Question

If 1.0 mole of \( \text{I}_2 \) is introduced into a 1.0 litre flask at 1000 K, at equilibrium (\( K_c = 10^{-6} \)), which one is correct?

• A. \( [ \text{I}_2 (g) ] > [ \text{I} (g) ] \)
• B. \( [ \text{I}_2 (g) ] < [ \text{I} (g) ] \)
• C. \( [ \text{I}_2 (g) ] = [ \text{I} (g) ] \)
• D. \( [ \text{I}_2 (g) ] = \frac{1}{2} [ \text{I} (g) ] \)

Hint:

For a reaction with a small equilibrium constant, the reactants (in this case, \( \text{I}_2 \)) are favored, and the products (in this case, \( \text{I} \)) will be present in much smaller amounts.

Answer 1

The Correct Option is B

Step 1: Analyze the equilibrium expression.
The equilibrium constant for the dissociation of iodine is given by: \[ K_c = \frac{[ \text{I} ]^2}{[ \text{I}_2 ]} \] Since \( K_c \) is very small, this means that most of the iodine remains in the molecular form, and the concentration of \( \text{I}_2 \) will be greater than that of \( \text{I} \).
Step 2: Conclusion.
At equilibrium, the concentration of \( \text{I}_2 \) will be less than that of \( \text{I} \). Final Answer: \[ \boxed{[ \text{I}_2 (g) ] < [ \text{I} (g) ]} \]