`I_(2)+I^(ɵ)hArrI_(3)^(ɵ)`
This reaction is set-up in aqueous medium. We start with `1` mol of `I_(2)` and `0.5` mol of `I^(ɵ)` in `1 L` flask. After equilibrium reached, excess of `AgNO_(3)` gave `0.25` mol of yellow precipitate. Equilibrium constant is

Options

A. 1.33
B. 2.66
C. `2.00`
D. `3.00`

Answer

To solve the problem step by step, we will analyze the reaction and the information given.

### Step 1: Write the Reaction and Initial Conditions
The reaction is:
\[ I_2 + I^- \rightleftharpoons I_3^- \]

Initially, we have:
- 1 mole of \( I_2 \)
- 0.5 moles of \( I^- \)
- Volume of the flask = 1 L

### Step 2: Set Up the Change in Concentrations
Let \( x \) be the amount of \( I_2 \) that reacts at equilibrium. Then, the changes in concentrations will be:
- \( I_2 \): \( 1 - x \)
- \( I^- \): \( 0.5 - x \)
- \( I_3^- \): \( x \)

### Step 3: Use the Information About AgNO3
After equilibrium is reached, excess \( AgNO_3 \) gives 0.25 moles of yellow precipitate \( AgI \). The formation of \( AgI \) indicates that 0.25 moles of \( I^- \) are present because the stoichiometry of the reaction \( Ag^+ + I^- \rightarrow AgI \) is 1:1.

### Step 4: Relate \( I^- \) to \( x \)
From the information about the precipitate:
\[ 0.5 - x = 0.25 \]
Solving for \( x \):
\[ x = 0.5 - 0.25 = 0.25 \]

### Step 5: Determine Equilibrium Concentrations
Now we can determine the equilibrium concentrations:
- Concentration of \( I_2 \):
\[ [I_2] = 1 - x = 1 - 0.25 = 0.75 \, \text{mol/L} \]
- Concentration of \( I^- \):
\[ [I^-] = 0.5 - x = 0.5 - 0.25 = 0.25 \, \text{mol/L} \]
- Concentration of \( I_3^- \):
\[ [I_3^-] = x = 0.25 \, \text{mol/L} \]

### Step 6: Write the Expression for the Equilibrium Constant \( K_c \)
The equilibrium constant \( K_c \) is given by:
\[ K_c = \frac{[I_3^-]}{[I_2][I^-]} \]

### Step 7: Substitute the Equilibrium Concentrations into the Expression
Substituting the values we found:
\[ K_c = \frac{0.25}{(0.75)(0.25)} \]

### Step 8: Calculate \( K_c \)
Calculating the value:
\[ K_c = \frac{0.25}{0.1875} = \frac{25}{18.75} \approx 1.33 \]

### Final Answer
Thus, the equilibrium constant \( K_c \) is approximately:
\[ K_c \approx 1.33 \]

---