Chapter 14.2, Problem 7PE

Interpretation Introduction

Interpretation:

For the given reaction, the equilibrium expression ${\text{K}}_{\text{c}}$ has to be written.

Concept introduction:

Law of mass action: The rate of chemical reaction is directly proportional to the product of concentrations of reactant to products.

$\begin{array}{l}\text{aA}\text{+}\text{bB}\to \text{cC}\text{+}\text{dD}\\ \\ {\text{K}}_{eq}\text{=}\frac{{\left[\text{C}\right]}^{\text{c}}{\left[\text{D}\right]}^{\text{d}}}{{\left[\text{A}\right]}^{\text{a}}{\left[\text{B}\right]}^{\text{b}}}\text{for}\text{aqueous}\\ \\ {\text{K}}_{\text{eq}}\text{=}\frac{{\left({\text{P}}_{\text{C}}\right)}^{\text{c}}{\left({\text{P}}_{\text{D}}\right)}^{\text{d}}}{{\left({\text{P}}_{\text{A}}\right)}^{\text{a}}{\left({\text{P}}_{\text{B}}\right)}^{\text{b}}}\text{for}\text{gases}\\ \end{array}$

Multiple equilibria: If a reaction can be expressed as the sum of two or more reactions, the equilibrium constant for the overall reaction is given by the product of the equilibrium constants of the individual reactions.

$\begin{array}{l}\underset{\_}{\begin{array}{l}\text{A}\text{+ B }\underset{}{\rightleftharpoons }{\text{ C + D K}}_{\text{c}}^{\text{'}}\\ \text{C}\text{+ D }\underset{}{\overset{}{\rightleftharpoons }}{\text{ E + F K}}_{\text{c}}^{\text{''}}\end{array}}\\ \text{overall}\text{reaction:}\underset{\_}{\text{A}\text{+ B }\underset{}{\rightleftharpoons }\text{ E + F}{\text{K}}_{\text{c}}}\end{array}$

The equilibrium constant for two separate equilibrium constants are,

${\text{K}}_{\text{c}}^{\text{'}}\text{=}\frac{\left[\text{C}\right]\left[\text{D}\right]}{\left[\text{A}\right]\left[\text{B}\right]}\text{and}{\text{K}}_{\text{c}}^{\text{''}}\text{=}\frac{\left[\text{E}\right]\left[\text{F}\right]}{\left[\text{C}\right]\left[\text{D}\right]}$

For overall reaction, the equilibrium constant ${\text{K}}_{\text{c}}$ is,

${\text{K}}_{\text{c}}^{\text{'}}{\text{K}}_{\text{c}}^{\text{''}}\text{=}\frac{\left[\text{C}\right]\left[\text{D}\right]}{\left[\text{A}\right]\left[\text{B}\right]}\times \frac{\left[\text{E}\right]\left[\text{F}\right]}{\left[\text{C}\right]\left[\text{D}\right]}=\frac{\left[\text{E}\right]\left[\text{F}\right]}{\left[\text{A}\right]\left[\text{B}\right]}$

Therefore, $\boxed{{\text{K}}_c\text{=}{\text{K}}_{\text{c}}^{\text{'}}\times {\text{K}}_{\text{c}}^{\text{''}}}$

Important rules for writing equilibrium constants:

• Reciprocal of X is 1/X:  when the equation for a reversible reaction is written in the opposite direction, the equilibrium constant becomes the reciprocal of the original equilibrium constant.
• Value of K also depends on how the equilibrium equation is balanced. Consider as,

$\begin{array}{l}\frac{\text{1}}{\text{2}}{\text{N}}_{\text{2}}{\text{O}}_{\text{4}}\left(\text{g}\right)\underset{}{\rightleftharpoons }{\text{NO}}_{\text{2}}\left(\text{g}\right){\text{K}}_{\text{c}}^{\text{'}}\text{=}\frac{\left[{\text{NO}}_{\text{2}}\right]}{{\left[{\text{N}}_{\text{2}}{\text{O}}_{\text{4}}\right]}^{\text{1/2}}}\\ \\ {\text{N}}_{\text{2}}{\text{O}}_{\text{4}}\left(\text{g}\right)\underset{}{\rightleftharpoons }{\text{2NO}}_{\text{2}}\left(\text{g}\right){\text{K}}_{\text{c}}\text{=}\frac{{\left[{\text{NO}}_{\text{2}}\right]}^{\text{2}}}{\left[{\text{N}}_{\text{2}}{\text{O}}_{\text{4}}\right]}\\ \text{Therefore,}\boxed{{\text{K}}_{\text{c}}^{\text{'}}\text{=}\sqrt{{\text{K}}_{\text{c}}}}\end{array}$