Chapter 16, Problem 72QRT

(a)

Interpretation Introduction

Interpretation:

The $K°$ value for ${\text{H}}_{\text{2}}\left(\text{g}\right)+{\text{I}}_{\text{2}}\left(\text{g}\right)\rightleftharpoons \text{2HI}\left(\text{g}\right)$ at $500\text{K}$ has to be calculated.

Concept Introduction:

The Gibbs free energy of a system is defined as the enthalpy of the system minus the product of the temperature times the entropy of the system.  The Gibbs free energy of the system is a state function as it is defined in terms of thermodynamic properties that are state functions.  The Gibbs free energy of a system is directly related to the equilibrium constant of a reaction.  The symbol for equilibrium constant is $K_{\text{P}}$.

Answer to Problem 72QRT

The $K°$ value for ${\text{H}}_{\text{2}}\left(\text{g}\right)+{\text{I}}_{\text{2}}\left(\text{g}\right)\rightleftharpoons \text{2HI}\left(\text{g}\right)$ at $500\text{K}$ is $\underset{\_}{134.8}$.

Explanation of Solution

The given reaction is shown below.

  ${\text{H}}_{\text{2}}\left(\text{g}\right)+{\text{I}}_{\text{2}}\left(\text{g}\right)\rightleftharpoons \text{2HI}\left(\text{g}\right)$

The formula to calculate ${\Delta }_{\text{r}}G°$ is shown below.

    ${\Delta }_{\text{r}}G°={\Delta }_{\text{r}}H°-T{\Delta }_{\text{r}}S°$        (1)

Where,

• ${\text{Δ}}_{\text{r}}G°$ is the change in standard Gibbs free energy of reaction.
• ${\Delta }_{\text{r}}H°$ is the change in standard enthalpy of reaction.
• ${\Delta }_{\text{r}}S°$ is the change in standard entropy of reaction.
• $T$ is the temperature.

The value of ${\Delta }_{\text{r}}H°$ is calculated by the formula shown below.

    ${\Delta }_{\text{r}}H°={\text{n}}_{\text{Products}}{\Delta }_{\text{f}}H°\left(\text{Products}\right)-{\text{n}}_{\text{Reactants}}{\Delta }_{\text{f}}H°\left(\text{Reactants}\right)$        (2)

Where,

• ${\Delta }_{\text{f}}H°$ is the change in standard enthalpy of formation.
• ${\text{n}}_{\text{Reactants}}$ is the number of moles of reactants.
• ${\text{n}}_{\text{Products}}$ is the number of moles of products.

The value of ${\Delta }_{\text{f}}H°$ for ${\text{H}}_2\left(\text{g}\right)$, ${\text{I}}_{\text{2}}\left(\text{g}\right)$ and $\text{HI}\left(\text{g}\right)$ is $0\text{kJ}/\text{mol},62.438\text{kJ}/\text{mol}$ and $26.48\text{kJ}/\text{mol}$ respectively.

Substitute the values in equation (2) as shown below.

    $\begin{array}{c}{\Delta }_{\text{r}}H°=\left(\begin{array}{c}\left({\text{n}}_{\text{HI}\left(\text{g}\right)}\times {\Delta }_{\text{f}}H°\left(\text{HI}\left(\text{g}\right)\right)\right)-\\ \left(\left({\text{n}}_{{\text{H}}_2\left(\text{g}\right)}\times {\Delta }_{\text{f}}H°\left({\text{H}}_2\left(\text{g}\right)\right)\right)+{\text{n}}_{{\text{I}}_{\text{2}}\left(\text{g}\right)}\times {\Delta }_{\text{f}}H°\left({\text{I}}_{\text{2}}\left(\text{g}\right)\right)\right)\end{array}\right)\\ =\left(\begin{array}{c}\left(2\times \left(26.48\text{kJ}/\text{mol}\right)\right)-\\ \left(\left(1\times 0\right)+\left(\text{1}\times 62.438\text{kJ}/\text{mol}\right)\right)\end{array}\right)\\ =-9.478\text{kJ}/\text{mol}\end{array}$

The value of ${\Delta }_{\text{r}}S°$ is calculated by the formula shown below.

