Chapter 16, Problem 49QRT

(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 49QRT

The value of ${\Delta }_{\text{r}}{S}_{}^{\text{ο}}$ is $\underset{\_}{-37.47J{K}^{-1}mo{l}^{-1}}$ and the value of ${\Delta }_{\text{r}}{H}_{}^{\text{ο}}$ is $\underset{\_}{-851.5kJmo{l}^{-1}}$.

Explanation of Solution

The given reaction is shown below.

    ${\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}\left(\text{s}\right)+2\text{Al}\left(\text{g}\right)\to \text{2Fe}\left(\text{s}\right)+{\text{Al}}_{\text{2}}{\text{O}}_{\text{3}}\left(\text{s}\right)$

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.

  ${\Delta }_{\text{r}}{H}_{}^{\text{ο}}=\left[2\cdot H_{\text{f}}^{\text{ο}}\left(\text{Fe}\left(\text{s}\right)\right)+1\cdot H_{\text{f}}^{\text{ο}}\left({\text{Al}}_{\text{2}}{\text{O}}_{\text{3}}\left(\text{s}\right)\right)\right]-\left[1\cdot H_{\text{f}}^{\text{ο}}\left({\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}\left(\text{s}\right)\right)+2\cdot H_{\text{f}}^{\text{ο}}\left(\text{Al}\left(\text{g}\right)\right)\right]$

The value of $H_{\text{f}}^{\text{ο}}\left({\text{Al}}_{\text{2}}{\text{O}}_{\text{3}}\left(\text{s}\right)\right)$ is $-1675.7{\text{ kJ mol}}^{-1}$.

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

The value of $H_{\text{f}}^{\text{ο}}\left({\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}\left(\text{s}\right)\right)$ is $-824.2{\text{ kJ mol}}^{-1}$.

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

Substitute the values in the above expression.

  $\begin{array}{c}{\Delta }_{\text{r}}{H}_{}^{\text{ο}}=\left[2\cdot H_{\text{f}}^{\text{ο}}\left(\text{Fe}\left(\text{s}\right)\right)+1\cdot H_{\text{f}}^{\text{ο}}\left({\text{Al}}_{\text{2}}{\text{O}}_{\text{3}}\left(\text{s}\right)\right)\right]-\left[1\cdot H_{\text{f}}^{\text{ο}}\left({\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}\left(\text{s}\right)\right)+2\cdot H_{\text{f}}^{\text{ο}}\left(\text{Al}\left(\text{g}\right)\right)\right]\\ =\left[2\times \left(0{\text{ kJ mol}}^{-1}\right)+1\times \left(-1675.7{\text{ kJ mol}}^{-1}\right)\right]-\left[1\times \left(-824.2{\text{ kJ mol}}^{-1}\right)+2\times \left(0{\text{ kJ mol}}^{-1}\right)\right]\\ =-1675.7{\text{ kJ mol}}^{-1}+824.2{\text{ kJ mol}}^{-1}\\ =\underset{\_}{-851.5kJmo{l}^{-1}}\end{array}$

The value of ${\Delta }_{\text{r}}{H}_{}^{\text{ο}}$ is $\underset{\_}{-851.5kJmo{l}^{-1}}$.

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.

  ${\Delta }_{\text{r}}{S}_{}^{\text{ο}}=\left[2\cdot S_{\text{f}}^{\text{ο}}\left(\text{Fe}\left(\text{s}\right)\right)+1\cdot S_{\text{f}}^{\text{ο}}\left({\text{Al}}_{\text{2}}{\text{O}}_{\text{3}}\left(\text{s}\right)\right)\right]-\left[1\cdot S_{\text{f}}^{\text{ο}}\left({\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}\left(\text{s}\right)\right)+2\cdot S_{\text{f}}^{\text{ο}}\left(\text{Al}\left(\text{g}\right)\right)\right]$

The value of $S_{\text{f}}^{\text{ο}}\left({\text{Al}}_{\text{2}}{\text{O}}_{\text{3}}\left(\text{s}\right)\right)$ is $50.92{\text{ J K}}^{-1}{\text{ mol}}^{-1}$

The value of $S_{\text{f}}^{\text{ο}}\left(\text{Fe}\left(\text{s}\right)\right)$ is $27.78{\text{ J K}}^{-1}{\text{ mol}}^{-1}$.

The value of $S_{\text{f}}^{\text{ο}}\left({\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}\left(\text{s}\right)\right)$ is $87.4{\text{ J K}}^{-1}{\text{ mol}}^{-1}$.

The value of $S_{\text{f}}^{\text{ο}}\left(\text{Al}\left(\text{g}\right)\right)$ is $28.275{\text{ kJ mol}}^{-1}$.

Substitute the values in the above expression.

