Chapter 16, Problem 119QRT

(a)

Interpretation Introduction

Interpretation:

The value of ${\Delta }_{\text{r}}G°$ for the reaction at $25°\text{C}$, ${\text{CH}}_4\left(\text{g}\right)+{\text{H}}_{\text{2}}\text{O}\left(\text{g}\right)\rightleftharpoons \text{CO}\left(\text{g}\right)+3{\text{H}}_{\text{2}}\left(\text{g}\right)$, 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 in terms of partial pressure is $K_p$.

Answer to Problem 119QRT

The value of ${\Delta }_{\text{r}}G°$ for the reaction at $25°\text{C}$, ${\text{CH}}_4\left(\text{g}\right)+{\text{H}}_{\text{2}}\text{O}\left(\text{g}\right)\rightleftharpoons \text{CO}\left(\text{g}\right)+3{\text{H}}_{\text{2}}\left(\text{g}\right)$, is $\underset{\_}{142.124kJmo{l}^{-1}}$.

Explanation of Solution

The value of standard Gibbs free energy of the formation ${\text{H}}_{\text{2}}\left(\text{g}\right)$ is $0\text{kJ}{\text{mol}}^{-1}$.

The value of standard Gibbs free energy of the formation $\text{CO}\left(\text{g}\right)$ is $-137.168\text{kJ}{\text{mol}}^{-1}$.

The value of standard Gibbs free energy of the formation of ${\text{CH}}_4\left(\text{g}\right)$ is $-50.72\text{kJ}{\text{mol}}^{-1}$.

The value of standard Gibbs free energy of the formation of ${\text{H}}_{\text{2}}\text{O}\left(\text{g}\right)$ is $-228.572\text{kJ}{\text{mol}}^{-1}$.

The given reaction is shown below.

  ${\text{CH}}_4\left(\text{g}\right)+{\text{H}}_{\text{2}}\text{O}\left(\text{g}\right)\rightleftharpoons \text{CO}\left(\text{g}\right)+3{\text{H}}_{\text{2}}\left(\text{g}\right)$

The formula to calculate the standard enthalpy change of the reaction is shown below.

    ${\Delta }_{\text{r}}G°={\Delta }_{\text{f}}G°\left(\text{CO}\left(\text{g}\right)\right)+3\times {\Delta }_{\text{f}}G°\left({\text{H}}_{\text{2}}\left(\text{g}\right)\right)-\left({\Delta }_{\text{f}}G°\left({\text{CH}}_4\left(\text{g}\right)\right)+{\Delta }_{\text{f}}G°\left({\text{H}}_{\text{2}}\text{O}\left(\text{g}\right)\right)\right)$

Where,

• ${\Delta }_{\text{r}}G°$ is the Gibbs free energy change of reaction.
• ${\Delta }_{\text{f}}G°\left({\text{H}}_{\text{2}}\left(\text{g}\right)\right)$ is the standard Gibbs free energy of formation of ${\text{H}}_{\text{2}}\left(\text{g}\right)$.
• ${\Delta }_{\text{f}}G°\left(\text{CO}\left(\text{g}\right)\right)$ is the standard Gibbs free energy of formation of $\text{CO}\left(\text{g}\right)$.
• ${\Delta }_{\text{f}}G°\left({\text{CH}}_4\left(\text{g}\right)\right)$ is the standard Gibbs free energy of formation of ${\text{CH}}_4\left(\text{g}\right)$.
• ${\Delta }_{\text{f}}G°\left({\text{H}}_{\text{2}}\text{O}\left(\text{g}\right)\right)$ is the standard Gibbs free energy of formation of ${\text{H}}_{\text{2}}\text{O}\left(\text{g}\right)$.

Substitute the value of ${\Delta }_{\text{f}}G°\left({\text{H}}_{\text{2}}\left(\text{g}\right)\right)$, ${\Delta }_{\text{f}}G°\left({\text{CH}}_4\left(\text{g}\right)\right)$, ${\Delta }_{\text{f}}G°\left({\text{H}}_{\text{2}}\text{O}\left(\text{g}\right)\right)$ and ${\Delta }_{\text{f}}G°\left(\text{CO}\left(\text{g}\right)\right)$ in the above equation.

