Chapter 16, Problem 71QRT

(a)

Interpretation Introduction

Interpretation:

The $K°$ value for ${\text{2H}}_2\left(\text{g}\right)+{\text{O}}_{\text{2}}\left(\text{g}\right)\rightleftarrows {\text{2H}}_2\text{O}\left(\text{g}\right)$ at $800\text{K}$ 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 71QRT

The $K°$ value for ${\text{2H}}_2\left(\text{g}\right)+{\text{O}}_{\text{2}}\left(\text{g}\right)\rightleftarrows {\text{2H}}_2\text{O}\left(\text{g}\right)$ at $800\text{K}$ is $\underset{\_}{8.7\times 10^{26}}$.

Explanation of Solution

The given reaction is shown below.

  ${\text{2H}}_2\left(\text{g}\right)+{\text{O}}_{\text{2}}\left(\text{g}\right)\rightleftarrows {\text{2H}}_2\text{O}\left(\text{g}\right)$

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 ${\text{H}}_2\left(\text{g}\right)$, ${\text{O}}_{\text{2}}\left(\text{g}\right)$ and ${\text{H}}_2\text{O}\left(\text{g}\right)$ is $0\text{kJ}/\text{mol},0\text{kJ}/\text{mol}$ and $-241.818\text{kJ}/\text{mol}$ respectively.

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

    $\begin{array}{c}{\Delta }_{\text{r}}H°=\left(\begin{array}{c}\left({\text{n}}_{{\text{H}}_2\text{O}\left(\text{g}\right)}\times {\Delta }_{\text{f}}H°\left({\text{H}}_2\text{O}\left(\text{g}\right)\right)\right)-\\ \left(\left({\text{n}}_{{\text{H}}_2\left(\text{g}\right)}\times {\Delta }_{\text{f}}H°\left({\text{H}}_2\left(\text{g}\right)\right)\right)+{\text{n}}_{{\text{O}}_{\text{2}}\left(\text{g}\right)}\times {\Delta }_{\text{f}}H°\left({\text{O}}_{\text{2}}\left(\text{g}\right)\right)\right)\end{array}\right)\\ =\left(\begin{array}{c}\left(2\times \left(-241.818\text{kJ}/\text{mol}\right)\right)-\\ \left(\left(2\times 0\right)+\text{1}\times 0\right)\end{array}\right)\\ =-483.636\text{kJ}/\text{mol}\end{array}$

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 ${\text{H}}_2\left(\text{g}\right)$, ${\text{O}}_{\text{2}}\left(\text{g}\right)$ and ${\text{H}}_2\text{O}\left(\text{g}\right)$ is $130.684\text{J}/\text{mol}\cdot \text{K},205.138\text{J}/\text{mol}\cdot \text{K}$ and $188.825\text{J}/\text{mol}\cdot \text{K}$ respectively.

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

    $\begin{array}{c}{\Delta }_{\text{r}}S°=\left(\begin{array}{c}\left({\text{n}}_{{\text{H}}_2\text{O}\left(\text{g}\right)}\times {\Delta }_{\text{f}}S°\left({\text{H}}_2\text{O}\left(\text{g}\right)\right)\right)-\\ \left(\left({\text{n}}_{{\text{H}}_2\left(\text{g}\right)}\times {\Delta }_{\text{f}}S°\left({\text{H}}_2\left(\text{g}\right)\right)\right)+{\text{n}}_{{\text{O}}_{\text{2}}\left(\text{g}\right)}\times {\Delta }_{\text{f}}S°\left({\text{O}}_{\text{2}}\left(\text{g}\right)\right)\right)\end{array}\right)\\ =\left(\begin{array}{c}\left(2\times \left(188.825\text{J}/\text{mol}\cdot \text{K}\right)\right)-\\ \left(\left(2\times 130.684\text{J}/\text{mol}\cdot \text{K}\right)+\left(\text{1}\times 205.138\text{J}/\text{mol}\cdot \text{K}\right)\right)\end{array}\right)\\ =-88.856\text{J}/\text{mol}\cdot \text{K}\end{array}$

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).

