At 1000K, the pressure of iodine gas is found to be 0.112 atm due to partial dissociation of `I_(2)(g)` into I(g). Had there been no dissociation, the pressure would have been 0.074 atm. Calculate the value of `K_(p)` for the reaction:
`I_(2)(g) hArr 2I(g)`.

Explanation

To calculate the value of \( K_p \) for the reaction \( I_2(g) \rightleftharpoons 2I(g) \) at 1000 K, we can follow these steps:

### Step 1: Identify the Initial and Equilibrium Pressures
- The initial pressure of \( I_2 \) without dissociation is given as \( P_{I_2} = 0.074 \, \text{atm} \).
- The pressure at equilibrium, after partial dissociation, is \( P_{total} = 0.112 \, \text{atm} \).

### Step 2: Define the Change in Pressure
Let \( x \) be the amount of \( I_2 \) that dissociates. According to the stoichiometry of the reaction:
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