Chapter 16, Problem 50QRT

(a)

Interpretation Introduction

Interpretation:

The values of ${\Delta }_{\text{r}}{H}_{}^{\text{ο}}$ and ${\Delta }_{\text{r}}{S}_{}^{\text{ο}}$ for the given reaction are to be calculated and the prediction about the nature of the reaction on the basis of temperature is to be stated.

Concept Introduction:

The term entropy is used to represent the randomness in a system.  When a system moves from an ordered arrangement to a less order arrangement, then the entropy of the system increases.

Answer to Problem 50QRT

The value of ${\Delta }_{\text{r}}{S}_{}^{\text{ο}}$ is $\underset{\_}{235.172J{K}^{-1}mo{l}^{-1}}$ and the value of ${\Delta }_{\text{r}}{H}_{}^{\text{ο}}$ is $\underset{\_}{-2801.58kJmo{l}^{-1}}$.

Explanation of Solution

The given reaction is shown below.

    ${\text{C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_{\text{6}}\left(\text{s}\right)+6{\text{O}}_2\left(\text{g}\right)\to 6{\text{CO}}_2\left(\text{g}\right)+{\text{6H}}_2\text{O}\left(\text{l}\right)$

The standard enthalpy change for the reaction (${\Delta }_{\text{r}}{H}_{}^{\text{ο}}$) is calculated by using the expression shown below.

  ${\Delta }_{\text{r}}{H}_{}^{\text{ο}}={\displaystyle \sum m\cdot H_{}^{\text{ο}}\left(\text{products}\right)}-{\displaystyle \sum n\cdot H_{}^{\text{ο}}\left(\text{reactants}\right)}$

Where,

• $H_{}^{\text{ο}}\left(\text{products}\right)$ is the standard heat of formation of products.
• $H_{}^{\text{ο}}\left(\text{reactants}\right)$ is the standard heat of formation of reactants.
• $m$ is the total moles of products.
• $n$ is the total moles of reactants.

For the given reaction, the standard enthalpy change is calculated by the expression shown below.

  ${\Delta }_{\text{r}}{H}_{}^{\text{ο}}=\left[6\cdot H_{\text{f}}^{\text{ο}}\left({\text{CO}}_2\left(\text{g}\right)\right)+6\cdot H_{\text{f}}^{\text{ο}}\left({\text{H}}_2\text{O}\left(\text{l}\right)\right)\right]-\left[1\cdot H_{\text{f}}^{\text{ο}}\left({\text{C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_{\text{6}}\left(\text{s}\right)\right)+6\cdot H_{\text{f}}^{\text{ο}}\left({\text{O}}_2\left(\text{g}\right)\right)\right]$

The value of $H_{\text{f}}^{\text{ο}}\left({\text{CO}}_2\left(\text{g}\right)\right)$ is $-393.509{\text{ kJ mol}}^{-1}$.

The value of $H_{\text{f}}^{\text{ο}}\left({\text{H}}_2\text{O}\left(\text{l}\right)\right)$ is $-285.83{\text{ kJ mol}}^{-1}$.

The value of $H_{\text{f}}^{\text{ο}}\left({\text{C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_{\text{6}}\left(\text{s}\right)\right)$ is $-1274.4{\text{ kJ mol}}^{-1}$.

The value of $H_{\text{f}}^{\text{ο}}\left({\text{O}}_2\left(\text{g}\right)\right)$ is $0{\text{ kJ mol}}^{-1}$.

Substitute the values in the above expression.

  $\begin{array}{c}{\Delta }_{\text{r}}{H}_{}^{\text{ο}}=\left[6\cdot H_{\text{f}}^{\text{ο}}\left({\text{CO}}_2\left(\text{g}\right)\right)+6\cdot H_{\text{f}}^{\text{ο}}\left({\text{H}}_2\text{O}\left(\text{l}\right)\right)\right]-\left[1\cdot H_{\text{f}}^{\text{ο}}\left({\text{C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_{\text{6}}\left(\text{s}\right)\right)+6\cdot H_{\text{f}}^{\text{ο}}\left({\text{O}}_2\left(\text{g}\right)\right)\right]\\ =\left[6\times \left(-393.509{\text{ kJ mol}}^{-1}\right)+6\times \left(-285.83{\text{ kJ mol}}^{-1}\right)\right]-\left[1\times \left(-1274.4{\text{ kJ mol}}^{-1}\right)+6\times \left(0{\text{ kJ mol}}^{-1}\right)\right]\\ =-4075.98{\text{ kJ mol}}^{-1}+1274.4{\text{ kJ mol}}^{-1}\\ =\underset{\_}{-2801.58kJmo{l}^{-1}}\end{array}$

The value of ${\Delta }_{\text{r}}{H}_{}^{\text{ο}}$ is $\underset{\_}{-2801.58kJmo{l}^{-1}}$.