    $\begin{array}{c}{\Delta }_{\text{r}}G°=\left(-137.168\text{kJ}{\text{mol}}^{-1}\right)+3\times \left(0\text{kJ}{\text{mol}}^{-1}\right)-\left(\begin{array}{l}-50.72\text{kJ}{\text{mol}}^{-1}\\ +\left(-228.572\text{kJ}{\text{mol}}^{-1}\right)\end{array}\right)\\ =\left(-137.168\text{kJ}{\text{mol}}^{-1}\right)+279.292\text{kJ}{\text{mol}}^{-1}\\ =\underset{\_}{142.124kJmo{l}^{-1}}\end{array}$

Therefore, the value of ${\Delta }_{\text{r}}G°$ for the reaction at $25°\text{C}$ is $\underset{\_}{142.124kJmo{l}^{-1}}$.

(b)

Interpretation Introduction

Interpretation:

The value of equilibrium costant $K_{\text{P}}$ for the reaction at $25°\text{C}$, ${\text{CH}}_4\left(\text{g}\right)+{\text{H}}_{\text{2}}\text{O}\left(\text{g}\right)\rightleftharpoons \text{CO}\left(\text{g}\right)+3{\text{H}}_{\text{2}}\left(\text{g}\right)$, has to be calculated.

Concept Introduction:

Refer to part (a).

Answer to Problem 119QRT

The value of equilibrium costant $K_{\text{P}}$ for the reaction at $25°\text{C}$, ${\text{CH}}_4\left(\text{g}\right)+{\text{H}}_{\text{2}}\text{O}\left(\text{g}\right)\rightleftharpoons \text{CO}\left(\text{g}\right)+3{\text{H}}_{\text{2}}\left(\text{g}\right)$ is $\underset{\_}{1.25\times 10^{-25}}$.

Explanation of Solution

The given reaction is shown below.

  ${\text{CH}}_4\left(\text{g}\right)+{\text{H}}_{\text{2}}\text{O}\left(\text{g}\right)\rightleftharpoons \text{CO}\left(\text{g}\right)+3{\text{H}}_{\text{2}}\left(\text{g}\right)$

The value of ${\Delta }_{\text{r}}G°$ for the reaction at $25°\text{C}$ is $142.124\text{kJ}{\text{mol}}^{-1}$.

The conversion of ${\Delta }_{\text{r}}G°$ from $\text{kJ}{\text{mol}}^{-1}$ to $\text{J}{\text{mol}}^{-1}$ is shown below.

    $\begin{array}{c}{\Delta }_{\text{r}}G°=\left(142.124\text{kJ}{\text{mol}}^{-1}\right)\left(\frac{1000\text{J}{\text{mol}}^{-1}}{1\text{kJ}{\text{mol}}^{-1}}\right)\\ =142.124\times {10}^3\text{J}{\text{mol}}^{-1}\end{array}$

The temperature of the system is $25°\text{C}$.

The conversion of temperature from $°\text{C}$ to $\text{K}$ is shown below.

    $\begin{array}{c}T=\left(25°\text{C}+273.15\right)\text{K}\\ =298.15\text{K}\end{array}$

The relation between the standard Gibbs free energy change of the reaction and standard equilibrium constant of the reaction is given by the expression as shown below.

  ${\Delta }_{\text{r}}G°=-RT\ln K°$

Where,

• ${\Delta }_{\text{r}}G°$ is the standard Gibbs free energy change of the reaction.
• $K°$ is the standard equilibrium constant of the reaction.
• $R$ is the gas constant $\left(8.314\text{J}{\text{K}}^{-1}{\text{mol}}^{-1}\right)$.
• $T$ is the temperature.

Rearrange the equation for the value of $K°$.

    $K°={\text{e}}^{\frac{-{\Delta }_{\text{r}}G°}{RT}}$

Substitute the value of ${\Delta }_{\text{r}}G°$, $R$ and $T$ in the above equation.

