`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. `2.0`
D. `3.0`

Answer

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

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

### Step 2: Set Up Initial Conditions
Initially, we have:
- \( I_2 = 1 \) mol
- \( I^- = 0.5 \) mol
- \( I_3^- = 0 \) mol

Since the reaction is in a 1 L flask, the initial concentrations are:
- \([I_2] = 1 \, \text{mol/L}\)
- \([I^-] = 0.5 \, \text{mol/L}\)
- \([I_3^-] = 0 \, \text{mol/L}\)

### Step 3: Define Changes at Equilibrium
Let \( x \) be the amount of \( I_2 \) that reacts to form \( I_3^- \). At equilibrium, we will have:
- \( [I_2] = 1 - x \)
- \( [I^-] = 0.5 - x \)
- \( [I_3^-] = x \)

### Step 4: Use Information from Precipitate Formation
It is given that after equilibrium is reached, excess \( AgNO_3 \) gave \( 0.25 \) mol of yellow precipitate. The yellow precipitate is \( AgI \), which forms from \( I^- \). Therefore, \( 0.25 \) mol of \( I^- \) reacted with \( AgNO_3 \).

This means:
\[ 0.5 - x = 0.25 \]

### Step 5: Solve for \( x \)
From the equation:
\[ 0.5 - x = 0.25 \]
We can solve for \( x \):
\[ x = 0.5 - 0.25 = 0.25 \]

### Step 6: Determine Equilibrium Concentrations
Now substituting \( x \) back into the equilibrium expressions:
- \( [I_2] = 1 - 0.25 = 0.75 \, \text{mol/L} \)
- \( [I^-] = 0.5 - 0.25 = 0.25 \, \text{mol/L} \)
- \( [I_3^-] = 0.25 \, \text{mol/L} \)

### Step 7: Calculate the Equilibrium Constant \( K_c \)
The equilibrium constant \( K_c \) is given by the expression:
\[ K_c = \frac{[I_3^-]}{[I_2][I^-]} \]
Substituting the equilibrium concentrations:
\[ K_c = \frac{0.25}{(0.75)(0.25)} \]

### Step 8: Simplify the Expression
Calculating the denominator:
\[ (0.75)(0.25) = 0.1875 \]
Now substituting back:
\[ K_c = \frac{0.25}{0.1875} = \frac{25}{18.75} = \frac{4}{3} \approx 1.33 \]

### Final Answer
The equilibrium constant \( K_c \) is approximately \( 1.33 \).

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