    $\begin{array}{c}{\Delta }_{\text{r}}G°={\Delta }_{\text{r}}H°-T{\Delta }_{\text{r}}S°\\ =-483.636\text{kJ}/\text{mol}-\left(\left(\text{800}\text{K}\right)\times \left(-88.856\text{J}/\text{mol}\cdot \text{K}\right)\right)\\ =-483.636\times {\text{10}}^{\text{3}}\text{J}/\text{mol}+71084.8\text{J}/\text{mol}\because \text{1}{\text{kJ=10}}^{\text{3}}\text{J}\\ =-412551.2\text{J}/\text{mol}\end{array}$

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).

    $\begin{array}{c}K°=e^{-\frac{-412551.2\text{J}/\text{mol}}{8.314\text{J}/\text{K}\cdot \text{mol}\times 800\text{K}}}\\ =e^{62.03}\\ =\underset{\_}{8.7\times 10^{26}}\end{array}$

(b)

Interpretation Introduction

Interpretation:

The $K°$ value for ${\text{2SO}}_{\text{2}}\left(\text{g}\right)+{\text{O}}_{\text{2}}\left(\text{g}\right)\rightleftarrows {\text{2SO}}_3\left(\text{g}\right)$ at $500\text{K}$ has to be calculated.

Concept Introduction:

Refer to part (a).

Answer to Problem 71QRT

The $K°$ value for ${\text{2SO}}_{\text{2}}\left(\text{g}\right)+{\text{O}}_{\text{2}}\left(\text{g}\right)\rightleftarrows {\text{2SO}}_3\left(\text{g}\right)$ at $500\text{K}$ is $\underset{\_}{7.0\times 10^{10}}$.

Explanation of Solution

The given reaction is shown below.

  ${\text{2SO}}_{\text{2}}\left(\text{g}\right)+{\text{O}}_{\text{2}}\left(\text{g}\right)\rightleftarrows {\text{2SO}}_3\left(\text{g}\right)$

The value of ${\Delta }_{\text{f}}H°$ for ${\text{SO}}_{\text{2}}\left(\text{g}\right)$, ${\text{O}}_{\text{2}}\left(\text{g}\right)$ and ${\text{SO}}_3\left(\text{g}\right)$ is $-296.830\text{kJ}/\text{mol},0\text{kJ}/\text{mol}$ and $-395.72\text{kJ}/\text{mol}$ respectively.

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

    $\begin{array}{c}{\Delta }_{\text{r}}H°=\left(\begin{array}{c}\left({\text{n}}_{{\text{SO}}_3\left(\text{g}\right)}\times {\Delta }_{\text{f}}H°\left({\text{SO}}_3\left(\text{g}\right)\right)\right)-\\ \left(\left({\text{n}}_{{\text{SO}}_{\text{2}}\left(\text{g}\right)}\times {\Delta }_{\text{f}}H°\left({\text{SO}}_{\text{2}}\left(\text{g}\right)\right)\right)+{\text{n}}_{{\text{O}}_{\text{2}}\left(\text{g}\right)}\times {\Delta }_{\text{f}}H°\left({\text{O}}_{\text{2}}\left(\text{g}\right)\right)\right)\end{array}\right)\\ =\left(\begin{array}{c}\left(2\times \left(-395.72\text{kJ}/\text{mol}\right)\right)-\\ \left(\left(2\times \left(-296.830\text{kJ}/\text{mol}\right)\right)+\text{1}\times 0\right)\end{array}\right)\\ =-197.78\text{kJ}/\text{mol}\end{array}$

The value of ${\Delta }_{\text{f}}S°$ for ${\text{SO}}_{\text{2}}\left(\text{g}\right)$, ${\text{O}}_{\text{2}}\left(\text{g}\right)$ and ${\text{SO}}_3\left(\text{g}\right)$ is $248.22\text{J}/\text{mol}\cdot \text{K},205.138\text{J}/\text{mol}\cdot \text{K}$ and $256.76\text{J}/\text{mol}\cdot \text{K}$ respectively.