    ${\Delta }_{\text{r}}S°={\text{n}}_{\text{Products}}{\Delta }_{\text{f}}S°\left(\text{Products}\right)-{\text{n}}_{\text{Reactants}}{\Delta }_{\text{f}}S°\left(\text{Reactants}\right)$        (3)

Where,

• ${\Delta }_{\text{f}}S°$ is the change in standard entropy of formation.
• ${\text{n}}_{\text{Reactants}}$ is the number of moles of reactants.
• ${\text{n}}_{\text{Products}}$ is the number of moles of products.

The value of ${\Delta }_{\text{f}}S°$ for ${\text{H}}_2\left(\text{g}\right)$, ${\text{I}}_{\text{2}}\left(\text{g}\right)$ and $\text{HI}\left(\text{g}\right)$ is $130.684\text{J}/\text{mol}\cdot \text{K},260.69\text{J}/\text{mol}\cdot \text{K}$ and $206.594\text{J}/\text{mol}\cdot \text{K}$ respectively.

Substitute the values in equation (3) as shown below.

    $\begin{array}{c}{\Delta }_{\text{r}}S°=\left(\begin{array}{c}\left({\text{n}}_{\text{HI}\left(\text{g}\right)}\times {\Delta }_{\text{f}}S°\left(\text{HI}\left(\text{g}\right)\right)\right)-\\ \left(\left({\text{n}}_{{\text{H}}_2\left(\text{g}\right)}\times {\Delta }_{\text{f}}S°\left({\text{H}}_2\left(\text{g}\right)\right)\right)+{\text{n}}_{{\text{I}}_{\text{2}}\left(\text{g}\right)}\times {\Delta }_{\text{f}}S°\left({\text{I}}_{\text{2}}\left(\text{g}\right)\right)\right)\end{array}\right)\\ =\left(\begin{array}{c}\left(2\times \left(206.594\text{J}/\text{mol}\cdot \text{K}\right)\right)-\\ \left(\left(1\times 130.684\text{J}/\text{mol}\cdot \text{K}\right)+\left(\text{1}\times 260.69\text{J}/\text{mol}\cdot \text{K}\right)\right)\end{array}\right)\\ =21.814\text{J}/\text{mol}\cdot \text{K}\end{array}$

The value of temperature is given as $500\text{K}$.

Substitute the values of ${\Delta }_{\text{r}}H°$, ${\Delta }_{\text{r}}S°$ and $T$ in equation (1).

    $\begin{array}{c}{\Delta }_{\text{r}}G°={\Delta }_{\text{r}}H°-T{\Delta }_{\text{r}}S°\\ =-9.478\text{kJ}/\text{mol}-\left(\left(\text{500}\text{K}\right)\times \left(21.814\text{J}/\text{mol}\cdot \text{K}\right)\right)\\ =-9.478\times {\text{10}}^{\text{3}}\text{J}/\text{mol}-10907\text{J}/\text{mol}\because \text{1}{\text{kJ=10}}^{\text{3}}\text{J}\\ =-20385\text{J}/\text{mol}\end{array}$

The relation between ${\Delta }_{\text{r}}G°$ and $K_{\text{P}}$ is shown below.

    $K_{\text{P}}=e^{-\frac{{\Delta }_{\text{r}}G°}{RT}}$        (4)

Where,

• ${\text{Δ}}_{\text{r}}G°$ is the change in standard Gibbs free energy of reaction.
• $R$ is the gas constant.
• $K_{\text{P}}$ is the equilibrium constant.
• $T$ is the temperature.

Substitute the values of ${\text{Δ}}_{\text{r}}G°$, $R$ and $T$ in equation (4).

    $\begin{array}{c}K°=e^{-\frac{-20385\text{J}/\text{mol}}{8.314\text{J}/\text{K}\cdot \text{mol}\times 500\text{K}}}\\ =\underset{\_}{134.8}\end{array}$

(b)

Interpretation Introduction

Interpretation:

The $K°$ value for ${\text{N}}_{\text{2}}\left(\text{g}\right)+3{\text{H}}_{\text{2}}\left(\text{g}\right)\rightleftharpoons {\text{2NH}}_3\left(\text{g}\right)$ at $400\text{K}$ has to be calculated.

