Chapter 5, Problem 5G.22E

(a)

Interpretation Introduction

Interpretation:

The equilibrium constant at ${\text{25}}^{\text{o}}\text{C}$ has to be determined for the given reaction.

  ${\text{CH}}_4{\text{(g)+Cl}}_{\text{2}}\text{(g)}\rightleftharpoons {\text{CH}}_3\text{Cl(g)+HCl(g)}$

Concept Introduction:

The equilibrium constant of a reaction can be calculated using the given expression,

  $\begin{array}{l}\text{lnK=}\frac{{\text{-ΔG}}_{\text{r}}{}^{\text{o}}}{\text{RT}}\\ \text{where,}\\ \text{R=Gas constant (8}\text{.314}{\text{JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\text{)}\\ {\text{ΔG}}_{\text{r}}^{\text{°}}\text{=}\text{Standard}\text{Gibb's}\text{free}\text{energy}\\ \text{T}\text{=}\text{Temperature}\\ \text{K= Equilibrium constant}\end{array}$

Answer to Problem 5G.22E

The equilibrium constant of the given reaction is $\text{2}\text{.0}\times {\text{10}}^{-1}$.

Explanation of Solution

Given reaction is

The synthesis of trichloromethane (chloroform) from natural gas (methane):

  ${\text{CH}}_4{\text{(g)+Cl}}_{\text{2}}\text{(g)}\rightleftharpoons {\text{CH}}_3\text{Cl(g)+HCl(g)}$

Temperature of the reaction is $\text{298}\text{K}$,

The ${\text{ΔG}}_{\text{f}}^{\text{°}}$ value of ${\text{CH}}_{\text{3}}\text{Cl}$ is $48.5\text{k}\text{J}\cdot {\text{mol}}^{\text{-1}}$

The ${\text{ΔG}}_{\text{f}}^{\text{°}}$ value of $\text{HCl}$ is $-95.30\text{k}\text{J}\cdot {\text{mol}}^{\text{-1}}$

The ${\text{ΔG}}_{\text{f}}^{\text{°}}$ value of ${\text{Cl}}_2$ is $0$

The ${\text{ΔG}}_{\text{f}}^{\text{°}}$ value of ${\text{CH}}_4$ is $-50.72\text{k}\text{J}\cdot {\text{mol}}^{\text{-1}}$

$\text{R=}\text{8}\text{.3145 J}\cdot {\text{K}}^{\text{-1}}\cdot {\text{mol}}^{\text{-1}}$

$\text{T=298K}$

Here, ${\text{ΔG}}_{\text{r}}^{\text{°}}$ is calculated using given formula,

  $\begin{array}{c}{\text{ΔG}}_{\text{r}}^{\text{°}}{\text{=[ΔG}}_{\text{f}}^{\text{°}}{\text{CH}}_{\text{3}}{\text{Cl(g)+ΔG}}_{\text{f}}^{\text{°}}{\text{HCl(g)]-[ΔG}}_{\text{f}}^{\text{°}}{\text{CH}}_4{\text{(g)+ΔG}}_{\text{f}}^{\text{°}}{\text{Cl}}_{\text{2}}\text{(g)]}\\ \text{=(48}\text{.5}\times {10}^3-95.30\times {10}^3)-(-50.72\times {10}^3-0)\\ =4.0\times {10}^3\end{array}$

Now, the equilibrium constant of the reaction is calculated as

  $\begin{array}{c}\text{lnK=}\frac{{\text{-ΔG}}_{\text{r}}{}^{\text{o}}}{\text{RT}}\\ \text{=-}\frac{4.0{\text{×10}}^{\text{3}}\text{J}\cdot {\text{mol}}^{\text{-1}}}{(8.3145\text{J}\cdot {\text{K}}^{\text{-1}}\cdot {\text{mol}}^{\text{-1}}\text{)×(298}\text{K)}}\\ =-1.61\end{array}$

To calculate $\text{K}$ value, inverse logarithm is taken for the above reaction.

  $\begin{array}{c}{\text{K=e}}^{-1.61}\\ \text{=2}\text{.0}\times {\text{10}}^{-1}\end{array}$

Therefore, the $\text{K}$ value of given reaction is $\text{2}\text{.0}\times {\text{10}}^{-1}$.

(b)

Interpretation Introduction

Interpretation:

The equilibrium constant at ${\text{25}}^{\text{o}}\text{C}$ has to be determined for the given reaction.

  ${\text{C}}_2{\text{H}}_2{\text{(g)+2H}}_{\text{2}}\text{(g)}\rightleftharpoons {\text{C}}_{\text{2}}{\text{H}}_{\text{6}}\text{(g)}$

Concept introduction:

Refer to part (a).

Answer to Problem 5G.22E

The equilibrium constant of the given reaction is.