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

    $\begin{array}{c}{\Delta }_{\text{r}}S°=\left(\begin{array}{c}\left({\text{n}}_{{\text{SO}}_3\left(\text{g}\right)}\times {\Delta }_{\text{f}}S°\left({\text{SO}}_3\left(\text{g}\right)\right)\right)-\\ \left(\left({\text{n}}_{{\text{SO}}_{\text{2}}\left(\text{g}\right)}\times {\Delta }_{\text{f}}S°\left({\text{SO}}_{\text{2}}\left(\text{g}\right)\right)\right)+\left({\text{n}}_{{\text{O}}_{\text{2}}\left(\text{g}\right)}\times {\Delta }_{\text{f}}S°\left({\text{O}}_{\text{2}}\left(\text{g}\right)\right)\right)\right)\end{array}\right)\\ =\left(\begin{array}{c}\left(2\times \left(256.76\text{J}/\text{mol}\cdot \text{K}\right)\right)-\\ \left(\left(2\times \left(248.22\text{J}/\text{mol}\cdot \text{K}\right)\right)+\left(\text{1}\times 205.138\text{J}/\text{mol}\cdot \text{K}\right)\right)\end{array}\right)\\ =-188.058\text{J}/\text{mol}\cdot \text{K}\end{array}$

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).

    $\begin{array}{c}{\Delta }_{\text{r}}G°={\Delta }_{\text{r}}H°-T{\Delta }_{\text{r}}S°\\ =-197.78\text{kJ}/\text{mol}-\left(\left(\text{500}\text{K}\right)\times \left(-188.058\text{J}/\text{mol}\cdot \text{K}\right)\right)\\ =-197.78\times {\text{10}}^{\text{3}}\text{J}/\text{mol}+94029\text{J}/\text{mol}\because \text{1}{\text{kJ=10}}^{\text{3}}\text{J}\\ =-103751\text{J}/\text{mol}\end{array}$

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

    $\begin{array}{c}K°=e^{-\frac{-103751\text{J}/\text{mol}}{8.314\text{J}/\text{K}\cdot \text{mol}\times 500\text{K}}}\\ =e^{24.96}\\ =6.92\times {10}^{10}\\ \approx \underset{\_}{7.0\times 10^{10}}\end{array}$

(c)

Interpretation Introduction

Interpretation:

The $K°$ value for $\text{2HF}\left(\text{g}\right)\rightleftarrows {\text{H}}_{\text{2}}\left(\text{g}\right)+{\text{F}}_{\text{2}}\left(\text{g}\right)$ at $2000\text{K}$ has to be calculated.

Concept Introduction:

Refer to part (a).

Answer to Problem 71QRT

The $K°$ value for $\text{2HF}\left(\text{g}\right)\rightleftarrows {\text{H}}_{\text{2}}\left(\text{g}\right)+{\text{F}}_{\text{2}}\left(\text{g}\right)$ at $2000\text{K}$ is $\underset{\_}{1.3\times 10^{-15}}$.

Explanation of Solution

The given reaction is shown below.

  $\text{2HF}\left(\text{g}\right)\rightleftarrows {\text{H}}_{\text{2}}\left(\text{g}\right)+{\text{F}}_{\text{2}}\left(\text{g}\right)$

The value of ${\Delta }_{\text{f}}H°$ for $\text{HF}\left(\text{g}\right)$, ${\text{H}}_{\text{2}}\left(\text{g}\right)$ and ${\text{F}}_{\text{2}}\left(\text{g}\right)$ is $-271.1\text{kJ}/\text{mol},0\text{kJ}/\text{mol}$ and $0\text{kJ}/\text{mol}$ respectively.

