In the dissociation of `I_(2)(g)` at 1000 K in a container of 1 litre :-
`I_(2)(g)rArr2I^(-)(g),K_(c)=10^(10)`
Select the correct relation :-

Options

A. `(a)[I_(2)]gt[I^(-)]`
B. (b)`[I_(2)]lt[I^(-)]`
C. (c)`[I_(2)]=[I^(-)]`
D. (d)`[I_(2)]=(1)/(2)[I^(-)]`

Answer

To solve the problem, we need to analyze the dissociation of \( I_2(g) \) into \( 2I^-(g) \) and understand the implications of the equilibrium constant \( K_c \).

### Step-by-Step Solution:

1. **Write the Reaction and Equilibrium Expression**:
The dissociation reaction is:
\[ I_2(g) \rightleftharpoons 2I^-(g) \]
The equilibrium constant \( K_c \) for this reaction is given by:
\[ K_c = \frac{[I^-]^2}{[I_2]} \]

2. **Substitute the Given Value of \( K_c \)**:
We know that \( K_c = 10^{10} \). Therefore, we can write:
\[ 10^{10} = \frac{[I^-]^2}{[I_2]} \]

3. **Analyze the Implications of \( K_c \)**:
A \( K_c \) value of \( 10^{10} \) indicates that at equilibrium, the concentration of products (in this case, \( I^- \)) is much greater than that of the reactant (\( I_2 \)). This suggests that the reaction favors the formation of products.

4. **Set Up Relationships Between Concentrations**:
Let's denote the concentration of \( I_2 \) at equilibrium as \( [I_2] \) and the concentration of \( I^- \) as \( [I^-] \). From the equilibrium expression:
\[ [I^-]^2 = 10^{10} \times [I_2] \]
This implies that:
\[ [I^-] = \sqrt{10^{10} \times [I_2]} = 10^5 \sqrt{[I_2]} \]

5. **Determine the Relationship Between \( [I^-] \) and \( [I_2] \)**:
Since \( K_c \) is very large, it indicates that \( [I^-] \) must be significantly greater than \( [I_2] \). Hence, we can conclude:
\[ [I^-] > [I_2] \]

6. **Evaluate the Options**:
Based on our analysis, we can evaluate the given options:
- \( [I_2] > [I^-] \) (False)
- \( [I^-] > [I_2] \) (True)
- \( [I_2] = [I^-] \) (False)
- \( [I_2] = \frac{1}{2}[I^-] \) (False)

The correct relation is \( [I^-] > [I_2] \).

### Conclusion:
The correct answer is that the concentration of \( I^- \) is greater than the concentration of \( I_2 \).

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