The standard entropy change for the reaction (${\Delta }_{\text{r}}{S}_{}^{\text{ο}}$) is calculated by using the expression shown below.

  ${\Delta }_{\text{r}}{S}_{}^{\text{ο}}={\displaystyle \sum m\cdot S_{}^{\text{ο}}\left(\text{products}\right)}-{\displaystyle \sum n\cdot S_{}^{\text{ο}}\left(\text{reactants}\right)}$

Where,

• $S_{}^{\text{ο}}\left(\text{products}\right)$ is the absolute molar entropy of products.
• $S_{}^{\text{ο}}\left(\text{reactants}\right)$ is the absolute molar entropy of reactants.
• $m$ is the total moles of products.
• $n$ is the total moles of reactants.

For the given reaction, the standard entropy change is calculated by the expression shown below.

  ${\Delta }_{\text{r}}{S}_{}^{\text{ο}}=\left[6\cdot S_{\text{f}}^{\text{ο}}\left({\text{CO}}_2\left(\text{g}\right)\right)+6\cdot S_{\text{f}}^{\text{ο}}\left({\text{H}}_2\text{O}\left(\text{l}\right)\right)\right]-\left[1\cdot S_{\text{f}}^{\text{ο}}\left({\text{C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_{\text{6}}\left(\text{s}\right)\right)+6\cdot S_{\text{f}}^{\text{ο}}\left({\text{O}}_2\left(\text{g}\right)\right)\right]$

The value of $S_{\text{f}}^{\text{ο}}\left({\text{CO}}_2\left(\text{g}\right)\right)$ is $213.74{\text{ J k}}^{-1}{\text{ mol}}^{-1}$.

The value of $S_{\text{f}}^{\text{ο}}\left({\text{H}}_2\text{O}\left(\text{l}\right)\right)$ is $69.91{\text{ J k}}^{-1}{\text{ mol}}^{-1}$.

The value of $S_{\text{f}}^{\text{ο}}\left({\text{C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_{\text{6}}\left(\text{s}\right)\right)$ is $\text{235}{\text{.9 J k}}^{-1}{\text{ mol}}^{-1}$.

The value of $S_{\text{f}}^{\text{ο}}\left({\text{O}}_2\left(\text{g}\right)\right)$ is $205.138{\text{ J k}}^{-1}{\text{ mol}}^{-1}$.

Substitute the values in the above expression.

  $\begin{array}{c}{\Delta }_{\text{r}}{S}_{}^{\text{ο}}=\left[6\cdot S_{\text{f}}^{\text{ο}}\left({\text{CO}}_2\left(\text{g}\right)\right)+6\cdot S_{\text{f}}^{\text{ο}}\left({\text{H}}_2\text{O}\left(\text{l}\right)\right)\right]-\left[1\cdot S_{\text{f}}^{\text{ο}}\left({\text{C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_{\text{6}}\left(\text{s}\right)\right)+6\cdot S_{\text{f}}^{\text{ο}}\left({\text{O}}_2\left(\text{g}\right)\right)\right]\\ =\left[6\times \left(213.74{\text{ J k}}^{-1}{\text{ mol}}^{-1}\right)+6\times \left(69.91{\text{ J k}}^{-1}{\text{ mol}}^{-1}\right)\right]-\left[1\times \left(\text{235}{\text{.9 J k}}^{-1}{\text{ mol}}^{-1}\right)+6\times \left(205.138{\text{ J k}}^{-1}{\text{ mol}}^{-1}\right)\right]\\ =1701.9{\text{ J K}}^{-1}{\text{ mol}}^{-1}-1466.728{\text{ J K}}^{-1}{\text{ mol}}^{-1}\\ =\underset{\_}{235.172J{K}^{-1}mo{l}^{-1}}\end{array}$

The value of ${\Delta }_{\text{r}}{S}_{}^{\text{ο}}$ is $\underset{\_}{235.172J{K}^{-1}mo{l}^{-1}}$.

