Question

Following reaction is set up in aqueous medium:
${{\text{I}}_2} + {{\text{I}}^ - } \rightleftharpoons {\text{I}}_3^ - $
We start with one mole of ${{\text{I}}_2}$ and $0.5{\text{ mol}}$ of ${{\text{I}}^ - }$ in $1{\text{ L}}$ flask. After equilibrium is reached, excess of ${\text{AgN}}{{\text{O}}_{\text{3}}}$ gave $0.25{\text{ mol}}$ of yellow precipitate. Equilibrium constant is
A) 1.33
B) 2.66
C) 2.00
D) 3.00

Answer 1

Hint:The yellow precipitate formed is silver iodide $\left( {{\text{AgI}}} \right)$. Silver iodide is formed by the reaction of ${{\text{I}}^ - }$ with ${\text{AgN}}{{\text{O}}_{\text{3}}}$. The equilibrium constant expresses the relationship between the amounts of products and the amounts of reactants that are present at equilibrium in a reversible reaction.

Complete solution:
We are given the reaction as follows:
${{\text{I}}_2} + {{\text{I}}^ - } \rightleftharpoons {\text{I}}_3^ - $
In the reaction, with one mole of ${{\text{I}}_2}$ and $0.5{\text{ mol}}$ of ${{\text{I}}^ - }$ react. Let the moles of ${\text{I}}_3^ - $ formed at equilibrium be x. Thus, at equilibrium the moles of ${{\text{I}}_2}$ will be $1 - x$, moles of ${{\text{I}}^ - }$ will be $0.5 - x$.
We are given that after equilibrium is reached, excess of ${\text{AgN}}{{\text{O}}_{\text{3}}}$ gave $0.25{\text{ mol}}$ of yellow precipitate. The yellow precipitate formed is silver iodide $\left( {{\text{AgI}}} \right)$. Silver iodide is formed by the reaction of ${{\text{I}}^ - }$ with ${\text{AgN}}{{\text{O}}_{\text{3}}}$.
From the molecular formula of silver iodide we can say that one mole of silver iodide corresponds to one mole of ${{\text{I}}^ - }$. Thus,
${\text{1 mol AgI}} = 1{\text{ mol }}{{\text{I}}^ - }$
We are given that $0.25{\text{ mol}}$ of yellow precipitate of silver iodide is produced. Thus,
$0.25{\text{ mol AgI}} = 0.5 - x$
 $x = 0.5 - 0.25$
 $x = 0.25$
The equilibrium constant expresses the relationship between the amounts of products and the amounts of reactants that are present at equilibrium in a reversible reaction. Thus,
$K = \dfrac{{[{\text{I}}_3^ - ]}}{{[{{\text{I}}_2}][{{\text{I}}^ - }]}}$
Where $K$ is the equilibrium constant.
$K = \dfrac{x}{{\left( {1 - x} \right)\left( {0.5 - x} \right)}}$
We have calculated that $x = 0.25$. Thus,
$K = \dfrac{{0.25}}{{\left( {1 - 0.25} \right)\left( {0.5 - 0.25} \right)}}$
$K = \dfrac{{0.25}}{{\left( {0.75} \right)\left( {0.25} \right)}}$
$K = 1.33$
Thus, the equilibrium constant is 1.33.

Thus, the correct option is (A) 1.33.

Note:Remember that the equilibrium constant expresses the relationship between the amounts of products and the amounts of reactants that are present at equilibrium in a reversible reaction. In simple words, equilibrium constant is the ratio of the concentration of products to the concentration of reactants.