Chapter 3, Problem 3D.1AE

(i)

Interpretation Introduction

Interpretation: The standard reaction enthalpy of the given reaction at $298\text{K}$ has to be predicted.  It has to be used to predict the standard reaction Gibbs energy at $298\text{K}$

Concept introduction: The standard reaction enthalpy of a given reaction represents the exothermic or endothermic property of a reaction.  If the value of standard reaction enthalpy is negative, the reaction is exothermic and releases heat.  If the value of standard reaction enthalpy is positive, the reaction is endothermic and requires heat.  The standard Gibbs energy is a criterion for determining the spontaneity of a reaction.

Answer to Problem 3D.1AE

The standard reaction enthalpy for the given reaction at $298\text{K}$ is $\underset{\_}{-636.62kJmo{l}^{-1}}$.  The standard reaction Gibbs energy for the given reaction at $298\text{K}$ is $\underset{\_}{-575.25kJmo{l}^{-1}}$.

Explanation of Solution

The given reaction is,

    $2{\text{CH}}_{\text{3}}\text{CHO}\left(\text{g}\right)+{\text{O}}_{\text{2}}\left(\text{g}\right)\to 2{\text{CH}}_{\text{3}}\text{COOH}\left(\text{l}\right)$

The standard reaction enthalpy of the reaction is calculated by the following formula,

    $\Delta H^{\Theta }{}_{\text{reaction}}={\Delta }_{\text{f}}{H}^{\Theta }{}_{\text{products}}-{\Delta }_{\text{f}}{H}^{\Theta }{}_{\text{reactants}}$

    $\Delta H^{\Theta }{}_{\text{reaction}}=\left(\begin{array}{l}\left(2\times {\Delta }_{\text{f}}{H}^{\Theta }\left({\text{CH}}_{\text{3}}\text{COOH}\left(\text{l}\right)\right)\right)\\ -\left(2\times {\Delta }_{\text{f}}{H}^{\Theta }\left({\text{CH}}_{\text{3}}\text{CHO}\left(\text{g}\right)\right)+{\Delta }_{\text{f}}{H}^{\Theta }\left({\text{O}}_{\text{2}}\left(\text{g}\right)\right)\right)\end{array}\right)$

The values of standard enthalpy of formations are:

• ${\Delta }_{\text{f}}{H}^{\Theta }\left({\text{CH}}_{\text{3}}\text{COOH}\left(\text{l}\right)\right)=-484.5\text{kJ}{\text{mol}}^{-1}$
• ${\Delta }_{\text{f}}{H}^{\Theta }\left({\text{CH}}_{\text{3}}\text{CHO}\left(\text{g}\right)\right)=-166.19\text{kJ}{\text{mol}}^{-1}$
• ${\Delta }_{\text{f}}{H}^{\Theta }\left({\text{O}}_{\text{2}}\left(\text{g}\right)\right)=0\text{kJ}{\text{mol}}^{-1}$

Substitute the corresponding values in the above equation.

    $\begin{array}{c}\Delta H^{\Theta }{}_{\text{reaction}}=\left(\begin{array}{c}\left(2\times -484.5\text{kJ}{\text{mol}}^{-1}\right)\\ -\left(\left(2\times -166.19\text{kJ}{\text{mol}}^{-1}\right)+\left(0\text{kJ}{\text{mol}}^{-1}\right)\right)\end{array}\right)\\ =\left(-969\text{kJ}{\text{mol}}^{-1}\right)-\left(-332.38\text{kJ}{\text{mol}}^{-1}\right)\\ =\underset{\_}{-636.62kJmo{l}^{-1}}\end{array}$

Hence, the standard reaction enthalpy for the given reaction at $298\text{K}$ is $\underset{\_}{-636.62kJmo{l}^{-1}}$.

