Chapter 15, Problem 15.67QP

a)

Interpretation Introduction

Interpretation:

An expression for ${\text{K}}_{\text{P}}$ in terms of $\alpha$ and P, total pressure for the given reaction has to be derived.

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.

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

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

Calculating equilibrium concentration:

• Express the equilibrium concentrations of all species in terms of the initial concentrations and a single unknown x, which represents the change in concentration.
• Write the equilibrium constant expression in terms of the equilibrium concentrations. Knowing the value of the equilibrium constant, solve for x.
• Having solved for x, calculate the equilibrium concentrations of all species.

Explanation of Solution

For the given reaction,

     ${\text{N}}_2{\text{O}}_4\left(\text{g}\right)\stackrel{}{\rightleftharpoons }2{\text{NO}}_2\left(\text{g}\right)$

Given: Initially $\text{1}\text{.0}\text{mole}\text{of}{\text{N}}_{\text{2}}{\text{O}}_{\text{4}}$; at equilibrium $\text{α mole}\text{of}{\text{N}}_{\text{2}}{\text{O}}_{\text{4}}$ has dissociated to form ${\text{NO}}_{\text{2}}$.

Construct ICE table as follows,

$\begin{array}{l}{\text{N}}_2{\text{O}}_4\left(\text{g}\right)\stackrel{}{\rightleftharpoons }2{\text{NO}}_2\left(\text{g}\right)\\ Initial:1M0\\ Change:-\alpha +2\alpha \\ Equilibrium:1-\alpha +2\alpha \end{array}$

The total moles in the system = $\left(\text{moles}\text{of}{\text{N}}_{\text{2}}{\text{O}}_{\text{4}}\text{+}\text{moles}\text{of}{\text{NO}}_{\text{2}}\right)$= $\left[\left(1-\alpha \right)+2\alpha \right]=1+\alpha$. If the total pressure in the system is P, then:

$P_{N_2{O}_4}=\frac{1-\alpha }{1+\alpha }Pand{P}_{N{O}_2}=\frac{2\alpha }{1+\alpha }P$

$K_P=\frac{\left(P_{N{O}_2}{}^2\right)}{P_{N_2{O}_4}}=\frac{{\left(\frac{2\alpha }{1+\alpha }\right)}^2{P}^2}{\left(\frac{1-\alpha }{1+\alpha }\right)P}$

$K_P=\frac{\left(P_{N{O}_2}{}^2\right)}{P_{N_2{O}_4}}=\frac{4{\alpha }^2}{1-{\alpha }^2}P$

 Hence, expression for ${\text{K}}_{\text{P}}$ in terms of $\alpha$ and P is $\frac{4{\alpha }^2}{1-{\alpha }^2}P$

b)

Interpretation Introduction

Interpretation:

The shift in equilibrium due to an increase in P has to be predicted and the prediction has to be agreed with Lee-Chatelier’s principle.

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.

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

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

Calculating equilibrium concentration:

• Express the equilibrium concentrations of all species in terms of the initial concentrations and a single unknown x, which represents the change in concentration.
• Write the equilibrium constant expression in terms of the equilibrium concentrations. Knowing the value of the equilibrium constant, solve for x.
• Having solved for x, calculate the equilibrium concentrations of all species.

Explanation of Solution

For the given reaction,

     ${\text{N}}_2{\text{O}}_4\left(\text{g}\right)\stackrel{}{\rightleftharpoons }2{\text{NO}}_2\left(\text{g}\right)$

Given: Initially $\text{1}\text{.0}\text{mole}\text{of}{\text{N}}_{\text{2}}{\text{O}}_{\text{4}}$; at equilibrium $\text{α mole}\text{of}{\text{N}}_{\text{2}}{\text{O}}_{\text{4}}$ has dissociated to form ${\text{NO}}_{\text{2}}$. The obtained expression for ${\text{K}}_{\text{P}}$ in terms of $\alpha$ and P is $\frac{4{\alpha }^2}{1-{\alpha }^2}P$

Rearranging ${\text{K}}_{\text{P}}$ expression:

$\begin{array}{l}{\text{K}}_{\text{P}}\text{=}\frac{{\text{4α}}^{\text{2}}}{{\text{1-α}}^{\text{2}}}\text{P}\\ \\ {\text{4α}}^{\text{2}}\text{P}\text{=}{\text{K}}_{\text{P}}\text{-}{\text{α}}^{\text{2}}{\text{K}}_{\text{P}}\\ \\ {\text{α}}^{\text{2}}\text{=}\frac{{\text{K}}_{\text{P}}}{\text{4P+}{\text{K}}_{\text{P}}}\\ \\ \text{α}\text{=}\sqrt{\frac{{\text{K}}_{\text{P}}}{{\text{4P+K}}_{\text{P}}}}\end{array}$

${\text{K}}_{\text{P}}$ is a constant (at constant temperature). Thus, as P increases $\alpha$ must decreases, indicating that the system shifts to the left. This argument would predict based on Le-Chatelier’s principle.