Explanation of Solution

Given reaction is

The hydrogenation of acetylene to ethane:

  ${\text{C}}_2{\text{H}}_2{\text{(g)+2H}}_{\text{2}}\text{(g)}\rightleftharpoons {\text{C}}_{\text{2}}{\text{H}}_{\text{6}}\text{(g)}$

Temperature of the reaction is $\text{298}\text{K}$,

The ${\text{ΔG}}_{\text{f}}^{\text{°}}$ value of ${\text{C}}_{\text{2}}{\text{H}}_{\text{6}}$ is $-32.82\text{k}\text{J}\cdot {\text{mol}}^{\text{-1}}$

The ${\text{ΔG}}_{\text{f}}^{\text{°}}$ value of ${\text{C}}_2{\text{H}}_{\text{2}}$ is $+209.20\text{k}\text{J}\cdot {\text{mol}}^{\text{-1}}$

The ${\text{ΔG}}_{\text{f}}^{\text{°}}$ value of ${\text{H}}_2$ is $0$

$\text{R=}\text{8}\text{.3145 J}\cdot {\text{K}}^{\text{-1}}\cdot {\text{mol}}^{\text{-1}}$

$\text{T=298K}$

Here, ${\text{ΔG}}_{\text{r}}^{\text{°}}$ is calculated using given formula,

  $\begin{array}{c}{\text{ΔG}}_{\text{r}}^{\text{°}}{\text{=ΔG}}_{\text{f}}^{\text{°}}{\text{(C}}_{\text{2}}{\text{H}}_{\text{6}}{\text{(g))-[ΔG}}_{\text{f}}^{\text{°}}{\text{(C}}_2{\text{H}}_2{\text{(g))+2ΔG}}_{\text{f}}^{\text{°}}{\text{(H}}_{\text{2}}\text{(g))]}\\ \text{=(-32}\text{.82}\text{×}{10}^3)-[209.20\times {10}^3-0]\\ =-2.42\times {10}^5\end{array}$

Now, the equilibrium constant of the reaction is calculated as

  $\begin{array}{c}\text{lnK=}\frac{{\text{-ΔG}}_{\text{r}}{}^{\text{o}}}{\text{RT}}\\ \text{=}\frac{2.42\times {10}^5\text{J}\cdot {\text{mol}}^{\text{-1}}}{(8.3145\text{J}\cdot {\text{K}}^{\text{-1}}\cdot {\text{mol}}^{\text{-1}}\text{)×(298}\text{K)}}\\ =97.7\end{array}$

To calculate $\text{K}$ value, inverse logarithm is taken for the above reaction.

  $\begin{array}{c}{\text{K=e}}^{97.7}\\ \text{=2}\text{.7}\times {\text{10}}^{42}\end{array}$

Therefore, the $\text{K}$ value of given reaction is $\text{1}\times {\text{10}}^{90}$.

(c)

Interpretation Introduction

Interpretation:

The equilibrium constant at ${\text{25}}^{\text{o}}\text{C}$ has to be determined for the given reaction.

  ${\text{3NO}}_{\text{2}}{\text{(g)+H}}_{\text{2}}\text{O(l)}\rightleftharpoons {\text{2HNO}}_{\text{3}}\text{(aq)+NO(g)}$

Concept introduction:

Refer to part (a).

Answer to Problem 5G.22E

The equilibrium constant of the given reaction is $\text{1}\text{.8}\times {\text{10}}^9$.

Explanation of Solution

Given reaction is

The final step in the industrial production of nitric acid:

  ${\text{3NO}}_{\text{2}}{\text{(g)+H}}_{\text{2}}\text{O(l)}\rightleftharpoons {\text{2HNO}}_{\text{3}}\text{(aq)+NO(g)}$

Temperature of the reaction is $\text{298}\text{K}$,

The ${\text{ΔG}}_{\text{f}}^{\text{°}}$ value of ${\text{HNO}}_3$ is $-111.25\text{k}\text{J}\cdot {\text{mol}}^{\text{-1}}$

The ${\text{ΔG}}_{\text{f}}^{\text{°}}$ value of $\text{NO}$ is $+86.55\text{k}\text{J}\cdot {\text{mol}}^{\text{-1}}$

The ${\text{ΔG}}_{\text{f}}^{\text{°}}$ value of ${\text{NO}}_2$ is $+51.31$

The ${\text{ΔG}}_{\text{f}}^{\text{°}}$ value of ${\text{H}}_{\text{2}}\text{O}$ is $-237.13\text{k}\text{J}\cdot {\text{mol}}^{\text{-1}}$

$\text{R=}\text{8}\text{.3145 J}\cdot {\text{K}}^{\text{-1}}\cdot {\text{mol}}^{\text{-1}}$

$\text{T=298K}$

Here, ${\text{ΔG}}_{\text{r}}^{\text{°}}$ is calculated using given formula,

  $\begin{array}{c}{\text{ΔG}}_{\text{r}}^{\text{°}}{\text{=[ΔG}}_{\text{f}}^{\text{°}}{\text{2HNO}}_{\text{3}}{\text{(aq)+ΔG}}_{\text{f}}^{\text{°}}{\text{NO(g)]-[3ΔG}}_{\text{f}}^{\text{°}}{\text{NO}}_{\text{2}}{\text{(g)+ΔG}}_{\text{f}}^{\text{°}}{\text{H}}_{\text{2}}\text{O(aq)]}\\ \text{=[2×(-111}\text{.25}\text{×}{10}^3)+86.55\text{×}{10}^3]-[3\text{×(51}\text{.31×}{10}^3)-237.13\text{×}{10}^3]\\ =-5.28\times {10}^4\end{array}$

