Question

In the process ${{\rm{I}}_2} + {{\rm{I}}^ - } \to {{\rm{I}}_{\rm{3}}}^ - $, initially there are 2 moles ${{\rm{I}}_2}$and 2 mole of ${\rm{I}}$. But at equilibrium, due to addition of ${\rm{AgN}}{{\rm{O}}_{\rm{3}}}\left( {aq} \right)$, 1.75 mole of yellow ppt. is obtained. ${K_c}$for the process is $\left( {{V_{flask}} = 1\;{\rm{d}}{{\rm{m}}^3}} \right)$nearly
A. 0.08
B. 0.02
C. 0.16
D. 0.12

Answer 1

Hint: First, we have to write the ICE table for the reaction, ${{\rm{I}}_2} + {{\rm{I}}^ - } \to {{\rm{I}}_{\rm{3}}}^ - $ using the values provided in the question. Then, we calculate the ${K_c}$value for the reaction.

Complete step-by-step answer:
Given the moles of ${{\rm{I}}_2}$is 2 and moles of I is 2 initially. Now, we write the ICE table. We take x as the change of moles of reactants.

For the reaction	${{\rm{I}}_2} + {{\rm{I}}^ - } \to {{\rm{I}}_{\rm{3}}}^ - $
	${{\rm{I}}_2}$	${{\rm{I}}^ - }$	${{\rm{I}}_{\rm{3}}}^ - $
At $t = 0$	2	2	0
At equilibrium	$2 - x$	$2 - x$	x

At equilibrium, addition of ${\rm{AgN}}{{\rm{O}}_{\rm{3}}}\left( {aq} \right)$ produces 1.75 mole of yellow ppt. Silver nitrate $\left( {{\rm{AgN}}{{\rm{O}}_{\rm{3}}}} \right)$ reacts with ${{\rm{I}}^ - }$, produces silver iodide $\left( {{\rm{AgI}}} \right)$which is yellow in color. So, the reaction is,

${\rm{AgN}}{{\rm{O}}_{\rm{3}}} + {{\rm{I}}^ - } \to {\rm{AgI}} \downarrow $

Given that the moles of silver iodide formed is 1.75 mole. So, the change of moles of ${{\rm{I}}_2}$ and ${{\rm{I}}^ - }$ is equal to 1.75 mole.

$\begin{array}{c}2 - x = 1.75\\x = 0.25\end{array}$

So, moles of ${{\rm{I}}_{\rm{3}}}^ - $ is 0.25.
Now, we calculate the moles of ${{\rm{I}}_2}$ at equilibrium using the value of x.

$\begin{array}{c}{\rm{Moles}}\,{\rm{at equilibrium}} = 2 - x\\ = 2 - 0.25\\ = 1.75\end{array}$

So, the moles at equilibrium of ${{\rm{I}}_2}$ and ${{\rm{I}}^ - }$ is 1.75.
Now, we calculate the equilibrium concentration.
$\begin{array}{c}{K_c} = \dfrac{{{\rm{Concentration}}\,{\rm{of}}\;{\rm{product}}}}{{{\rm{Concentration}}\,{\rm{of}}\;{\rm{reactant}}}}\\ = \dfrac{{\dfrac{{{\rm{mole of product}}}}{{{\rm{Volume}}}}}}{{\dfrac{{{\rm{mole}}\;{\rm{of product}}}}{{{\rm{Volume}}}}}}\end{array}$

Now, we write the equilibrium concentration expression for the ${{\rm{I}}_2} + {{\rm{I}}^ - } \to {{\rm{I}}_{\rm{3}}}^ - $.
${K_c} = \dfrac{{\left[ {{{\rm{I}}_{\rm{3}}}^ - } \right]}}{{\left[ {{{\rm{I}}_2}} \right]\left[ {{{\rm{I}}^ - }} \right]}}$ …… (1)

Moles of ${{\rm{I}}_2}$ and ${{\rm{I}}^ - }$ is 1.75 and moles of ${{\rm{I}}_{\rm{3}}}^ - $ is 0.25 and volume given is $1\;{\rm{d}}{{\rm{m}}^3}$.

$\begin{array}{c}{K_c} = \dfrac{{\dfrac{{0.25\;{\rm{mol}}}}{{1\;{\rm{d}}{{\rm{m}}^3}}}}}{{\dfrac{{1.75\;{\rm{mol}}}}{{1\;{\rm{d}}{{\rm{m}}^3}}} \times \dfrac{{1.75\;{\rm{mol}}}}{{1\;{\rm{d}}{{\rm{m}}^3}}}}}\\ = 0.082\end{array}$
Hence, the correct option is A.

Note: From the given mole of ppt. formed of silver iodide at equilibrium, we get to know about the change of moles of ${{\rm{I}}^ - }$ and ${{\rm{I}}_2}$. Putting the value of moles of ${{\rm{I}}^ - }$, ${{\rm{I}}_2}$, ${{\rm{I}}_{\rm{3}}}^ - $and volume in the expression of equilibrium concentration gives the value of ${K_c}$.