Chapter 5, Problem 5.8P

(a)

Interpretation Introduction

Interpretation:

For the given reaction, whether the equilibrium constant value $\text{K}\text{>}\text{}\text{1}$ or not has to be decided.

Concept Introduction:

Standard Gibbs energy and equilibrium constant:

The relationship between Gibbs energy and equilibrium constant is given by,

$\begin{array}{l}{\text{Δ}}_{\text{r}}{\text{G}}^{\sout{\text{0}}}\text{=}\text{-}\text{R}\text{T}\text{ln}\text{K}\\ \text{where,}\\ {\text{Δ}}_{\text{r}}{\text{G}}^{\sout{\text{0}}}\text{-}\text{reaction}\text{Gibbs}\text{energy}\\ \text{R}\text{-}\text{gas}\text{constant}\\ \text{T}\text{-}\text{temperature}\\ \text{K}\text{-}\text{equilibrium}\text{constant}\end{array}$

Explanation of Solution

Given reaction,

$\text{2}{\text{CH}}_{\text{3}}{\text{CHO}}_{\text{(g)}}\text{+}{\text{O}}_{\text{2(g)}}\stackrel{}{\rightleftharpoons }\text{2}{\text{CH}}_{\text{3}}{\text{COOH}}_{\left(\text{l}\right)}$

Calculate the reaction Gibbs energy for the given reaction,

$\begin{array}{l}{\text{Δ}}_{\text{r}}{\text{G}}^{\sout{\text{0}}}\text{=}{\displaystyle \sum _{\text{products}}\text{v}{\text{G}}_{\text{f}}{}^{\sout{\text{0}}}}\text{-}{\displaystyle \sum _{\text{reactants}}{\text{vG}}_{\text{f}}{}^{\sout{\text{0}}}}\\ \text{=}\left[\left(\text{2}\text{×}\text{-}\text{389}\text{.9}\right)\text{-}\left(\left(\text{2}\text{×}\text{-}\text{132}\text{.8}\right)\text{+}\text{0}\right)\right]\text{kJ}{\text{mol}}^{\text{-1}}\\ \text{=}\left(\text{-}\text{779}\text{.8}\text{+}\text{265}\text{.6}\right)\text{kJ}{\text{mol}}^{\text{-1}}\text{=}\text{-}\text{514}\text{.2}\text{kJ}{\text{mol}}^{\text{-1}}\end{array}$

Calculate the equilibrium constant value,

$\begin{array}{l}{\text{Δ}}_{\text{r}}{\text{G}}^{\sout{\text{0}}}\text{=}\text{-}\text{R}\text{T}\text{ln}\text{K}\\ \text{ln}\text{K}\text{=}\text{-}\frac{{\text{Δ}}_{\text{r}}{\text{G}}^{\sout{\text{0}}}}{\text{R}\text{T}}\text{=}\text{-}\frac{\text{-}\text{514}\text{.2}\text{kJ}{\text{mol}}^{\text{-1}}}{\left(\text{8}\text{.3145}\text{J}{\text{K}}^{\text{-1}}{\text{mol}}^{\text{-1}}\right)\left(\text{298}\text{K}\right)}\\ \\ \text{=}\frac{\text{514}\text{.2}\text{kJ}{\text{mol}}^{\text{-1}}}{\text{2}\text{.477}\text{kJ}{\text{mol}}^{\text{-1}}}\text{=}\text{207}\text{.59}\\ \text{therefore,}\\ \text{K}\text{=}{\text{e}}^{\text{207}\text{.59}}\text{=}\text{1}\text{.43}\text{×}{\text{10}}^{\text{90}}\end{array}$

Therefore, the equilibrium constant value $\text{K}\text{>}\text{}\text{1}$ for the given reaction.

(b)

Interpretation Introduction

Interpretation:

For the given reaction, whether the equilibrium constant value $\text{K}\text{>}\text{}\text{1}$ or not has to be decided.