  $\begin{array}{c}{\Delta }_{\text{r}}{S}_{}^{\text{ο}}=\left[2\cdot S_{\text{f}}^{\text{ο}}\left(\text{Fe}\left(\text{s}\right)\right)+1\cdot S_{\text{f}}^{\text{ο}}\left({\text{Al}}_{\text{2}}{\text{O}}_{\text{3}}\left(\text{s}\right)\right)\right]-\left[1\cdot S_{\text{f}}^{\text{ο}}\left({\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}\left(\text{s}\right)\right)+2\cdot S_{\text{f}}^{\text{ο}}\left(\text{Al}\left(\text{g}\right)\right)\right]\\ =\left[2\times \left(27.78{\text{ J K}}^{-1}{\text{ mol}}^{-1}\right)+1\times \left(50.92{\text{ J K}}^{-1}{\text{ mol}}^{-1}\right)\right]-\left[1\times \left(84.7{\text{ J K}}^{-1}{\text{ mol}}^{-1}\right)+2\times \left(28.275{\text{ kJ mol}}^{-1}\right)\right]\\ =106.48{\text{ J K}}^{-1}{\text{ mol}}^{-1}-143.95{\text{ J K}}^{-1}{\text{ mol}}^{-1}\\ =\underset{\_}{-37.47J{K}^{-1}mo{l}^{-1}}\end{array}$

The value of ${\Delta }_{\text{r}}{S}_{}^{\text{ο}}$ is $\underset{\_}{-37.47J{K}^{-1}mo{l}^{-1}}$.

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

(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 49QRT

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.

    ${\text{N}}_2\left(\text{g}\right)+2{\text{O}}_2\left(\text{g}\right)\to {\text{2NO}}_{\text{2}}\left(\text{g}\right)$

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.

  ${\Delta }_{\text{r}}{H}_{}^{\text{ο}}=\left[2\cdot H_{\text{f}}^{\text{ο}}\left({\text{NO}}_{\text{2}}\left(\text{g}\right)\right)\right]-\left[1\cdot H_{\text{f}}^{\text{ο}}\left({\text{N}}_2\left(\text{g}\right)\right)+2\cdot H_{\text{f}}^{\text{ο}}\left({\text{O}}_2\left(\text{g}\right)\right)\right]$

The value of $H_{\text{f}}^{\text{ο}}\left({\text{NO}}_{\text{2}}\left(\text{g}\right)\right)$ is $83.18{\text{ kJ mol}}^{-1}$.

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

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.

  $\begin{array}{c}{\Delta }_{\text{r}}{H}_{}^{\text{ο}}=\left[2\cdot H_{\text{f}}^{\text{ο}}\left({\text{NO}}_{\text{2}}\left(\text{g}\right)\right)\right]-\left[1\cdot H_{\text{f}}^{\text{ο}}\left({\text{N}}_2\left(\text{g}\right)\right)+2\cdot H_{\text{f}}^{\text{ο}}\left({\text{O}}_2\left(\text{g}\right)\right)\right]\\ =\left[2\times \left(83.18{\text{ kJ mol}}^{-1}\right)\right]-\left[1\times \left(0{\text{ kJ mol}}^{-1}\right)+2\times \left(0{\text{ kJ mol}}^{-1}\right)\right]\\ =\underset{\_}{166.36kJmo{l}^{-1}}\end{array}$

The value of ${\Delta }_{\text{r}}{H}_{}^{\text{ο}}$ is $\underset{\_}{166.36kJmo{l}^{-1}}$.

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.

  ${\Delta }_{\text{r}}{S}_{}^{\text{ο}}=\left[2\cdot S_{\text{f}}^{\text{ο}}\left({\text{NO}}_{\text{2}}\left(\text{g}\right)\right)\right]-\left[1\cdot S_{\text{f}}^{\text{ο}}\left({\text{N}}_2\left(\text{g}\right)\right)+2\cdot S_{\text{f}}^{\text{ο}}\left({\text{O}}_2\left(\text{g}\right)\right)\right]$

The value of $S_{\text{f}}^{\text{ο}}\left({\text{NO}}_{\text{2}}\left(\text{g}\right)\right)$ is $240.06{\text{ J k}}^{-1}{\text{ mol}}^{-1}$.

The value of $S_{\text{f}}^{\text{ο}}\left({\text{N}}_2\left(\text{g}\right)\right)$ is $191.16{\text{ J k}}^{-1}{\text{ mol}}^{-1}$.

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.

  $\begin{array}{c}{\Delta }_{\text{r}}{S}_{}^{\text{ο}}=\left[2\cdot S_{\text{f}}^{\text{ο}}\left({\text{NO}}_{\text{2}}\left(\text{g}\right)\right)\right]-\left[1\cdot S_{\text{f}}^{\text{ο}}\left({\text{N}}_2\left(\text{g}\right)\right)+2\cdot S_{\text{f}}^{\text{ο}}\left({\text{O}}_2\left(\text{g}\right)\right)\right]\\ =\left[2\times \left(240.06{\text{ J k}}^{-1}{\text{ mol}}^{-1}\right)\right]-\left[1\times \left(191.16{\text{ J k}}^{-1}{\text{ mol}}^{-1}\right)+2\times \left(205.138{\text{ J k}}^{-1}{\text{ mol}}^{-1}\right)\right]\\ =480.12{\text{ J K}}^{-1}{\text{ mol}}^{-1}-601.436{\text{ J K}}^{-1}{\text{ mol}}^{-1}\\ =\underset{\_}{-121.316J{K}^{-1}mo{l}^{-1}}\end{array}$

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

The value of entropy is negative; therefore, the term $T{\Delta }_{\text{r}}{S}_{}^{\text{ο}}$ will be positive.  Enthalpy is positive.  Therefore, the Gibbs free energy will be always positive.  Hence, the reaction will never be product-favored.