    $\begin{array}{c}K°={\text{e}}^{\frac{-\left(142.124\times {10}^3\text{J}{\text{mol}}^{-1}\right)}{\left(8.314\text{J}{\text{K}}^{-1}{\text{mol}}^{-1}\right)\left(298.15\text{K}\right)}}\\ ={\text{e}}^{-\frac{\left(40.684\times {10}^3\right)}{\left(2478.82\right)}}\\ ={\text{e}}^{-57.34}\\ =\underset{\_}{1.25\times 10^{-25}}\end{array}$

For the complete gaseous system. The value of $K°$ corresponds to $K_{\text{P}}$.

Therefore, the value of equilibrium costant $K_{\text{P}}$ for the given reaction is $\underset{\_}{1.25\times 10^{-25}}$.

(c)

Interpretation Introduction

Interpretation:

Whether the reaction at $25°\text{C}$, ${\text{CH}}_4\left(\text{g}\right)+{\text{H}}_{\text{2}}\text{O}\left(\text{g}\right)\rightleftharpoons \text{CO}\left(\text{g}\right)+3{\text{H}}_{\text{2}}\left(\text{g}\right)$, is product favored or not has to be stated. The temperature at which the reaction would changes from reactant-favored to product-favored has to be stated.

Concept Introduction:

Refer to part (a).

Answer to Problem 119QRT

The given reaction is reactant favored. the temperature at which the reaction would changes from reactant-favored to product-favored is $\underset{\_}{960.23K}$.

Explanation of Solution

The given reaction is shown below.

  ${\text{CH}}_4\left(\text{g}\right)+{\text{H}}_{\text{2}}\text{O}\left(\text{g}\right)\rightleftharpoons \text{CO}\left(\text{g}\right)+3{\text{H}}_{\text{2}}\left(\text{g}\right)$

The value of ${\Delta }_{\text{r}}G°$ for the reaction at $25°\text{C}$ is $142.124\text{kJ}{\text{mol}}^{-1}$.

The value of ${\Delta }_{\text{r}}G°$ of the reaction reaction is positive. The sign of positive sign of ${\Delta }_{\text{r}}G°$ of the reaction indicates that the reaction reactant favored. Therefore, the given reaction is reactant favored.

The value of standard entropy of ${\text{H}}_{\text{2}}\left(\text{g}\right)$ is $130.684\text{J}{\text{K}}^{-1}{\text{mol}}^{-1}$.

The value of standard entropy of $\text{CO}\left(\text{g}\right)$ is $197.674\text{J}{\text{K}}^{-1}{\text{mol}}^{-1}$.

The value of standard entropy of ${\text{CH}}_4\left(\text{g}\right)$ is $186.264\text{J}{\text{K}}^{-1}{\text{mol}}^{-1}$.

The value of standard entropy of ${\text{H}}_{\text{2}}\text{O}\left(\text{g}\right)$ is $188.825\text{J}{\text{K}}^{-1}{\text{mol}}^{-1}$.

The formula to calculate the standard enthalpy change of the reaction is shown below.

    ${\Delta }_{\text{r}}S°=S°\left(\text{CO}\left(\text{g}\right)\right)+3\times S°\left({\text{H}}_{\text{2}}\left(\text{g}\right)\right)-\left(S°\left({\text{CH}}_4\left(\text{g}\right)\right)+S°\left({\text{H}}_{\text{2}}\text{O}\left(\text{g}\right)\right)\right)$

Where,

• ${\Delta }_{\text{r}}S°$ is the entropy change of reaction.
• $S°\left({\text{H}}_{\text{2}}\left(\text{g}\right)\right)$ is the standard entropy of ${\text{H}}_{\text{2}}\left(\text{g}\right)$.
• $S°\left(\text{CO}\left(\text{g}\right)\right)$ is the standard entropy of $\text{CO}\left(\text{g}\right)$.
• $S°\left({\text{CH}}_4\left(\text{g}\right)\right)$ is the standard entropy of ${\text{CH}}_4\left(\text{g}\right)$.
• $S°\left({\text{H}}_{\text{2}}\text{O}\left(\text{g}\right)\right)$ is the standard entropy of ${\text{H}}_{\text{2}}\text{O}\left(\text{g}\right)$.