Concept Introduction:

Standard Gibbs energy and equilibrium constant:

The relationship between Gibbs energy and equilibrium constant is given by,

$\begin{array}{l}{\text{Δ}}_{\text{r}}{\text{G}}^{\sout{\text{0}}}\text{=}\text{-}\text{R}\text{T}\text{ln}\text{K}\\ \text{where,}\\ {\text{Δ}}_{\text{r}}{\text{G}}^{\sout{\text{0}}}\text{-}\text{reaction}\text{Gibbs}\text{energy}\\ \text{R}\text{-}\text{gas}\text{constant}\\ \text{T}\text{-}\text{temperature}\\ \text{K}\text{-}\text{equilibrium}\text{constant}\end{array}$

Explanation of Solution

Given reaction,

$\text{2}{\text{AgCl}}_{\text{(s)}}\text{+}{\text{Br}}_{\text{2(l)}}\stackrel{}{\rightleftharpoons }\text{2}{\text{AgBr}}_{\left(\text{s}\right)}\text{+}{\text{Cl}}_{\text{2}}{}_{\left(\text{g}\right)}$

Calculate the reaction Gibbs energy for the given reaction,

$\begin{array}{l}{\text{Δ}}_{\text{r}}{\text{G}}^{\sout{\text{0}}}\text{=}{\displaystyle \sum _{\text{products}}\text{v}{\text{G}}_{\text{f}}{}^{\sout{\text{0}}}}\text{-}{\displaystyle \sum _{\text{reactants}}{\text{vG}}_{\text{f}}{}^{\sout{\text{0}}}}\\ \text{=}\left[\left(\left(\text{2}\text{×}\text{-}\text{96}\text{.90}\right)\text{+}\text{0}\right)\text{-}\left(\left(\text{2}\text{×}\text{-}\text{109}\text{.79}\right)\text{+}\text{0}\right)\right]\text{kJ}{\text{mol}}^{\text{-1}}\\ \text{=}\left(\text{-}\text{193}\text{.8}\text{+}\text{219}\text{.58}\right)\text{kJ}{\text{mol}}^{\text{-1}}\text{=}\text{25}\text{.78}\text{kJ}{\text{mol}}^{\text{-1}}\end{array}$

Calculate the equilibrium constant value,

$\begin{array}{l}{\text{Δ}}_{\text{r}}{\text{G}}^{\sout{\text{0}}}\text{=}\text{-}\text{R}\text{T}\text{ln}\text{K}\\ \text{ln}\text{K}\text{=}\text{-}\frac{{\text{Δ}}_{\text{r}}{\text{G}}^{\sout{\text{0}}}}{\text{R}\text{T}}\text{=}\text{-}\frac{\text{53}\text{.4}\text{kJ}{\text{mol}}^{\text{-1}}}{\left(\text{8}\text{.3145}\text{J}{\text{K}}^{\text{-1}}{\text{mol}}^{\text{-1}}\right)\left(\text{298}\text{K}\right)}\\ \\ \text{=}\text{-}\frac{\text{25}\text{.78}\text{kJ}{\text{mol}}^{\text{-1}}}{\text{2}\text{.477}\text{kJ}{\text{mol}}^{\text{-1}}}\text{=}\text{-}\text{10}\text{.41}\\ \text{therefore,}\\ \text{K}\text{=}{\text{e}}^{\text{-}\text{21}\text{.56}}\text{=}\text{3}\text{×}{\text{10}}^{\text{-5}}\end{array}$

Therefore, the equilibrium constant value $\text{K}\text{<}\text{}\text{1}$ for the given reaction.

(c)

Interpretation Introduction

Interpretation:

For the given reaction, whether the equilibrium constant value $\text{K}\text{>}\text{}\text{1}$ or not has to be decided.

Concept Introduction:

Standard Gibbs energy and equilibrium constant:

The relationship between Gibbs energy and equilibrium constant is given by,

$\begin{array}{l}{\text{Δ}}_{\text{r}}{\text{G}}^{\sout{\text{0}}}\text{=}\text{-}\text{R}\text{T}\text{ln}\text{K}\\ \text{where,}\\ {\text{Δ}}_{\text{r}}{\text{G}}^{\sout{\text{0}}}\text{-}\text{reaction}\text{Gibbs}\text{energy}\\ \text{R}\text{-}\text{gas}\text{constant}\\ \text{T}\text{-}\text{temperature}\\ \text{K}\text{-}\text{equilibrium}\text{constant}\end{array}$

Explanation of Solution

Given reaction,

${\text{Hg}}_{\text{(l)}}\text{+}{\text{Cl}}_{\text{2(g)}}\stackrel{}{\rightleftharpoons }{\text{HgCl}}_{\text{2}}{}_{\left(\text{s}\right)}$

Calculate the reaction Gibbs energy for the given reaction,

$\begin{array}{l}{\text{Δ}}_{\text{r}}{\text{G}}^{\sout{\text{0}}}\text{=}{\displaystyle \sum _{\text{products}}\text{v}{\text{G}}_{\text{f}}{}^{\sout{\text{0}}}}\text{-}{\displaystyle \sum _{\text{reactants}}{\text{vG}}_{\text{f}}{}^{\sout{\text{0}}}}\\ \text{=}\left[\left(\text{-}\text{178}\text{.6}\right)\text{-}\left(\text{0}\text{+}\text{0}\right)\right]\text{kJ}{\text{mol}}^{\text{-1}}\\ \text{=}\text{-}\text{178}\text{.6}\text{kJ}{\text{mol}}^{\text{-1}}\end{array}$

