Chapter 15.3, Problem 1PE

Interpretation Introduction

Interpretation:

The direction of the given system proceed to reach equilibrium has to be predicted.

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.

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

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

Relationship between Chemical kinetics and Chemical Equilibrium:

For any reversible reaction occurs via in a mechanism consisting of a single elementary step in both the forward and reverse directions:

$\text{A}\text{+}\text{2B}\underset{{\text{k}}_{\text{r}}}{\overset{{\text{k}}_{\text{f}}}{\rightleftharpoons }}{\text{AB}}_{\text{2}}$

The forward rate is given by

${\text{rate}}_{\text{f}}\text{=}{\text{k}}_{\text{f}}\left[\text{A}\right]{\left[\text{B}\right]}^{\text{2}}$

The reverse rate is given by

${\text{rate}}_r\text{=}{\text{k}}_r\left[{\text{AB}}_2\right]$

Where ${\text{k}}_r$ and ${\text{k}}_f$ are the rate constants for the forward and reverse directions. At equilibrium, when no net change occurs, the two rates must be equal:

${\text{rate}}_{\text{f}}\text{=}{\text{rate}}_r$

Or

Because both ${\text{k}}_r$ and ${\text{k}}_f$ are constants at a given temperature, their ratio is also a constant, which is equal to the equilibrium constant ${\text{K}}_c$

        $\frac{{\text{k}}_{\text{f}}}{{\text{k}}_{\text{r}}}\text{=}{\text{K}}_{\text{c}}\text{=}\frac{\left[{\text{AB}}_{\text{2}}\right]}{\left[\text{A}\right]{\left[\text{B}\right]}^2}$