The pressure of iodine gas at `1273 K` is found to be `0.112` atm whereas the expected pressure is `0.074` atm. The increased pressure is due to dissociation `I_(2) hArr 2I`. Calculate `K_(p)`.

Explanation

To calculate the equilibrium constant \( K_p \) for the dissociation of iodine gas (\( I_2 \)) into iodine atoms (\( 2I \)) at a temperature of 1273 K, we can follow these steps:

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

### Step 2: Define initial and equilibrium conditions
Let:
- At time \( t = 0 \), the initial amount of \( I_2 \) is 1 mole and no \( I \) is present.
- At equilibrium, let \( x \) be the degree of dissociation of \( I_2 \).

Thus, at equilibrium:
- Moles of \( I_2 \) = \( 1 - x \)
- Moles of \( I \) = \( 2x \)

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

### Step 4: Relate experimental and expected pressures
Given:
- Experimental pressure \( P_{\text{exp}} = 0.112 \) atm
- Expected pressure \( P_{\text{expected}} = 0.074 \) atm

Using the relationship between pressure and moles:
\[ \frac{P_{\text{exp}}}{P_{\text{expected}}} = \frac{\text{Total moles at equilibrium}}{\text{Initial moles}} \]
Substituting the known values:
\[ \frac{0.112}{0.074} = \frac{1 + x}{1} \]

### Step 5: Solve for \( x \)
Calculating the left side:
\[ \frac{0.112}{0.074} \approx 1.5135 \]
Thus, we have:
\[ 1 + x = 1.5135 \implies x \approx 0.5135 \]

### Step 6: Write the expression for \( K_p \)
The equilibrium constant \( K_p \) for the reaction is given by:
\[ K_p = \frac{(P_I)^2}{P_{I_2}} \]
Where:
- \( P_I \) is the partial pressure of iodine atoms
- \( P_{I_2} \) is the partial pressure of iodine gas

### Step 7: Calculate partial pressures
Using the mole fractions:
- Partial pressure of \( I \):
\[ P_I = \left(\frac{2x}{1 + x}\right) \times P_{\text{total}} = \left(\frac{2 \times 0.5135}{1 + 0.5135}\right) \times 0.112 \]
Calculating this:
\[ P_I = \left(\frac{1.027}{1.5135}\right) \times 0.112 \approx 0.074 \]

- Partial pressure of \( I_2 \):
\[ P_{I_2} = \left(\frac{1 - x}{1 + x}\right) \times P_{\text{total}} = \left(\frac{1 - 0.5135}{1 + 0.5135}\right) \times 0.112 \]
Calculating this:
\[ P_{I_2} = \left(\frac{0.4865}{1.5135}\right) \times 0.112 \approx 0.042 \]

### Step 8: Substitute into \( K_p \) expression
Now substituting the values into the \( K_p \) expression:
\[ K_p = \frac{(0.074)^2}{0.042} \]
Calculating this gives:
\[ K_p \approx 0.1575 \text{ atm} \]

### Final Answer
Thus, the equilibrium constant \( K_p \) is:
\[ \boxed{0.1575 \text{ atm}} \]