Chapter 5, Problem 5G.21E

(a)

Interpretation Introduction

Interpretation:

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

  ${\text{2H}}_{\text{2}}{\text{(g)+O}}_{\text{2}}\text{(g)}\rightleftharpoons \text{2}{\text{H}}_{\text{2}}\text{O(g)}$

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.21E

The equilibrium constant of the given reaction is $\text{2}\text{.0}\times {\text{10}}^{80}$.

Explanation of Solution

Given reaction is

The combustion of hydrogen:

  ${\text{2H}}_{\text{2}}{\text{(g)+O}}_{\text{2}}\text{(g)}\rightleftharpoons \text{2}{\text{H}}_{\text{2}}\text{O(g)}$

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

The ${\text{ΔG}}_{\text{f}}^{\text{°}}$ value of ${\text{H}}_{\text{2}}\text{O}$ is $-228.57\text{k}\text{J}\cdot {\text{mol}}^{\text{-1}}$

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

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

  $\begin{array}{l}{\text{ΔG}}_{\text{r}}^{\text{°}}{\text{=2ΔG}}_{\text{f}}^{\text{°}}{\text{(H}}_2{\text{O(g))-[2ΔG}}_{\text{f}}^{\text{°}}{\text{(H}}_{\text{2}}{\text{(g))+ΔG}}_{\text{f}}^{\text{°}}{\text{(O}}_{\text{2}}\text{(g))]}\\ \text{=2}\times (-228.57\times {10}^3)-0\\ =-457.14\times {10}^3\end{array}$

Now, the equilibrium constant of the reaction is calculated as

  $\begin{array}{c}\text{lnK=}\frac{{\text{-ΔG}}_{\text{r}}{}^{\text{o}}}{\text{RT}}\\ \text{=}\frac{457.14{\text{×10}}^{\text{3}}\text{J}\cdot {\text{mol}}^{\text{-1}}}{(8.3145\text{J}\cdot {\text{K}}^{\text{-1}}\cdot {\text{mol}}^{\text{-1}}\text{)×(298}\text{K)}}\\ =184.9\end{array}$

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

  $\begin{array}{c}{\text{K=e}}^{184.9}\\ \text{=2}\text{.0}\times {\text{10}}^{80}\end{array}$

Therefore, the $\text{K}$ value of given reaction is $\text{2}\text{.0}\times {\text{10}}^{80}$.

(b)

Interpretation Introduction

Interpretation:

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

  ${\text{2CO(g)+O}}_{\text{2}}\text{(g)}\rightleftharpoons \text{2}{\text{CO}}_2\text{(g)}$

Concept introduction:

Refer to part (a).

Answer to Problem 5G.21E

The equilibrium constant of the given reaction is $\text{1}\times {\text{10}}^{90}$.

Explanation of Solution

Given reaction is

The oxidation of carbon monoxide:

  ${\text{2CO(g)+O}}_{\text{2}}\text{(g)}\rightleftharpoons \text{2}{\text{CO}}_2\text{(g)}$

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

The ${\text{ΔG}}_{\text{f}}^{\text{°}}$ value of $\text{CO}$ is $-137.17\text{k}\text{J}\cdot {\text{mol}}^{\text{-1}}$

The ${\text{ΔG}}_{\text{f}}^{\text{°}}$ value of ${\text{CO}}_{\text{2}}$ is $-394.36\text{k}\text{J}\cdot {\text{mol}}^{\text{-1}}$

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

  $\begin{array}{l}{\text{ΔG}}_{\text{r}}^{\text{°}}{\text{=2ΔG}}_{\text{f}}^{\text{°}}\text{(}C{\text{O}}_2{\text{(g))-[2ΔG}}_{\text{f}}^{\text{°}}{\text{(CO(g))+ΔG}}_{\text{f}}^{\text{°}}{\text{(O}}_{\text{2}}\text{(g))]}\\ \text{=2}\times (-394.36\times {10}^3)-[2\times (-137.17\times {10}^3)-0]\\ =-5.14\times {10}^5\end{array}$

Now, the equilibrium constant of the reaction is calculated as

  $\begin{array}{c}\text{lnK=}\frac{{\text{-ΔG}}_{\text{r}}{}^{\text{o}}}{\text{RT}}\\ \text{=}\frac{5.14\times {10}^5\text{J}\cdot {\text{mol}}^{\text{-1}}}{(8.3145\text{J}\cdot {\text{K}}^{\text{-1}}\cdot {\text{mol}}^{\text{-1}}\text{)×(298}\text{K)}}\\ =207.44\end{array}$

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

  $\begin{array}{c}{\text{K=e}}^{207.44}\\ \text{=1}\times {\text{10}}^{90}\end{array}$

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.

  ${\text{CaCO}}_{\text{3}}\text{(s)}\rightleftharpoons {\text{CaO(s)+CO}}_{\text{2}}\text{(g)}$

Concept introduction:

Refer to part (a).

Answer to Problem 5G.21E

The equilibrium constant of the given reaction is $\text{6}\text{.1}\times {\text{10}}^{22}$.

Explanation of Solution

Given reaction is

The decomposition of limestone:

  ${\text{CaCO}}_{\text{3}}\text{(s)}\rightleftharpoons {\text{CaO(s)+CO}}_{\text{2}}\text{(g)}$

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

The ${\text{ΔG}}_{\text{f}}^{\text{°}}$ value of ${\text{CO}}_{\text{2}}$ is $-394.36\text{k}\text{J}\cdot {\text{mol}}^{\text{-1}}$

The ${\text{ΔG}}_{\text{f}}^{\text{°}}$ value of $\text{CaO}$ is $-604.03\text{k}\text{J}\cdot {\text{mol}}^{\text{-1}}$

The ${\text{ΔG}}_{\text{f}}^{\text{°}}$ value of ${\text{CaCO}}_3$ is $-1127.8\text{k}\text{J}\cdot {\text{mol}}^{\text{-1}}$

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

$\text{T=298K}$

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

  $\begin{array}{l}{\text{ΔG}}_{\text{r}}^{\text{°}}{\text{=[ΔG}}_{\text{f}}^{\text{°}}\text{CaO(s}){\text{+ΔG}}_{\text{f}}^{\text{°}}{\text{CO}}_{\text{2}}\text{(g)}]{\text{-ΔG}}_{\text{f}}^{\text{°}}{\text{CaCO}}_3\text{(s)}\\ \text{=}(-604.03\times {10}^3-394.36\times {10}^3)-(-1127.8\times {10}^3)\\ =1.30\times {10}^5\end{array}$

Now, the equilibrium constant of the reaction is calculated as

  $\begin{array}{c}\text{lnK=}\frac{{\text{-ΔG}}_{\text{r}}{}^{\text{o}}}{\text{RT}}\\ \text{=}\frac{1.30\times {10}^5\text{J}\cdot {\text{mol}}^{\text{-1}}}{(8.3145\text{J}\cdot {\text{K}}^{\text{-1}}\cdot {\text{mol}}^{\text{-1}}\text{)×(298}\text{K)}}\\ =52.47\end{array}$

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

  $\begin{array}{c}{\text{K=e}}^{52.47}\\ \text{=6}\text{.1}\times {\text{10}}^{22}\end{array}$

Therefore, the $\text{K}$ value of given reaction is $\text{6}\text{.1}\times {\text{10}}^{22}$.