Calculate the equilibrium constant value,

$\begin{array}{l}{\text{Δ}}_{\text{r}}{\text{G}}^{\sout{\text{0}}}\text{=}\text{-}\text{R}\text{T}\text{ln}\text{K}\\ \text{ln}\text{K}\text{=}\text{-}\frac{{\text{Δ}}_{\text{r}}{\text{G}}^{\sout{\text{0}}}}{\text{R}\text{T}}\text{=}\text{-}\frac{\text{-}\text{178}\text{.6}\text{kJ}{\text{mol}}^{\text{-1}}}{\left(\text{8}\text{.3145}\text{J}{\text{K}}^{\text{-1}}{\text{mol}}^{\text{-1}}\right)\left(\text{298}\text{K}\right)}\\ \\ \text{=}\frac{\text{178}\text{.6}\text{kJ}{\text{mol}}^{\text{-1}}}{\text{2}\text{.477}\text{kJ}{\text{mol}}^{\text{-1}}}\text{=}\text{72}\text{.1}\\ \text{therefore,}\\ \text{K}\text{=}{\text{e}}^{\text{72}\text{.1}}\text{=}\text{2}\text{.05}\text{×}{\text{10}}^{\text{31}}\end{array}$

Therefore, the equilibrium constant value $\text{K}\text{>}\text{}\text{1}$ for the given reaction.

(d)

Interpretation Introduction

Interpretation:

For the given reaction, whether the equilibrium constant value $\text{K}\text{>}\text{}\text{1}$ or not has to be decided.

Concept Introduction:

Standard Gibbs energy and equilibrium constant:

The relationship between Gibbs energy and equilibrium constant is given by,

$\begin{array}{l}{\text{Δ}}_{\text{r}}{\text{G}}^{\sout{\text{0}}}\text{=}\text{-}\text{R}\text{T}\text{ln}\text{K}\\ \text{where,}\\ {\text{Δ}}_{\text{r}}{\text{G}}^{\sout{\text{0}}}\text{-}\text{reaction}\text{Gibbs}\text{energy}\\ \text{R}\text{-}\text{gas}\text{constant}\\ \text{T}\text{-}\text{temperature}\\ \text{K}\text{-}\text{equilibrium}\text{constant}\end{array}$

Explanation of Solution

Given reaction,

${\text{Zn}}_{\text{(s)}}\text{+}{\text{Cu}}^{\text{2+}}{}_{\left(\text{aq}\right)}\stackrel{}{\rightleftharpoons }{\text{Zn}}^{\text{2+}}{}_{\left(\text{aq}\right)}\text{+}{\text{Cu}}_{\left(\text{s}\right)}$

Calculate the reaction Gibbs energy for the given reaction,

$\begin{array}{l}{\text{Δ}}_{\text{r}}{\text{G}}^{\sout{\text{0}}}\text{=}{\displaystyle \sum _{\text{products}}\text{v}{\text{G}}_{\text{f}}{}^{\sout{\text{0}}}}\text{-}{\displaystyle \sum _{\text{reactants}}{\text{vG}}_{\text{f}}{}^{\sout{\text{0}}}}\\ \text{=}\left[\left(\text{-}\text{147}\text{.06}\text{+}\text{0}\right)\text{-}\left(\text{0}\text{+}\text{65}\text{.49}\right)\right]\text{kJ}{\text{mol}}^{\text{-1}}\\ \text{=}\text{-}\text{212}\text{.55}\text{kJ}{\text{mol}}^{\text{-1}}\end{array}$

Calculate the equilibrium constant value,

$\begin{array}{l}{\text{Δ}}_{\text{r}}{\text{G}}^{\sout{\text{0}}}\text{=}\text{-}\text{R}\text{T}\text{ln}\text{K}\\ \text{ln}\text{K}\text{=}\text{-}\frac{{\text{Δ}}_{\text{r}}{\text{G}}^{\sout{\text{0}}}}{\text{R}\text{T}}\text{=}\text{-}\frac{\text{-}\text{212}\text{.55}\text{kJ}{\text{mol}}^{\text{-1}}}{\left(\text{8}\text{.3145}\text{J}{\text{K}}^{\text{-1}}{\text{mol}}^{\text{-1}}\right)\left(\text{298}\text{K}\right)}\\ \\ \text{=}\frac{\text{212}\text{.55}\text{kJ}{\text{mol}}^{\text{-1}}}{\text{2}\text{.477}\text{kJ}{\text{mol}}^{\text{-1}}}\text{=}\text{85}\text{.81}\\ \text{therefore,}\\ \text{K}\text{=}{\text{e}}^{\text{85}\text{.81}}\text{=}\text{1}\text{.8}\text{×}{\text{10}}^{\text{37}}\end{array}$

Therefore, the equilibrium constant value $\text{K}\text{>}\text{}\text{1}$ for the given reaction.