The value of entropy is positive; therefore, the term $T{\Delta }_{\text{r}}{S}_{}^{\text{ο}}$ will be negative.  Enthalpy is also negative.  Therefore, the Gibbs free energy will be negative.  Hence, the reaction will be product-favored.

(b)

Interpretation Introduction

Interpretation:

The values of ${\Delta }_{\text{r}}{H}_{}^{\text{ο}}$ and ${\Delta }_{\text{r}}{S}_{}^{\text{ο}}$ for the given reaction are to be calculated and the prediction about the nature of the reaction on the basis of temperature is to be stated.

Concept Introduction:

Refer to part (a)

Answer to Problem 50QRT

The value of ${\Delta }_{\text{r}}{S}_{}^{\text{ο}}$ is $\underset{\_}{-121.316J{K}^{-1}mo{l}^{-1}}$ and the value of ${\Delta }_{\text{r}}{H}_{}^{\text{ο}}$ is $\underset{\_}{166.36kJmo{l}^{-1}}$.

Explanation of Solution

The given reaction is shown below.

    $\text{MgO}\left(\text{s}\right)+\text{C}\left(\text{s,graphite}\right)\to \text{Mg}\left(\text{s}\right)+\text{CO}\left(\text{g}\right)$

The standard enthalpy change for the reaction (${\Delta }_{\text{r}}{H}_{}^{\text{ο}}$) is calculated by using the expression shown below.

  ${\Delta }_{\text{r}}{H}_{}^{\text{ο}}={\displaystyle \sum m\cdot H_{}^{\text{ο}}\left(\text{products}\right)}-{\displaystyle \sum n\cdot H_{}^{\text{ο}}\left(\text{reactants}\right)}$

Where,

• $H_{}^{\text{ο}}\left(\text{products}\right)$ is the standard heat of formation of products.
• $H_{}^{\text{ο}}\left(\text{reactants}\right)$ is the standard heat of formation of reactants.
• $m$ is the total moles of products.
• $n$ is the total moles of reactants.

For the given reaction, the standard enthalpy change is calculated by the expression shown below.

  ${\Delta }_{\text{r}}{H}_{}^{\text{ο}}=\left[1\cdot H_{\text{f}}^{\text{ο}}\left(\text{Mg}\left(\text{s}\right)\right)+1\cdot H_{\text{f}}^{\text{ο}}\left(\text{CO}\left(\text{g}\right)\right)\right]-\left[1\cdot H_{\text{f}}^{\text{ο}}\left(\text{MgO}\left(\text{s}\right)\right)+1\cdot H_{\text{f}}^{\text{ο}}\left(\text{C}\left(\text{s}\right)\right)\right]$

The value of $H_{\text{f}}^{\text{ο}}\left(\text{MgO}\left(\text{s}\right)\right)$ is $-601.7{\text{ kJ mol}}^{-1}$.

The value of $H_{\text{f}}^{\text{ο}}\left(\text{C}\left(\text{s}\right)\right)$ is $0{\text{ kJ mol}}^{-1}$.

The value of $H_{\text{f}}^{\text{ο}}\left(\text{Mg}\left(\text{s}\right)\right)$ is $0{\text{ kJ mol}}^{-1}$.

The value of $H_{\text{f}}^{\text{ο}}\left(\text{CO}\left(\text{g}\right)\right)$ is $-110.525{\text{ kJ mol}}^{-1}$.

Substitute the values in the above expression.

  $\begin{array}{c}{\Delta }_{\text{r}}{H}_{}^{\text{ο}}=\left[1\cdot H_{\text{f}}^{\text{ο}}\left(\text{Mg}\left(\text{s}\right)\right)+1\cdot H_{\text{f}}^{\text{ο}}\left(\text{CO}\left(\text{g}\right)\right)\right]-\left[1\cdot H_{\text{f}}^{\text{ο}}\left(\text{MgO}\left(\text{s}\right)\right)+1\cdot H_{\text{f}}^{\text{ο}}\left(\text{C}\left(\text{s}\right)\right)\right]\\ =\left[1\times \left(0{\text{ kJ mol}}^{-1}\right)+1\times \left(-110.525{\text{ kJ mol}}^{-1}\right)\right]-\left[1\times \left(-601.7{\text{ kJ mol}}^{-1}\right)+1\times \left(0{\text{ kJ mol}}^{-1}\right)\right]\\ =\underset{\_}{491.175kJmo{l}^{-1}}\end{array}$

The value of ${\Delta }_{\text{r}}{H}_{}^{\text{ο}}$ is $\underset{\_}{491.175kJmo{l}^{-1}}$.