Concept Introduction:

Refer to part (a).

Answer to Problem 72QRT

The $K°$ value for ${\text{N}}_{\text{2}}\left(\text{g}\right)+3{\text{H}}_{\text{2}}\left(\text{g}\right)\rightleftharpoons {\text{2NH}}_3\left(\text{g}\right)$ at $400\text{K}$ is $\underset{\_}{45.8}$.

Explanation of Solution

The given reaction is shown below.

  ${\text{N}}_{\text{2}}\left(\text{g}\right)+3{\text{H}}_{\text{2}}\left(\text{g}\right)\rightleftharpoons {\text{2NH}}_3\left(\text{g}\right)$

The value of ${\Delta }_{\text{f}}H°$ for ${\text{N}}_{\text{2}}\left(\text{g}\right)$, ${\text{H}}_{\text{2}}\left(\text{g}\right)$ and ${\text{NH}}_3\left(\text{g}\right)$ is $0\text{kJ}/\text{mol},0\text{kJ}/\text{mol}$ and $-46.11\text{kJ}/\text{mol}$ respectively.

Substitute the values in equation (2) as shown below.

    $\begin{array}{c}{\Delta }_{\text{r}}H°=\left(\begin{array}{c}\left({\text{n}}_{{\text{NH}}_3\left(\text{g}\right)}\times {\Delta }_{\text{f}}H°\left({\text{NH}}_3\left(\text{g}\right)\right)\right)-\\ \left(\left({\text{n}}_{{\text{N}}_{\text{2}}\left(\text{g}\right)}\times {\Delta }_{\text{f}}H°\left({\text{N}}_{\text{2}}\left(\text{g}\right)\right)\right)+{\text{n}}_{{\text{H}}_{\text{2}}\left(\text{g}\right)}\times {\Delta }_{\text{f}}H°\left({\text{H}}_{\text{2}}\left(\text{g}\right)\right)\right)\end{array}\right)\\ =\left(\begin{array}{c}\left(2\times \left(-46.11\text{kJ}/\text{mol}\right)\right)-\\ \left(\left(1\times 0\right)+\left(3\times 0\right)\right)\end{array}\right)\\ =-92.22\text{kJ}/\text{mol}\end{array}$

The value of ${\Delta }_{\text{f}}S°$ for ${\text{N}}_{\text{2}}\left(\text{g}\right)$, ${\text{H}}_{\text{2}}\left(\text{g}\right)$ and ${\text{NH}}_3\left(\text{g}\right)$ is $191.61\text{J}/\text{mol}\cdot \text{K},130.684\text{J}/\text{mol}\cdot \text{K}$ and $192.45\text{J}/\text{mol}\cdot \text{K}$ respectively.

Substitute the values in equation (3) as shown below.

    $\begin{array}{c}{\Delta }_{\text{r}}S°=\left(\begin{array}{c}\left({\text{n}}_{{\text{NH}}_3\left(\text{g}\right)}\times {\Delta }_{\text{f}}S°\left({\text{NH}}_3\left(\text{g}\right)\right)\right)-\\ \left(\left({\text{n}}_{{\text{N}}_{\text{2}}\left(\text{g}\right)}\times {\Delta }_{\text{f}}S°\left({\text{N}}_{\text{2}}\left(\text{g}\right)\right)\right)+{\text{n}}_{{\text{H}}_{\text{2}}\left(\text{g}\right)}\times {\Delta }_{\text{f}}S°\left({\text{H}}_{\text{2}}\left(\text{g}\right)\right)\right)\end{array}\right)\\ =\left(\begin{array}{c}\left(2\times \left(192.45\text{J}/\text{mol}\cdot \text{K}\right)\right)-\\ \left(\left(1\times 191.61\text{J}/\text{mol}\cdot \text{K}\right)+\left(3\times 130.684\text{J}/\text{mol}\cdot \text{K}\right)\right)\end{array}\right)\\ =-198.762\text{J}/\text{mol}\cdot \text{K}\end{array}$

The value of temperature is given as $400\text{K}$.