The reaction Gibbs energy $\left(\Delta G^{\Theta }{}_{\text{reaction}}\right)$ is given as,

    $\Delta G^{\Theta }{}_{\text{reaction}}=\Delta H^{\Theta }{}_{\text{reaction}}-T\Delta S^{\Theta }{}_{\text{reaction}}$                                                               (1)

Where,

• $T$ is the temperature of the reaction
• $\Delta S^{\Theta }{}_{\text{reaction}}$ is the standard reaction entropy of the reaction.

The standard reaction entropy of the given reaction, $\Delta S^{\Theta }{}_{\text{reaction}}$ is $-\text{205}\text{.938}\text{J}{\text{K}}^{\text{-1}}{\text{mol}}^{\text{-1}}$.

The temperature of the reaction, $T$ is $298\text{K}$.

Substitute the corresponding values in equation (1).

    $\begin{array}{c}\Delta G^{\Theta }{}_{\text{reaction}}=-\text{636}\text{.62}\times {10}^3\text{J}{\text{mol}}^{-1}-\left(298\text{K}\times -\text{205}\text{.938}\text{J}{\text{K}}^{\text{-1}}{\text{mol}}^{\text{-1}}\right)\\ =-575250.476\text{J}{\text{mol}}^{-1}\\ =-575250.476\text{J}{\text{mol}}^{-1}\times \frac{\text{1}\text{kJ}}{1000\text{J}}\\ =\underset{\_}{-575.25kJmo{l}^{-1}}\end{array}$

Hence, the standard reaction Gibbs energy for the given reaction at $298\text{K}$ is $\underset{\_}{-575.25kJmo{l}^{-1}}$.

(ii)

Interpretation Introduction

Interpretation: The standard reaction enthalpy of the given reaction at $298\text{K}$ has to be predicted.  It has to be used to predict the standard reaction Gibbs energy at $298\text{K}$.

Concept introduction: The standard reaction enthalpy of a given reaction represents the exothermic or endothermic property of a reaction.  If the value of standard reaction enthalpy is negative, the reaction is exothermic and releases heat.  If the value of standard reaction enthalpy is positive, the reaction is endothermic and requires heat.  The standard Gibbs energy is a criterion for determining the spontaneity of a reaction.

Answer to Problem 3D.1AE

The standard reaction enthalpy for the given reaction at $298\text{K}$ is $\underset{\_}{53.4kJmol}$.  The standard reaction Gibbs energy for the given reaction at $298\text{K}$ is $\underset{\_}{25.79328kJmo{l}^{-1}}$.

Explanation of Solution

The given reaction is,

    $\text{2AgCl}\left(\text{s}\right)+{\text{Br}}_{\text{2}}\left(\text{l}\right)\to \text{2AgBr}\left(\text{s}\right)+{\text{Cl}}_{\text{2}}\left(\text{g}\right)$

The standard reaction enthalpy of the reaction is calculated by the following formula,

    $\begin{array}{c}\Delta H^{\Theta }{}_{\text{reaction}}={\Delta }_{\text{f}}{H}^{\Theta }{}_{\text{products}}-{\Delta }_{\text{f}}{H}^{\Theta }{}_{\text{reactants}}\\ =\left(\begin{array}{l}\left(2\times {\Delta }_{\text{f}}{H}^{\Theta }\left(\text{AgBr}\left(\text{s}\right)\right)+{\text{Δ}}_{\text{f}}{H}^{\text{Θ}}\left({\text{Cl}}_{\text{2}}\left(\text{g}\right)\right)\right)\\ -\left(2\times {\Delta }_{\text{f}}{H}^{\Theta }\left(\text{AgCl}\left(\text{s}\right)\right)+{\Delta }_{\text{f}}{H}^{\Theta }\left({\text{Br}}_{\text{2}}\left(\text{l}\right)\right)\right)\end{array}\right)\end{array}$

The values of standard enthalpy of formations are:

