The degree of dissociation of `I_(2)` "mole"cule 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

Answer

To solve the problem regarding the degree of dissociation of \( I_2 \) at \( 1000^\circ C \) and under \( 1.0 \, \text{atm} \), we will follow these steps:

### Step 1: Write the dissociation equation
The dissociation of iodine can be represented as:
\[ I_2 \rightleftharpoons 2I \]

### Step 2: Define initial conditions
Assume we start with 1 mole of \( I_2 \).

### Step 3: Determine moles at equilibrium for 40% dissociation
If the degree of dissociation is 40%, then:
- \( x = 0.4 \) (where \( x \) is the fraction of \( I_2 \) that dissociates)

At equilibrium:
- Moles of \( I_2 \) left = \( 1 - x = 1 - 0.4 = 0.6 \)
- Moles of \( I \) formed = \( 2x = 2 \times 0.4 = 0.8 \)

Total moles at equilibrium:
\[ \text{Total moles} = 0.6 + 0.8 = 1.4 \]

### Step 4: Calculate partial pressures
The total pressure is given as \( 1.0 \, \text{atm} \). The partial pressures can be calculated as follows:
- Partial pressure of \( I_2 \):
\[ P_{I_2} = \left( \frac{0.6}{1.4} \right) \times 1.0 = \frac{0.6}{1.4} \approx 0.4286 \, \text{atm} \]
- Partial pressure of \( I \):
\[ P_I = \left( \frac{0.8}{1.4} \right) \times 1.0 = \frac{0.8}{1.4} \approx 0.5714 \, \text{atm} \]

### Step 5: Calculate \( K_p \)
Using the expression for \( K_p \):
\[ K_p = \frac{(P_I)^2}{P_{I_2}} = \frac{(0.5714)^2}{0.4286} \]
Calculating this gives:
\[ K_p = \frac{0.3265}{0.4286} \approx 0.7618 \]

### Step 6: Determine moles at equilibrium for 20% dissociation
Now, if the degree of dissociation is reduced to 20%, then:
- \( x' = 0.2 \)

At equilibrium:
- Moles of \( I_2 \) left = \( 1 - x' = 1 - 0.2 = 0.8 \)
- Moles of \( I \) formed = \( 2x' = 2 \times 0.2 = 0.4 \)

Total moles at equilibrium:
\[ \text{Total moles} = 0.8 + 0.4 = 1.2 \]

### Step 7: Set up the equation for \( K_p \) at 20% dissociation
Using the same \( K_p \):
\[ K_p = \frac{(P_I')^2}{P_{I_2}'} = \frac{(P_I')^2}{P_{I_2}'} \]
Where:
- \( P_{I_2}' = \left( \frac{0.8}{1.2} \right) P \)
- \( P_I' = \left( \frac{0.4}{1.2} \right) P \)

Substituting these into the \( K_p \) expression:
\[ K_p = \frac{\left( \frac{0.4}{1.2} P \right)^2}{\left( \frac{0.8}{1.2} P \right)} = \frac{0.16 P^2}{0.8} \cdot \frac{1}{1.44} = \frac{0.16 P^2}{0.8 \cdot 1.2} \]

### Step 8: Solve for the new pressure \( P \)
Setting the two expressions for \( K_p \) equal:
\[ 0.7618 = \frac{0.16 P^2}{0.96} \]
Solving for \( P \):
\[ P^2 = \frac{0.7618 \cdot 0.96}{0.16} \]
\[ P^2 = 4.57 \quad \Rightarrow \quad P \approx 2.14 \, \text{atm} \]

### Final Answer
The total equilibrium pressure on the gas when the dissociation is reduced to 20% is approximately \( 2.14 \, \text{atm} \).