Chapter 3, Problem 3D.1BE

(i)

Interpretation Introduction

Interpretation: Under the given conditions, the standard reaction enthalpy and the standard Gibbs free energy for the reaction has to be calculated.

Concept introduction: The change in enthalpy during the formation of products from reactants at standard conditions is known as standard reaction enthalpy.  The standard reaction enthalpy is the subtraction of enthalpy of formation of the products and the enthalpy of formation of the reactants

Answer to Problem 3D.1BE

The standard enthalpy of the reaction is $\underset{\_}{-218.66kJmo{l}^{-1}}$ and the standard Gibbs free energy for the reaction is $\underset{\_}{-212.437kJmo{l}^{-1}}$.

Explanation of Solution

The given reaction is,

  $\text{Zn}\left(\text{s}\right)+{\text{Cu}}^{\text{2+}}\left(\text{aq}\right)\to {\text{Zn}}^{\text{2+}}\left(\text{aq}\right)+\text{Cu}\left(\text{s}\right)$                                                         (1)

The standard enthalpy $\left({\Delta }_{\text{r}}{H}^{\text{ο}}\right)$ for a chemical reaction is expressed as,

  ${\Delta }_{\text{r}}{H}^{\text{ο}}=\left({\Delta }_{\text{f}}{H}_{{\text{Zn}}^{\text{2+}}\left(\text{aq}\right)}^{\text{ο}}+{\Delta }_{\text{f}}{H}_{\text{Cu}\left(\text{s}\right)}^{\text{ο}}-{\Delta }_{\text{f}}{H}_{\text{Zn}\left(\text{s}\right)}^{\text{ο}}-{\Delta }_{\text{f}}{H}_{{\text{Cu}}^{\text{2+}}\left(\text{aq}\right)}^{\text{ο}}\right)$                                (2)

It is given that,

The standard entropy of the above chemical reaction $\left({\Delta }_{\text{r}}{S}^{\text{ο}}\right)$ is $-20.88\text{J}{\text{K}}^{-1}{\text{mol}}^{-1}$

The standard enthalpy of formation of $\text{Zn}\left(\text{s}\right)$ is $0\text{kJ}{\text{mol}}^{-1}$.

The standard enthalpy of formation of ${\text{Zn}}^{\text{2+}}\left(\text{aq}\right)$ is $-153.89\text{kJ}{\text{mol}}^{-1}$.

The standard enthalpy of formation of $\text{Cu}\left(\text{s}\right)$ is $0\text{kJ}{\text{mol}}^{-1}$.

The standard enthalpy of formation of ${\text{Cu}}^{\text{2+}}\left(\text{aq}\right)$ is $64.77\text{kJ}{\text{mol}}^{-1}$.

The conversion of $\text{J}{\text{K}}^{-1}{\text{mol}}^{-1}$ to $\text{kJ}{\text{K}}^{-1}{\text{mol}}^{-1}$ is done as,

  $\text{1J}{\text{K}}^{-1}{\text{mol}}^{-1}={10}^{-3}\text{}\text{kJ}{\text{K}}^{-1}{\text{mol}}^{-1}$

Therefore, the conversion of $-20.88\text{J}{\text{K}}^{-1}{\text{mol}}^{-1}$ to $\text{kJ}{\text{mol}}^{-1}$ is done as,

  $-20.88\text{J}{\text{K}}^{-1}{\text{mol}}^{-1}=-\text{0}\text{.0208}\text{kJ}{\text{K}}^{-1}{\text{mol}}^{-1}$

Substitute the values of ${\Delta }_{\text{f}}{H}_{{\text{Zn}}^{\text{2+}}\left(\text{aq}\right)}^{\text{ο}},{\Delta }_{\text{f}}{H}_{\text{Cu}\left(\text{s}\right)}^{\text{ο}},{\Delta }_{\text{f}}{H}_{\text{Zn}\left(\text{s}\right)}^{\text{ο}}$ and ${\Delta }_{\text{f}}{H}_{{\text{Cu}}^{\text{2+}}\left(\text{aq}\right)}^{\text{ο}}$ in equation (2).

