What is the value of `k_(c)` for the reaction at 1473 K
`I_(2)(g) hArr 2I(g)`
when one mode of iodic gas is introduced into an evacuated one litre flask so that only 5% of it gets dissociated ?

Options

A. 0.0105
B. 0.046
C. 0.034
D. 0.5

Explanation

To find the value of \( K_c \) for the reaction \( I_2(g) \rightleftharpoons 2I(g) \) at 1473 K, we follow these steps:

### Step 1: Understand the Reaction
The reaction involves the dissociation of 1 mole of iodine gas (\( I_2 \)) into 2 moles of iodine atoms (\( I \)).

### Step 2: Initial Conditions
We start with 1 mole of \( I_2 \) in a 1-liter flask. Therefore, the initial concentration of \( I_2 \) is:
\[ [I_2] = 1 \text{ mole/L} \]
And the initial concentration of \( I \) is:
\[ [I] = 0 \text{ mole/L} \]

### Step 3: Degree of Dissociation
Given that 5% of \( I_2 \) dissociates, we can express this as:
\[ \alpha = 0.05 \]
This means that 5% of 1 mole of \( I_2 \) dissociates.

### Step 4: Calculate Concentrations at Equilibrium
After dissociation, the concentration of \( I_2 \) at equilibrium will be:
\[ [I_2] = 1 - \alpha = 1 - 0.05 = 0.95 \text{ moles/L} \]

The amount of \( I_2 \) that dissociates is \( 0.05 \) moles, which produces:
\[ 2 \times 0.05 = 0.1 \text{ moles of } I \]
Thus, the concentration of \( I \) at equilibrium will be:
\[ [I] = 0.1 \text{ moles/L} \]

### Step 5: Write the Expression for \( K_c \)
The equilibrium constant \( K_c \) for the reaction is given by:
\[ K_c = \frac{[I]^2}{[I_2]} \]

### Step 6: Substitute the Equilibrium Concentrations
Substituting the equilibrium concentrations into the expression:
\[ K_c = \frac{(0.1)^2}{0.95} \]

### Step 7: Calculate \( K_c \)
Calculating the value:
\[ K_c = \frac{0.01}{0.95} \approx 0.010526 \]

### Step 8: Round the Result
Rounding the result gives:
\[ K_c \approx 0.0105 \]

### Conclusion
The value of \( K_c \) for the reaction at 1473 K is approximately \( 0.0105 \).