The standard entropy change for the reaction (${\Delta }_{\text{r}}{S}_{}^{\text{ο}}$) is calculated by using the expression shown below.

  ${\Delta }_{\text{r}}{S}_{}^{\text{ο}}={\displaystyle \sum m\cdot S_{}^{\text{ο}}\left(\text{products}\right)}-{\displaystyle \sum n\cdot S_{}^{\text{ο}}\left(\text{reactants}\right)}$

Where,

• $S_{}^{\text{ο}}\left(\text{products}\right)$ is the absolute molar entropy of products.
• $S_{}^{\text{ο}}\left(\text{reactants}\right)$ is the absolute molar entropy of reactants.
• $m$ is the total moles of products.
• $n$ is the total moles of reactants.

For the given reaction, the standard entropy change is calculated by the expression shown below.

  ${\Delta }_{\text{r}}{S}_{}^{\text{ο}}=\left[1\cdot S_{\text{f}}^{\text{ο}}\left(\text{Mg}\left(\text{s}\right)\right)+1\cdot S_{\text{f}}^{\text{ο}}\left(\text{CO}\left(\text{g}\right)\right)\right]-\left[1\cdot S_{\text{f}}^{\text{ο}}\left(\text{MgO}\left(\text{s}\right)\right)+1\cdot S_{\text{f}}^{\text{ο}}\left(\text{C}\left(\text{s}\right)\right)\right]$

The value of $S_{\text{f}}^{\text{ο}}\left(\text{MgO}\left(\text{s}\right)\right)$ is $26.94{\text{ J K}}^{-1}{\text{ mol}}^{-1}$.

The value of $S_{\text{f}}^{\text{ο}}\left(\text{C}\left(\text{s}\right)\right)$ is $5.74{\text{ J K}}^{-1}{\text{ mol}}^{-1}$.

The value of $S_{\text{f}}^{\text{ο}}\left(\text{Mg}\left(\text{s}\right)\right)$ is $32.68{\text{ J K}}^{-1}{\text{ mol}}^{-1}$.

The value of $S_{\text{f}}^{\text{ο}}\left(\text{CO}\left(\text{g}\right)\right)$ is $\text{197}{\text{.674 J K}}^{-1}{\text{ mol}}^{-1}$.

Substitute the values in the above expression.

  $\begin{array}{c}{\Delta }_{\text{r}}{S}_{}^{\text{ο}}=\left[1\cdot S_{\text{f}}^{\text{ο}}\left(\text{Mg}\left(\text{s}\right)\right)+1\cdot S_{\text{f}}^{\text{ο}}\left(\text{CO}\left(\text{g}\right)\right)\right]-\left[1\cdot S_{\text{f}}^{\text{ο}}\left(\text{MgO}\left(\text{s}\right)\right)+1\cdot S_{\text{f}}^{\text{ο}}\left(\text{C}\left(\text{s}\right)\right)\right]\\ =\left[1\times \left(32.68{\text{ J K}}^{-1}{\text{ mol}}^{-1}\right)+1\times \left(\text{197}{\text{.674 J K}}^{-1}{\text{ mol}}^{-1}\right)\right]-\left[1\times \left(26.94{\text{ J K}}^{-1}{\text{ mol}}^{-1}\right)+1\times \left(5.74{\text{ J K}}^{-1}{\text{ mol}}^{-1}\right)\right]\\ =230.354{\text{ J K}}^{-1}{\text{ mol}}^{-1}-32.68{\text{ J K}}^{-1}{\text{ mol}}^{-1}\\ =\underset{\_}{197.674J{K}^{-1}mo{l}^{-1}}\end{array}$

The value of ${\Delta }_{\text{r}}{S}_{}^{\text{ο}}$ is $\underset{\_}{197.674J{K}^{-1}mo{l}^{-1}}$.

The value of entropy is positive; therefore, the term $T{\Delta }_{\text{r}}{S}_{}^{\text{ο}}$ will be negative.  Enthalpy is positive.  Therefore, the Gibbs free energy will negative at high temperatures.  Hence, the reaction will be product-favored at high-temperatures.