Question

Given,\[{{\text{I}}_{\text{2}}}\left( {{\text{aq}}} \right){\text{ + }}{{\text{I}}^{\text{ - }}}\left( {{\text{aq}}} \right) \rightleftarrows {{\text{I}}^{\text{ - }}}_{\text{3}}\left( {{\text{aq}}} \right)\]. We started with \[1\] mole of \[{{\text{I}}_{\text{2}}}\] and \[0.5\] mole of \[{{\text{I}}^{\text{ - }}}\] in one litre flask. After equilibrium is reached excess of \[{\text{AgN}}{{\text{O}}_{\text{3}}}\] gave \[0.25\] mole of yellow precipitate. Equilibrium constant is
A) \[1.33\]
B) \[2.66\]
C) \[2.0\]
D) \[3.0\]

Answer 1

Hint: We have to know that the mole is one of the important units in chemistry. All the chemical reactions are expressed in terms of the moles of the reactant and product. If one reaction is completed or not is also determined by the moles of the reactant and product. Moles of the molecules are also important for the manufacture of chemicals. Iodine is one of the halogens in the periodic table. Equilibrium constant is an important term in equilibrium chemistry.
Formula used:
When reactions attain the state of equilibrium, the equilibrium constant is an important term. Equilibrium constant is nothing but the relationship between the concentration of the product and reactant at equilibrium. The equilibrium constant is equal to the ratio of the product of the active concentration of the product in the stoichiometric of moles to the moles of reactant. The symbol of equilibrium constant is \[{{\text{K}}_C}\].
The formula of equilibrium constant is
\[{{\text{K}}_C}{\text{ = }}\dfrac{{{\text{number of moles of product at equilibrium}}}}{{{\text{number of moe lsof reactant at equilibrium}}}}\]

Complete answer:
Given chemical reaction,
\[{{\text{I}}_{\text{2}}}\left( {{\text{aq}}} \right){\text{ + }}{{\text{I}}^{\text{ - }}}\left( {{\text{aq}}} \right) \rightleftarrows {{\text{I}}^{\text{ - }}}_{\text{3}}\left( {{\text{aq}}} \right)\]
For the tabulation for initial, final and equilibrium state in moles of reactant and product.

	\[{{\text{I}}_{\text{2}}}\]	\[{{\text{I}}^{\text{ - }}}\]	\[{{\text{I}}^{\text{ - }}}_3\]
Initial moles	\[1\]	\[0.50\]	\[ - \]
After equilibrium	\[1 - 0.25\]	\[0.50 - 0.25\]	\[0.25\]
At equilibrium	\[0.75\]	\[0.25\]	\[0.25\]

For the calculation of equilibrium constant at above condition is
\[{{\text{K}}_C}{\text{ = }}\dfrac{{{\text{number of moles of product at equilibrium}}}}{{{\text{number of moe lsof reactant at equilibrium}}}}\]
Now we can substitute the known values we get,
\[ = \dfrac{{0.25}}{{0.25 \times 0.75}}\]
\[ = \dfrac{{0.25}}{{0.1875}}\]
On simplification we get,
\[{K_c} = 1.33\]
Form the above calculation, we find the equilibrium constant is \[1.33\]

Hence, option A is correct.

Note:
Depending on the equilibrium constant, we conclude the reaction is moving in which direction. The equilibrium constant at non equilibrium is $Q$ . The relationship between the equilibrium constants of ${K_C}$ and $Q$ . If one reaction is in equilibrium state means both values are equal. If the reaction moves forward, the formation of the product means \[Q > {{\text{K}}_C}\]. If the reaction moves forward, the formation of the product means \[Q < {{\text{K}}_C}\].