Substitute the value of $S°\left({\text{H}}_{\text{2}}\left(\text{g}\right)\right)$, $S°\left(\text{CO}\left(\text{g}\right)\right)$, $S°\left({\text{CH}}_4\left(\text{g}\right)\right)$ and $S°\left({\text{H}}_{\text{2}}\text{O}\left(\text{g}\right)\right)$ in the above equation.

    $\begin{array}{c}{\Delta }_{\text{r}}S°=197.674\text{J}{\text{K}}^{-1}{\text{mol}}^{-1}+3\times \left(130.684\text{J}{\text{K}}^{-1}{\text{mol}}^{-1}\right)-\left(\begin{array}{l}186.264\text{J}{\text{K}}^{-1}{\text{mol}}^{-1}\\ +188.825\text{J}{\text{K}}^{-1}{\text{mol}}^{-1}\end{array}\right)\\ =197.674\text{J}{\text{K}}^{-1}{\text{mol}}^{-1}+\left(392.052\text{J}{\text{K}}^{-1}{\text{mol}}^{-1}\right)-\left(375.089\text{J}{\text{K}}^{-1}{\text{mol}}^{-1}\right)\\ =214.637\text{J}{\text{K}}^{-1}{\text{mol}}^{-1}\end{array}$

Therefore, the entropy change of the reaction is $214.637\text{J}{\text{K}}^{-1}{\text{mol}}^{-1}$.

The value of standard enthalpy of the formation ${\text{H}}_{\text{2}}\left(\text{g}\right)$ is $0\text{kJ}{\text{mol}}^{-1}$.

The value of standard enthalpy of the formation $\text{CO}\left(\text{g}\right)$ is $-110.525\text{kJ}{\text{mol}}^{-1}$.

The value of standard enthalpy of the formation of ${\text{CH}}_4\left(\text{g}\right)$ is $-74.81\text{kJ}{\text{mol}}^{-1}$.

The value of standard enthalpy of the formation of ${\text{H}}_{\text{2}}\text{O}\left(\text{g}\right)$ is $-241.818\text{kJ}{\text{mol}}^{-1}$.

The given reaction is shown below.

  ${\text{CH}}_4\left(\text{g}\right)+{\text{H}}_{\text{2}}\text{O}\left(\text{g}\right)\rightleftharpoons \text{CO}\left(\text{g}\right)+3{\text{H}}_{\text{2}}\left(\text{g}\right)$

The formula to calculate the standard enthalpy change of the reaction is shown below.

    ${\Delta }_{\text{r}}H°={\Delta }_{\text{f}}H°\left(\text{CO}\left(\text{g}\right)\right)+3\times {\Delta }_{\text{f}}H°\left({\text{H}}_{\text{2}}\left(\text{g}\right)\right)-\left({\Delta }_{\text{f}}H°\left({\text{CH}}_4\left(\text{g}\right)\right)+{\Delta }_{\text{f}}H°\left({\text{H}}_{\text{2}}\text{O}\left(\text{g}\right)\right)\right)$

Where,

• ${\Delta }_{\text{r}}H°$ is the enthalpy change of reaction.
• ${\Delta }_{\text{f}}H°\left({\text{H}}_{\text{2}}\left(\text{g}\right)\right)$ is the standard enthalpy of formation of ${\text{H}}_{\text{2}}\left(\text{g}\right)$.
• ${\Delta }_{\text{f}}H°\left(\text{CO}\left(\text{g}\right)\right)$ is the standard enthalpy of formation of $\text{CO}\left(\text{g}\right)$.
• ${\Delta }_{\text{f}}H°\left({\text{CH}}_4\left(\text{g}\right)\right)$ is the standard enthalpy of formation of ${\text{CH}}_4\left(\text{g}\right)$.
• ${\Delta }_{\text{f}}H°\left({\text{H}}_{\text{2}}\text{O}\left(\text{g}\right)\right)$ is the standard enthalpy of formation of ${\text{H}}_{\text{2}}\text{O}\left(\text{g}\right)$.

