P'^2}{0.8 P'} = \frac{0.16 P'}{0.8} = \frac{0.2}{1} P' \]

Since \( K_p \) remains constant:
\[ \frac{32}{30} P = \frac{0.2}{1} P' \]

Solving for \( P' \):
\[ P' = \frac{32}{30} P \times \frac{1}{0.2} = \frac{32}{30} \times 5 = \frac{32}{6} = 5.33 \, \text{atm} \]

### Final Answer:
The total equilibrium pressure when the degree of dissociation is reduced to 20% is approximately \( 5.33 \, \text{atm} \).