Now, the equilibrium constant of the reaction is calculated as

  $\begin{array}{c}\text{lnK=}\frac{{\text{-ΔG}}_{\text{r}}{}^{\text{o}}}{\text{RT}}\\ \text{=}\frac{5.28\times {10}^4\text{J}\cdot {\text{mol}}^{\text{-1}}}{(8.3145\text{J}\cdot {\text{K}}^{\text{-1}}\cdot {\text{mol}}^{\text{-1}}\text{)×(298}\text{K)}}\\ =21.31\end{array}$

To calculate $\text{K}$ value, inverse logarithm is taken for the above reaction.

  $\begin{array}{c}{\text{K=e}}^{21.31}\\ \text{=1}\text{.8}\times {\text{10}}^9\end{array}$

Therefore, the $\text{K}$ value of given reaction is $\text{1}\text{.8}\times {\text{10}}^9$.

(d)

Interpretation Introduction

Interpretation:

The equilibrium constant at ${\text{25}}^{\text{o}}\text{C}$ has to be determined for the given reaction.

  ${\text{N}}_2{\text{H}}_4{\text{(l)+O}}_2\text{(g)}\rightleftharpoons {\text{N}}_2{\text{(g)+2H}}_2\text{O(l)}$

Concept introduction:

Refer to part (a).

Answer to Problem 5G.22E

The equilibrium constant of the given reaction is $\text{2}\text{.4}\times {\text{10}}^{109}$.

Explanation of Solution

Given reaction is

The reaction of hydrazine and oxygen in a rocket:

  ${\text{N}}_2{\text{H}}_4{\text{(l)+O}}_2\text{(g)}\rightleftharpoons {\text{N}}_2{\text{(g)+2H}}_2\text{O(l)}$

The ${\text{ΔG}}_{\text{f}}^{\text{°}}$ value of ${\text{N}}_{\text{2}}$ is $0$

The ${\text{ΔG}}_{\text{f}}^{\text{°}}$ value of ${\text{H}}_{\text{2}}\text{O}$ is $-237.13\text{k}\text{J}\cdot {\text{mol}}^{\text{-1}}$

The ${\text{ΔG}}_{\text{f}}^{\text{°}}$ value of ${\text{N}}_{\text{2}}{\text{H}}_{\text{4}}$ is $+149.34\text{k}\text{J}\cdot {\text{mol}}^{\text{-1}}$

The ${\text{ΔG}}_{\text{f}}^{\text{°}}$ value of ${\text{O}}_2$ is $0$

Temperature of the reaction is $\text{298}\text{K}$

$\text{R=}\text{8}\text{.3145 J}\cdot {\text{K}}^{\text{-1}}\cdot {\text{mol}}^{\text{-1}}$

$\text{T=298K}$

Here, ${\text{ΔG}}_{\text{r}}^{\text{°}}$ is calculated using given formula,

  $\begin{array}{c}{\text{ΔG}}_{\text{r}}^{\text{°}}{\text{=[ΔG}}_{\text{f}}^{\text{°}}{\text{N}}_{\text{2}}{\text{(g)+2ΔG}}_{\text{f}}^{\text{°}}{\text{H}}_{\text{2}}{\text{O(l)]-[ΔG}}_{\text{f}}^{\text{°}}{\text{N}}_{\text{2}}{\text{H}}_{\text{4}}{\text{(l)+ΔG}}_{\text{f}}^{\text{°}}{\text{O}}_{\text{2}}\text{(g)]}\\ \text{=[0}+2\text{×}(-237.13\text{×}{10}^3)]-[149.34\text{×}{10}^3)-0]\\ =-6.24\times {10}^5\end{array}$

Now, the equilibrium constant of the reaction is calculated as

  $\begin{array}{c}\text{lnK=}\frac{{\text{-ΔG}}_{\text{r}}{}^{\text{o}}}{\text{RT}}\\ \text{=}\frac{6.24\times {10}^5\text{J}\cdot {\text{mol}}^{\text{-1}}}{(8.3145\text{J}\cdot {\text{K}}^{\text{-1}}\cdot {\text{mol}}^{\text{-1}}\text{)×(298}\text{K)}}\\ =251.84\end{array}$

To calculate $\text{K}$ value, inverse logarithm is taken for the above reaction.

  $\begin{array}{c}{\text{K=e}}^{251.84}\\ \text{=2}\text{.4}\times {\text{10}}^{109}\end{array}$

Therefore, the $\text{K}$ value of given reaction is $\text{2}\text{.4}\times {\text{10}}^{109}$.