  $\begin{array}{c}{\Delta }_{\text{r}}{H}^{\text{ο}}=\left(-153.89\text{kJ}{\text{mol}}^{-1}+0\text{kJ}{\text{mol}}^{-1}-0\text{kJ}{\text{mol}}^{-1}-64.77\text{kJ}{\text{mol}}^{-1}\right)\\ =\underset{\_}{-218.66kJmo{l}^{-1}}\end{array}$

Therefore, the standard enthalpy of the reaction is $\underset{\_}{-218.66kJmo{l}^{-1}}$.

The standard Gibbs free energy of the reaction is calculated as,

  ${\Delta }_{\text{r}}{G}^{\text{ο}}={\Delta }_{\text{r}}{H}^{\text{ο}}-T{\Delta }_{\text{r}}{S}^{\text{ο}}$                                                                                   (3)

Where,

• ${\Delta }_{\text{r}}{G}^{\text{ο}}$ is the standard Gibbs energy of reaction.
• ${\Delta }_{\text{r}}{H}^{\text{ο}}$ is the standard enthalpy of reaction.
• ${\Delta }_{\text{r}}{S}^{\text{ο}}$ is the standard entropy of reaction.
• $T$ is the standard temperature $\left(298\text{K}\right)$.

Substitute the values of ${\Delta }_{\text{r}}{H}^{\text{ο}}$, ${\Delta }_{\text{r}}{S}^{\text{ο}}$ and $T$ in equation (3).

  $\begin{array}{c}{\Delta }_{\text{r}}{G}^{\text{ο}}=-218.66\text{}\text{}\text{kJ}{\text{mol}}^{-1}-298\text{K}\left(-\text{0}\text{.02088}\text{kJ}{\text{K}}^{-1}{\text{mol}}^{-1}\right)\\ =-218.66\text{kJ}{\text{mol}}^{-1}+6.222\text{kJ}{\text{mol}}^{-1}\\ =\underset{\_}{-212.437kJmo{l}^{-1}}\end{array}$

Therefore, the standard Gibbs free energy for the reaction is $\underset{\_}{-212.437kJmo{l}^{-1}}$.

(ii)

Interpretation Introduction

Interpretation: Under the given conditions, the standard reaction enthalpy and the standard Gibbs free energy for the reaction has to be calculated.

Concept introduction: The change in enthalpy during the formation of products from reactants at standard conditions is known as standard reaction enthalpy.  The standard reaction enthalpy is the subtraction of enthalpy of formation of the products and the enthalpy of formation of the reactants.

Answer to Problem 3D.1BE

The standard enthalpy of the reaction is $\underset{\_}{-5644.25kJmo{l}^{-1}}$ and the standard Gibbs free energy for the reaction is $\underset{\_}{-5796.826kJmo{l}^{-1}}$.

Explanation of Solution

The given reaction is,

  ${\text{C}}_{12}{\text{H}}_{22}{\text{O}}_{\text{11}}\left(\text{s}\right)+1{\text{2O}}_{\text{2}}\left(\text{g}\right)\to 12{\text{CO}}_{\text{2}}\left(\text{g}\right)+11{\text{H}}_{\text{2}}\text{O}\left(\text{l}\right)$                                      (1)

The standard enthalpy $\left({\Delta }_{\text{r}}{H}^{\text{ο}}\right)$ for this chemical reaction is expressed as,

  ${\Delta }_{\text{r}}{H}^{\text{ο}}=\left(12{\Delta }_{\text{f}}{H}_{{\text{CO}}_{\text{2}}\left(\text{g}\right)}^{\text{ο}}+11{\Delta }_{\text{f}}{H}_{{\text{H}}_{\text{2}}\text{O}\left(\text{l}\right)}^{\text{ο}}-{\Delta }_{\text{f}}{H}_{{\text{C}}_{12}{\text{H}}_{22}{\text{O}}_{\text{11}}\left(\text{s}\right)}^{\text{ο}}-12{\Delta }_{\text{f}}{H}_{{\text{O}}_{\text{2}}\left(\text{g}\right)}^{\text{ο}}\right)$                 (2)

It is given that,

The standard entropy of the above chemical reaction $\left({\Delta }_{\text{r}}{S}^{\text{ο}}\right)$ is $512.034\text{J}{\text{K}}^{-1}{\text{mol}}^{-1}$

The standard enthalpy of formation of sucrose is $-2222\text{kJ}{\text{mol}}^{-1}$.