Substitute the value of ${\Delta }_{\text{f}}H°\left({\text{H}}_{\text{2}}\left(\text{g}\right)\right)$, ${\Delta }_{\text{f}}H°\left(\text{CO}\left(\text{g}\right)\right)$, ${\Delta }_{\text{f}}H°\left({\text{CH}}_4\left(\text{g}\right)\right)$ and ${\Delta }_{\text{f}}H°\left({\text{H}}_{\text{2}}\text{O}\left(\text{g}\right)\right)$ in the above equation.

    $\begin{array}{c}{\Delta }_{\text{r}}H°=-110.525\text{kJ}{\text{mol}}^{-1}+3\times \left(0\text{kJ}{\text{mol}}^{-1}\right)-\left(\begin{array}{l}-74.81\text{kJ}{\text{mol}}^{-1}\\ +\left(-241.818\text{kJ}{\text{mol}}^{-1}\right)\end{array}\right)\\ =-110.525\text{kJ}{\text{mol}}^{-1}-\left(-316.628\text{kJ}{\text{mol}}^{-1}\right)\\ =206.103\text{kJ}{\text{mol}}^{-1}\end{array}$

The conversion of ${\Delta }_{\text{r}}H°$ from $\text{kJ}{\text{mol}}^{-1}$ to $\text{J}{\text{mol}}^{-1}$ is shown below.

    $\begin{array}{c}{\Delta }_{\text{r}}H°=\left(206.103\text{kJ}{\text{mol}}^{-1}\right)\left(\frac{1000\text{J}{\text{mol}}^{-1}}{1\text{kJ}{\text{mol}}^{-1}}\right)\\ =206.103\times {10}^3\text{J}{\text{mol}}^{-1}\end{array}$

The temperature at which the reactant favored reaction becomes product favored reaction is calculated by the formula as shown below.

    $T=\frac{{\Delta }_{\text{r}}H°}{{\Delta }_{\text{r}}S°}$

Where,

• ${\Delta }_{\text{r}}H°$ is the enthalpy of change of reaction.
• ${\Delta }_{\text{r}}S°$ is the entropy change of reaction.
• $T$ is the temperature.

Substitute the value of ${\Delta }_{\text{r}}H°$ and ${\Delta }_{\text{r}}S°$ in the equation.

    $\begin{array}{c}T=\frac{206.103\times {10}^3\text{J}{\text{mol}}^{-1}}{214.637\text{J}{\text{K}}^{-1}{\text{mol}}^{-1}}\\ =\underset{\_}{960.23K}\end{array}$

Therefore, the temperature at which the reaction would changes from reactant-favored to product-favored is $\underset{\_}{960.23K}$.

(d)

Interpretation Introduction

Interpretation:

The value of equilibrium costant $K_{\text{c}}$ for the reaction at $1000\text{K}$, ${\text{CH}}_4\left(\text{g}\right)+{\text{H}}_{\text{2}}\text{O}\left(\text{g}\right)\rightleftharpoons \text{CO}\left(\text{g}\right)+3{\text{H}}_{\text{2}}\left(\text{g}\right)$, has to be calculated.

Concept Introduction:

Refer to part (a).

Answer to Problem 119QRT

The value of equilibrium costant $K_{\text{c}}$ for the reaction at $25°\text{C}$, ${\text{CH}}_4\left(\text{g}\right)+{\text{H}}_{\text{2}}\text{O}\left(\text{g}\right)\rightleftharpoons \text{CO}\left(\text{g}\right)+3{\text{H}}_{\text{2}}\left(\text{g}\right)$, at $1000\text{K}$ is $\underset{\_}{4.139\times 10^{-4}mo{l}^2/L^{2}at{m}^2}$.

Explanation of Solution

The given reaction is shown  below.

  ${\text{CH}}_4\left(\text{g}\right)+{\text{H}}_{\text{2}}\text{O}\left(\text{g}\right)\rightleftharpoons \text{CO}\left(\text{g}\right)+3{\text{H}}_{\text{2}}\left(\text{g}\right)$

The entropy change of the reaction is $214.637\text{J}{\text{K}}^{-1}{\text{mol}}^{-1}$.

The enthalpy of change of reaction is $206.103\times {10}^3\text{J}{\text{mol}}^{-1}$.