(e)

Interpretation Introduction

Interpretation:

For the given reaction, whether the equilibrium constant value $\text{K}\text{>}\text{}\text{1}$ or not has to be decided.

Concept Introduction:

Standard Gibbs energy and equilibrium constant:

The relationship between Gibbs energy and equilibrium constant is given by,

$\begin{array}{l}{\text{Δ}}_{\text{r}}{\text{G}}^{\sout{\text{0}}}\text{=}\text{-}\text{R}\text{T}\text{ln}\text{K}\\ \text{where,}\\ {\text{Δ}}_{\text{r}}{\text{G}}^{\sout{\text{0}}}\text{-}\text{reaction}\text{Gibbs}\text{energy}\\ \text{R}\text{-}\text{gas}\text{constant}\\ \text{T}\text{-}\text{temperature}\\ \text{K}\text{-}\text{equilibrium}\text{constant}\end{array}$

Explanation of Solution

Given reaction,

${\text{C}}_{\text{12}}{\text{H}}_{\text{22}}{\text{O}}_{\text{11}}{}_{\text{(s)}}\text{+}\text{12}{\text{O}}_{\text{2(g)}}\stackrel{}{\rightleftharpoons }\text{2}{\text{CO}}_{\text{2}}{}_{\left(\text{g}\right)}\text{+}\text{11}{\text{H}}_{\text{2}}{\text{O}}_{\left(\text{l}\right)}$

Calculate the reaction Gibbs energy for the given reaction,

$\begin{array}{l}{\text{Δ}}_{\text{r}}{\text{G}}^{\sout{\text{0}}}\text{=}{\displaystyle \sum _{\text{products}}\text{v}{\text{G}}_{\text{f}}{}^{\sout{\text{0}}}}\text{-}{\displaystyle \sum _{\text{reactants}}{\text{vG}}_{\text{f}}{}^{\sout{\text{0}}}}\\ \text{=}\left[\left(\left(\text{2}\text{×}\text{-}\text{394}\text{.36}\right)\text{+}\left(\text{11}\text{×}\text{-}\text{237}\text{.13}\right)\right)\text{-}\left(\left(\text{-}\text{1543}\right)\text{+}\left(\text{12}\text{×}\text{0}\right)\right)\right]\text{kJ}{\text{mol}}^{\text{-1}}\\ \text{=}\left(\text{-}\text{3397}\text{.15}\text{+}\text{1543}\right)\text{kJ}{\text{mol}}^{\text{-1}}\text{=}\text{-}\text{1854}\text{.15}\text{kJ}{\text{mol}}^{\text{-1}}\end{array}$

Calculate the equilibrium constant value,

$\begin{array}{l}{\text{Δ}}_{\text{r}}{\text{G}}^{\sout{\text{0}}}\text{=}\text{-}\text{R}\text{T}\text{ln}\text{K}\\ \text{ln}\text{K}\text{=}\text{-}\frac{{\text{Δ}}_{\text{r}}{\text{G}}^{\sout{\text{0}}}}{\text{R}\text{T}}\text{=}\text{-}\frac{\text{-}\text{1854}\text{.15}\text{kJ}{\text{mol}}^{\text{-1}}}{\left(\text{8}\text{.3145}\text{J}{\text{K}}^{\text{-1}}{\text{mol}}^{\text{-1}}\right)\left(\text{298}\text{K}\right)}\\ \\ \text{=}\frac{\text{1854}\text{.15}\text{kJ}{\text{mol}}^{\text{-1}}}{\text{2}\text{.477}\text{kJ}{\text{mol}}^{\text{-1}}}\text{=}\text{748}\text{.55}\\ \text{therefore,}\\ \text{K}\text{=}{\text{e}}^{\text{748}\text{.55}}\text{=}\text{infinity}\end{array}$

Therefore, the equilibrium constant value $\text{K}\text{>}\text{}\text{1}$ for the given reaction.