Question

The dissociation constants for acetic acid and ${\text{HCN}}$ at ${\text{2}}{{\text{5}}^{\text{0}}}{\text{C}}$ are $1.5 \times {10^{ - 5}}$ and $4.5 \times {10^{ - 10}}$, respectively. The equilibrium constant for equilibrium will be:
${\text{C}}{{\text{N}}^{\text{ - }}}{\text{ + C}}{{\text{H}}_{\text{3}}}{\text{COOH}} \rightleftharpoons {\text{HCN + C}}{{\text{H}}_{\text{3}}}{\text{CO}}{{\text{O}}^{\text{ - }}}$
(A)- $3.0 \times {10^5}$
(B)- $3.0 \times {10^{ - 5}}$
(C)- $3.0 \times {10^{ - 4}}$
(D)- $3.0 \times {10^4}$

Answer 1

Hint: Equilibrium constant is the measurement of the concentration of reactants and products in the equilibrium condition, it is also calculated in the terms of dissociation constant.

Complete answer:
It is given that,
-Dissociation constant (${{\text{K}}_{\text{1}}}$) for acetic acid = $1.5 \times {10^{ - 5}}$ ……….(i)
And dissociation reaction for acetic acid is as follow:
${\text{C}}{{\text{H}}_{\text{3}}}{\text{COOH}} \rightleftharpoons {\text{C}}{{\text{H}}_{\text{3}}}{\text{CO}}{{\text{O}}^{\text{ - }}}{\text{ + }}{{\text{H}}^{\text{ + }}}$
-Dissociation constant (${{\text{K}}_2}$) for ${\text{HCN}}$ = $4.5 \times {10^{ - 10}}$ …………(ii)
And dissociation reaction for ${\text{HCN}}$ is as follow:
\[{\text{HCN}} \rightleftharpoons {{\text{H}}^{\text{ + }}}{\text{ + C}}{{\text{N}}^{\text{ - }}}\]
-Overall chemical reaction of acetic acid & ${\text{HCN}}$ at an equilibrium condition is given as follow:
${\text{C}}{{\text{N}}^{\text{ - }}}{\text{ + C}}{{\text{H}}_{\text{3}}}{\text{COOH}} \rightleftharpoons {\text{HCN + C}}{{\text{H}}_{\text{3}}}{\text{CO}}{{\text{O}}^{\text{ - }}}$………..(iii)
(Which is obtained by subtracting the equation (ii) from equation (i))
-Equilibrium constant for the above reaction which is given in equation (iii) is calculated as follow:
${{\text{K}}_{{\text{eq}}}}{\text{ = }}\dfrac{{{{\text{K}}_{\text{1}}}}}{{{{\text{K}}_{\text{2}}}}}$
Now, putting the values of ${{\text{K}}_{\text{1}}}$ and ${{\text{K}}_2}$ from the equations (i) & (ii) to the above equation we get
${{\text{K}}_{{\text{eq}}}}{\text{ = }}\dfrac{{1.5 \times {{10}^{ - 5}}}}{{4.5 \times {{10}^{ - 10}}}}$
${{\text{K}}_{{\text{eq}}}}{\text{ = }}\dfrac{{{{10}^5}}}{3}$
${{\text{K}}_{{\text{eq}}}}{\text{ = 3}}{\text{.33}} \times {\text{1}}{{\text{0}}^4}$
So, an equilibrium constant for the given chemical reaction between acetic acid and ${\text{HCN}}$ is ${\text{3}}{\text{.33}} \times {\text{1}}{{\text{0}}^4}$ or equal to ${\text{3}}{\text{.0}} \times {\text{1}}{{\text{0}}^4}$.

Hence, option (D) is correct.

Additional information:
Dissociation constant is sometimes also known as an equilibrium constant in the reversible reaction at an equilibrium condition, because they give an idea about the species undergoing splitting for the formation of a product and it is applicable to those reactions where only one species is present. In this question two molecules are involved that’s why we calculate equilibrium constant by using dissociation constant.

Note:
Here some of you may do wrong calculation by dividing the dissociation constant of ${\text{HCN}}$ from acetic acid i.e. ${{\text{K}}_{{\text{eq}}}}{\text{ = }}\dfrac{{{{\text{K}}_2}}}{{{{\text{K}}_1}}}$, but that will be a wrong because in given reaction in the forward condition initially, the dissociation of acetic acid takes place that’s why we used this formula ${{\text{K}}_{{\text{eq}}}}{\text{ = }}\dfrac{{{{\text{K}}_{\text{1}}}}}{{{{\text{K}}_{\text{2}}}}}$.