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

    $\begin{array}{c}{\Delta }_{\text{r}}H°=\left(\begin{array}{c}\left(\left({\text{n}}_{{\text{H}}_{\text{2}}\left(\text{g}\right)}\times {\Delta }_{\text{f}}H°\left({\text{H}}_{\text{2}}\left(\text{g}\right)\right)\right)+\left({\text{n}}_{{\text{F}}_{\text{2}}\left(\text{g}\right)}\times {\Delta }_{\text{f}}H°\left({\text{F}}_{\text{2}}\left(\text{g}\right)\right)\right)\right)\\ -\left({\text{n}}_{\text{HF}\left(\text{g}\right)}\times {\Delta }_{\text{f}}H°\left(\text{HF}\left(\text{g}\right)\right)\right)\end{array}\right)\\ =\left(\begin{array}{c}\left(\left(\text{1}\times 0\right)+\left(\text{1}\times 0\right)\right)\\ -\left(2\times \left(-271.1\text{kJ}/\text{mol}\right)\right)\end{array}\right)\\ =542.2\text{kJ}/\text{mol}\end{array}$

The value of ${\Delta }_{\text{f}}S°$ for $\text{HF}\left(\text{g}\right)$, ${\text{H}}_{\text{2}}\left(\text{g}\right)$ and ${\text{F}}_{\text{2}}\left(\text{g}\right)$ is $173.779\text{J}/\text{mol}\cdot \text{K},130.684\text{J}/\text{mol}\cdot \text{K}$ and $202.78\text{J}/\text{mol}\cdot \text{K}$ respectively.

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

    $\begin{array}{c}{\Delta }_{\text{r}}S°=\left(\begin{array}{c}\left(\left({\text{n}}_{{\text{H}}_{\text{2}}\left(\text{g}\right)}\times {\Delta }_{\text{f}}S°\left({\text{H}}_{\text{2}}\left(\text{g}\right)\right)\right)+\left({\text{n}}_{{\text{F}}_{\text{2}}\left(\text{g}\right)}\times {\Delta }_{\text{f}}S°\left({\text{F}}_{\text{2}}\left(\text{g}\right)\right)\right)\right)\\ -\left({\text{n}}_{\text{HF}\left(\text{g}\right)}\times {\Delta }_{\text{f}}S°\left(\text{HF}\left(\text{g}\right)\right)\right)\end{array}\right)\\ =\left(\begin{array}{c}\left(\left(\text{1}\times 130.684\text{J}/\text{mol}\cdot \text{K}\right)+\left(\text{1}\times 202.78\text{J}/\text{mol}\cdot \text{K}\right)\right)\\ -\left(2\times \left(173.779\text{J}/\text{mol}\cdot \text{K}\right)\right)\end{array}\right)\\ =-14.094\text{J}/\text{mol}\cdot \text{K}\end{array}$

The value of temperature is given as $2000\text{K}$.

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

    $\begin{array}{c}{\Delta }_{\text{r}}G°={\Delta }_{\text{r}}H°-T{\Delta }_{\text{r}}S°\\ =542.2\text{kJ}/\text{mol}-\left(\left(\text{2000}\text{K}\right)\times \left(-14.094\text{J}/\text{mol}\cdot \text{K}\right)\right)\\ =542.2\times {\text{10}}^{\text{3}}\text{J}/\text{mol}+28188\text{J}/\text{mol}\because \text{1}{\text{kJ=10}}^{\text{3}}\text{J}\\ =570388\text{J}/\text{mol}\end{array}$

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

    $\begin{array}{c}K°=e^{-\frac{570388\text{J}/\text{mol}}{8.314\text{J}/\text{K}\cdot \text{mol}\times 2000\text{K}}}\\ =e^{-34.30}\\ =1.27\times {10}^{-15}\\ \approx \underset{\_}{1.3\times 10^{-15}}\end{array}$