Substitute the values of ${\Delta }_{\text{r}}H°$, ${\Delta }_{\text{r}}S°$ and $T$ in equation (1).

    $\begin{array}{c}{\Delta }_{\text{r}}G°={\Delta }_{\text{r}}H°-T{\Delta }_{\text{r}}S°\\ =-92.22\text{kJ}/\text{mol}-\left(\left(\text{400}\text{K}\right)\times \left(-198.762\text{J}/\text{mol}\cdot \text{K}\right)\right)\\ =-92.22\times {\text{10}}^{\text{3}}\text{J}/\text{mol}+79504.8\text{J}/\text{mol}\because \text{1}{\text{kJ=10}}^{\text{3}}\text{J}\\ =-12715.2\text{J}/\text{mol}\end{array}$

Substitute the values of ${\text{Δ}}_{\text{r}}G°$, $R$ and $T$ in equation (4).

    $\begin{array}{c}K°=e^{-\frac{-12715.2\text{J}/\text{mol}}{8.314\text{J}/\text{K}\cdot \text{mol}\times 400\text{K}}}\\ =\underset{\_}{45.8}\end{array}$

(c)

Interpretation Introduction

Interpretation:

The $K°$ value for $\text{CO}\left(\text{g}\right)+3{\text{H}}_{\text{2}}\left(\text{g}\right)\rightleftharpoons {\text{CH}}_{\text{4}}\left(\text{g}\right)+{\text{H}}_{\text{2}}\text{O}\left(\text{g}\right)$ at $800\text{K}$ has to be calculated.

Concept Introduction:

Refer to part (a).

Answer to Problem 72QRT

The $K°$ value for $\text{CO}\left(\text{g}\right)+3{\text{H}}_{\text{2}}\left(\text{g}\right)\rightleftharpoons {\text{CH}}_{\text{4}}\left(\text{g}\right)+{\text{H}}_{\text{2}}\text{O}\left(\text{g}\right)$ at $800\text{K}$ is $\underset{\_}{176.1}$.

Explanation of Solution

The given reaction is shown below.

  $\text{CO}\left(\text{g}\right)+3{\text{H}}_{\text{2}}\left(\text{g}\right)\rightleftharpoons {\text{CH}}_{\text{4}}\left(\text{g}\right)+{\text{H}}_{\text{2}}\text{O}\left(\text{g}\right)$

The value of ${\Delta }_{\text{f}}H°$ for $\text{CO}\left(\text{g}\right)$, ${\text{H}}_{\text{2}}\left(\text{g}\right)$, ${\text{CH}}_{\text{4}}\left(\text{g}\right)$ and ${\text{H}}_{\text{2}}\text{O}\left(\text{g}\right)$ is $-110.525\text{kJ}/\text{mol},0\text{kJ}/\text{mol},-74.81\text{kJ}/\text{mol}$ and $-241.818\text{kJ}/\text{mol}$ respectively.

Substitute the values in equation (2) as shown below.

    $\begin{array}{c}{\Delta }_{\text{r}}H°=\left(\begin{array}{c}\left(\left({\text{n}}_{{\text{H}}_{\text{2}}\text{O}\left(\text{g}\right)}\times {\Delta }_{\text{f}}H°\left({\text{H}}_{\text{2}}\text{O}\left(\text{g}\right)\right)\right)+\left({\text{n}}_{{\text{CH}}_{\text{4}}\left(\text{g}\right)}\times {\Delta }_{\text{f}}H°\left({\text{CH}}_{\text{4}}\left(\text{g}\right)\right)\right)\right)\\ -\left(\left({\text{n}}_{\text{CO}\left(\text{g}\right)}\times {\Delta }_{\text{f}}H°\left(\text{CO}\left(\text{g}\right)\right)\right)+\left({\text{n}}_{{\text{H}}_{\text{2}}\left(\text{g}\right)}\times {\Delta }_{\text{f}}H°\left({\text{H}}_{\text{2}}\left(\text{g}\right)\right)\right)\right)\end{array}\right)\\ =\left(\begin{array}{c}\left(\left(\text{1}\times \left(-241.818\text{kJ}/\text{mol}\right)\right)+\left(\text{1}\times \left(-74.81\text{kJ}/\text{mol}\right)\right)\right)\\ -\left(1\times \left(-110.525\text{kJ}/\text{mol}\right)+\left(3\times 0\right)\right)\end{array}\right)\\ =-206.103\text{kJ}/\text{mol}\end{array}$

