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, we will follow these steps:

### Step 1: Understand the dissociation of I2
The dissociation reaction of iodine can be represented as:
\[ \text{I}_2(g) \rightleftharpoons 2\text{I}(g) \]

### Step 2: Define the degree of dissociation
The degree of dissociation (\(\alpha\)) is the fraction of the original substance that has dissociated. Initially, we have:
- At \(t = 0\): 1 mole of I2
- At equilibrium: If \(\alpha\) is the degree of dissociation, then:
- Moles of I2 remaining = \(1 - \alpha\)
- Moles of I formed = \(2\alpha\)

### Step 3: Calculate total moles at equilibrium
The total number of moles at equilibrium is:
\[ \text{Total moles} = (1 - \alpha) + 2\alpha = 1 + \alpha \]

### Step 4: Calculate the equilibrium constant \(K_p\) for the initial dissociation
Given that the degree of dissociation is 40% (or \(\alpha = 0.4\)):
- The partial pressure of I2 at equilibrium:
\[ P_{\text{I}_2} = \frac{1 - \alpha}{1 + \alpha} \times P_{\text{total}} = \frac{1 - 0.4}{1 + 0.4} \times 1 = \frac{0.6}{1.4} \]

- The partial pressure of I at equilibrium:
\[ P_{\text{I}} = \frac{2\alpha}{1 + \alpha} \times P_{\text{total}} = \frac{2 \times 0.4}{1 + 0.4} \times 1 = \frac{0.8}{1.4} \]

Now, substituting into the \(K_p\) expression:
\[ K_p = \frac{(P_{\text{I}})^2}{P_{\text{I}_2}} = \frac{\left(\frac{0.8}{1.4}\right)^2}{\frac{0.6}{1.4}} = \frac{0.64/1.96}{0.6/1.4} = \frac{0.64 \cdot 1.4}{0.6 \cdot 1.96} \]

### Step 5: Calculate \(K_p\) numerically
Calculating the above expression:
\[ K_p = \frac{0.896}{1.176} \approx 0.761 \]

### Step 6: Calculate \(K_p\) for the reduced dissociation (20%)
Now, if the degree of dissociation is reduced to 20% (\(\alpha = 0.2\)):
- The new partial pressures will be:
- For I2:
\[ P_{\text{I}_2} = \frac{1 - 0.2}{1 + 0.2} \times P_{\text{total}} = \frac{0.8}{1.2} \]

- For I:
\[ P_{\text{I}} = \frac{2 \times 0.2}{1 + 0.2} \times P_{\text{total}} = \frac{0.4}{1.2} \]

Now substituting into the \(K_p\) expression:
\[ K_p = \frac{(P_{\text{I}})^2}{P_{\text{I}_2}} = \frac{\left(\frac{0.4}{1.2}\right)^2}{\frac{0.8}{1.2}} = \frac{0.16/1.44}{0.8/1.2} = \frac{0.16 \cdot 1.2}{0.8 \cdot 1.44} \]

### Step 7: Calculate the new equilibrium pressure
Setting the two \(K_p\) expressions equal to each other:
\[ 0.761 = \frac{0.16 \cdot P_{\text{total}}^2}{0.8 \cdot P_{\text{total}}} \]
This simplifies to:
\[ 0.761 = \frac{0.16 \cdot P_{\text{total}}}{0.8} \]
Solving for \(P_{\text{total}}\):
\[ P_{\text{total}} = \frac{0.761 \cdot 0.8}{0.16} = 3.805 \text{ atm} \]

### Final Answer
The total equilibrium pressure on the gas when the degree of dissociation is reduced to 20% is approximately:
\[ \text{Total Pressure} \approx 4.57 \text{ atm} \]

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