The given temperature is $1000\text{K}$.

The relation between ${\Delta }_{\text{r}}H°$ and ${\Delta }_{\text{r}}\text{S}°$ is given by the expression as shown below.

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

Where,

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

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

    $\begin{array}{c}{\Delta }_{\text{r}}G°=206.103\times {10}^3\text{J}{\text{mol}}^{-1}-\left(1000\text{K}\right)\left(214.637\text{J}{\text{K}}^{-1}{\text{mol}}^{-1}\right)\\ =206.103\times {10}^3\text{J}{\text{mol}}^{-1}-214.637\times {10}^3\text{J}{\text{mol}}^{-1}\\ =-8.534\times {10}^3\text{J}{\text{mol}}^{-1}\end{array}$

The relation between the standard Gibbs free energy change of the reaction and standard equilibrium constant of the reaction is given by the expression as shown below.

  ${\Delta }_{\text{r}}G°=-RT\ln K°$

Where,

• ${\Delta }_{\text{r}}G°$ is the standard Gibbs free energy change of the reaction.
• $K°$ is the standard equilibrium constant of the reaction.
• $R$ is the gas constant $\left(8.314\text{J}{\text{K}}^{-1}{\text{mol}}^{-1}\right)$.
• $T$ is the temperature.

Rearrange the equation for the value of $K°$.

    $K°={\text{e}}^{\frac{-{\Delta }_{\text{r}}G°}{RT}}$

Substitute the value of ${\Delta }_{\text{r}}G°$, $R$ and $T$ in the above equation.

    $\begin{array}{c}K°={\text{e}}^{\frac{-\left(-8.534\times {10}^3\text{J}{\text{mol}}^{-1}\right)}{\left(8.314\text{J}{\text{K}}^{-1}{\text{mol}}^{-1}\right)\left(1000\text{K}\right)}}\\ ={\text{e}}^{-\frac{\left(-8.534\times {10}^3\right)}{\left(8314\right)}}\\ ={\text{e}}^{1.026}\\ =2.79\end{array}$

For the complete gaseous system. The value of $K°$ corresponds to $K_{\text{P}}$.

Therefore, the value of equilibrium costant $K_{\text{P}}$ for the given reaction is $2.79$.

The number of moles of the gaseous molecule on the product side is $4$.

The number of moles of the gaseous molecule on the reactant side is $2$.

The relation between equilibrium constants $K_{\text{c}}$ and $K_{\text{P}}$ is shown below.

  $K_{\text{c}}=\frac{K_{\text{P}}}{{\left(RT\right)}^{\left(n_{\text{p}}-n_{\text{r}}\right)}}$

Where,

• $R$ is the gas constant $\left(0.0821\text{L}\text{atm}/\text{mol}\text{K}\right)$.
• $T$ is the temperature.
• $n_{\text{p}}$ is the number of moles of the gaseous molecule on the product side.
• $n_{\text{r}}$ is the number of moles of the gaseous molecule on the reactant side.
• $K_{\text{P}}$ is the equilibrium constant in terms of partial pressure.
• $K_{\text{c}}$ is the equilibrium constant in terms of concentration.

Substitute the value of $R$, $T$, $n_{\text{p}}$, and $n_{\text{r}}$ in the above equation.

  $\begin{array}{c}{K}_{\text{c}}=\frac{2.79}{{\left(\left(0.0821\text{L}\text{atm}/\text{mol}\text{K}\right)\left(1000\text{K}\right)\right)}^{\left(4-2\right)}}\\ =\frac{2.79}{{\left(\left(82.1\text{L}\text{atm}/\text{mol}\right)\right)}^{\left(2\right)}}\\ =\frac{2.79}{6740.41{\text{L}}^2{\text{atm}}^2/{\text{mol}}^2}\\ =\underset{\_}{4.139\times 10^{-4}mo{l}^2/L^{2}at{m}^2}\end{array}$

The equilibrium constants are unitless quantities.

Therefore, the value of $K_{\text{c}}$ is $\underset{\_}{4.139\times 10^{-4}mo{l}^2/L^{2}at{m}^2}$.