The value of ${\Delta }_{\text{f}}S°$ for $\text{CO}\left(\text{g}\right)$, ${\text{H}}_{\text{2}}\left(\text{g}\right)$, ${\text{CH}}_{\text{4}}\left(\text{g}\right)$ and ${\text{H}}_{\text{2}}\text{O}\left(\text{g}\right)$ is $197.674\text{J}/\text{mol}\cdot \text{K},130.684\text{J}/\text{mol}\cdot \text{K},186.264\text{J}/\text{mol}\cdot \text{K}$ and $188.825\text{J}/\text{mol}\cdot \text{K}$ respectively.

Substitute the values in equation (3) as shown below.

    $\begin{array}{c}{\Delta }_{\text{r}}S°=\left(\begin{array}{c}\left(\left({\text{n}}_{{\text{H}}_{\text{2}}\text{O}\left(\text{g}\right)}\times {\Delta }_{\text{f}}S°\left({\text{H}}_{\text{2}}\text{O}\left(\text{g}\right)\right)\right)+\left({\text{n}}_{{\text{CH}}_{\text{4}}\left(\text{g}\right)}\times {\Delta }_{\text{f}}S°\left({\text{CH}}_{\text{4}}\left(\text{g}\right)\right)\right)\right)\\ -\left(\left({\text{n}}_{\text{CO}\left(\text{g}\right)}\times {\Delta }_{\text{f}}S°\left(\text{CO}\left(\text{g}\right)\right)\right)+\left({\text{n}}_{{\text{H}}_{\text{2}}\left(\text{g}\right)}\times {\Delta }_{\text{f}}S°\left({\text{H}}_{\text{2}}\left(\text{g}\right)\right)\right)\right)\end{array}\right)\\ =\left(\begin{array}{c}\left(\left(\text{1}\times \left(188.825\text{J}/\text{mol}\cdot \text{K}\right)\right)+\left(\text{1}\times \left(186.264\text{J}/\text{mol}\cdot \text{K}\right)\right)\right)\\ -\left(1\times \left(197.674\text{J}/\text{mol}\cdot \text{K}\right)+\left(3\times 130.684\text{J}/\text{mol}\cdot \text{K}\right)\right)\end{array}\right)\\ =-214.637\text{J}/\text{mol}\cdot \text{K}\end{array}$

The value of temperature is given as $800\text{K}$.

Substitute the values of ${\Delta }_{\text{r}}H°$, ${\Delta }_{\text{r}}S°$ and $T$ in equation (1).

    $\begin{array}{c}{\Delta }_{\text{r}}G°={\Delta }_{\text{r}}H°-T{\Delta }_{\text{r}}S°\\ =-206.103\text{kJ}/\text{mol}-\left(\left(800\text{K}\right)\times \left(-214.637\text{J}/\text{mol}\cdot \text{K}\right)\right)\\ =-206.103\times {\text{10}}^{\text{3}}\text{J}/\text{mol}+171709.6\text{J}/\text{mol}\because \text{1}{\text{kJ=10}}^{\text{3}}\text{J}\\ =-34393.4\text{J}/\text{mol}\end{array}$

Substitute the values of ${\text{Δ}}_{\text{r}}G°$, $R$ and $T$ in equation (4).

    $\begin{array}{c}K°=e^{-\frac{-34393.4\text{J}/\text{mol}}{8.314\text{J}/\text{K}\cdot \text{mol}\times 800\text{K}}}\\ =\underset{\_}{176.1}\end{array}$