The standard enthalpy of formation of ${\text{O}}_{\text{2}}$ is $0\text{kJ}{\text{mol}}^{-1}$.

The standard enthalpy of formation of ${\text{CO}}_{\text{2}}$ is $-393.51\text{kJ}{\text{mol}}^{-1}$.

The standard enthalpy of formation of ${\text{H}}_{\text{2}}\text{O}$ is $-285.83\text{kJ}{\text{mol}}^{-1}$.

The conversion of $\text{J}{\text{K}}^{-1}{\text{mol}}^{-1}$ to $\text{kJ}{\text{K}}^{-1}{\text{mol}}^{-1}$ is done as,

  $\text{1J}{\text{K}}^{-1}{\text{mol}}^{-1}={10}^{-3}\text{}\text{kJ}{\text{K}}^{-1}{\text{mol}}^{-1}$

Therefore, the conversion of $512.034\text{J}{\text{K}}^{-1}{\text{mol}}^{-1}$ to $\text{kJ}{\text{K}}^{-1}{\text{mol}}^{-1}$ is done as,

  $512.034\text{J}{\text{K}}^{-1}{\text{mol}}^{-1}=0.5120\text{kJ}{\text{K}}^{-1}{\text{mol}}^{-1}$

Substitute the values of ${\Delta }_{\text{f}}{H}_{{\text{CO}}_{\text{2}}\left(\text{g}\right)}^{\text{ο}},{\Delta }_{\text{f}}{H}_{{\text{H}}_{\text{2}}\text{O}\left(\text{l}\right)}^{\text{ο}},{\Delta }_{\text{f}}{H}_{{\text{C}}_{12}{\text{H}}_{22}{\text{O}}_{\text{11}}\left(\text{s}\right)}^{\text{ο}}$ and ${\Delta }_{\text{f}}{H}_{{\text{O}}_{\text{2}}\left(\text{g}\right)}^{\text{ο}}$ in equation (2).

  $\begin{array}{c}{\Delta }_{\text{r}}{H}^{\text{ο}}=\left(\begin{array}{l}12\times \left(-393.51\text{kJ}{\text{mol}}^{-1}\right)+11\times \left(-285.83\text{kJ}{\text{mol}}^{-1}\right)-\\ \left(-2222\text{kJ}{\text{mol}}^{-1}\right)-12\times \text{0}\text{kJ}{\text{mol}}^{-1}\end{array}\right)\\ =-7866.25\text{kJ}{\text{mol}}^{-1}+2222\text{kJ}{\text{mol}}^{-1}\\ =\underset{\_}{-5644.25kJmo{l}^{-1}}\end{array}$

Therefore, the standard enthalpy of the reaction is $\underset{\_}{-5644.25kJmo{l}^{-1}}$.

The standard Gibbs free energy of the reaction is calculated as,

  ${\Delta }_{\text{r}}{G}^{\text{ο}}={\Delta }_{\text{r}}{H}^{\text{ο}}-T{\Delta }_{\text{r}}{S}^{\text{ο}}$                                                                                   (3)

Where,

• ${\Delta }_{\text{r}}{G}^{\text{ο}}$ is the standard Gibbs energy of reaction.
• ${\Delta }_{\text{r}}{H}^{\text{ο}}$ is the standard enthalpy of reaction.
• ${\Delta }_{\text{r}}{S}^{\text{ο}}$ is the standard entropy of reaction.
• $T$ is the standard temperature $\left(298\text{K}\right)$.

Substitute the values of ${\Delta }_{\text{r}}{H}^{\text{ο}}$, ${\Delta }_{\text{r}}{S}^{\text{ο}}$ and $T$ in equation (3).

  $\begin{array}{c}{\Delta }_{\text{r}}{G}^{\text{ο}}=-\text{5644}\text{.25}\text{kJ}{\text{mol}}^{-\text{1}}-298\text{K}\left(0.5120\text{kJ}{\text{K}}^{-1}{\text{mol}}^{-1}\right)\\ =-\text{5644}\text{.25}\text{kJ}{\text{mol}}^{-\text{1}}-152.576\text{kJ}{\text{mol}}^{-1}\\ =\underset{\_}{-5796.826kJmo{l}^{-1}}\end{array}$

Therefore, the standard Gibbs free energy for the reaction is $\underset{\_}{-5796.826kJmo{l}^{-1}}$.