• ${\Delta }_{\text{f}}{H}^{\Theta }\left(\text{AgBr}\right)=-100.37\text{kJ}{\text{mol}}^{\text{-1}}$
• ${\Delta }_{\text{f}}{H}^{\Theta }\left({\text{Cl}}_{\text{2}}\left(\text{g}\right)\right)=0\text{kJ}{\text{mol}}^{\text{-1}}$
• ${\Delta }_{\text{f}}{H}^{\Theta }\left({\text{Br}}_{\text{2}}\left(\text{l}\right)\right)=0\text{kJ}{\text{mol}}^{\text{-1}}$
• ${\Delta }_{\text{f}}{H}^{\Theta }\left(\text{AgCl}\left(\text{s}\right)\right)=-127.07\text{kJ}{\text{mol}}^{\text{-1}}$

Substitute the corresponding values in the above equation $\begin{array}{c}\Delta H^{\Theta }{}_{\text{reaction}}=\left(\left(2\times \left(-100.37\text{kJ}{\text{mol}}^{\text{-1}}\right)+0\right)-\left(2\times \left(-127.07\text{kJ}{\text{mol}}^{\text{-1}}\right)+0\right)\right)\\ ={\underset{\_}{53.4kJmol}}^{\text{-1}}\end{array}$

Hence, the standard reaction enthalpy for the given reaction at $298\text{K}$ $\underset{\_}{53.4kJmol}$.

The reaction Gibbs energy $\left(\Delta G^{\Theta }{}_{\text{reaction}}\right)$ is given as,

    $\Delta G^{\Theta }{}_{\text{reaction}}=\Delta H^{\Theta }{}_{\text{reaction}}-T\Delta S^{\Theta }{}_{\text{reaction}}$                                                               (1)

Where,

• $T$ is the temperature of the reaction
• $\Delta S^{\Theta }{}_{\text{reaction}}$ is the standard reaction entropy of the reaction.

The standard reaction entropy of the given reaction, $\Delta S^{\Theta }{}_{\text{reaction}}$ is $92.64\text{J}{\text{K}}^{\text{-1}}{\text{mol}}^{\text{-1}}$.

The temperature of the reaction, $T$ is $298\text{K}$.

Substitute the corresponding values in equation (1).

    $\begin{array}{c}\Delta G^{\Theta }{}_{\text{reaction}}=\text{53}\text{.4}\times {10}^3\text{J}{\text{mol}}^{-1}-\left(298\text{K}\times 92.64\text{J}{\text{K}}^{\text{-1}}{\text{mol}}^{\text{-1}}\right)\\ =25793.28\text{J}{\text{mol}}^{-1}\\ =25793.28\text{J}{\text{mol}}^{-1}\times \frac{\text{1}\text{kJ}}{1000\text{J}}\\ =\underset{\_}{25.79328kJmo{l}^{-1}}\end{array}$

Hence, the standard reaction Gibbs energy for the given reaction at $298\text{K}$ is $\underset{\_}{25.79328kJmo{l}^{-1}}$.

(iii)

Interpretation Introduction

Interpretation: The standard reaction enthalpy of the given reaction at $298\text{K}$ has to be predicted.  It has to be used to predict the standard reaction Gibbs energy at $298\text{K}$.

Concept introduction: The standard reaction enthalpy of a given reaction represents the exothermic or endothermic property of a reaction.  If the value of standard reaction enthalpy is negative, the reaction is exothermic and releases heat.  If the value of standard reaction enthalpy is positive, the reaction is endothermic and requires heat.  The standard Gibbs energy is a criterion for determining the spontaneity of a reaction.

Answer to Problem 3D.1AE

The standard reaction enthalpy for the given reaction at $298\text{K}$ is $\underset{\_}{-224.3kJmo{l}^{\text{-1}}}$.  The standard reaction Gibbs energy for the given reaction at $298\text{K}$ is $-\underset{\_}{178.67918kJmo{l}^{-1}}$.

Explanation of Solution

The given reaction is,

    $\text{Hg}\left(\text{l}\right)+{\text{Cl}}_{\text{2}}\left(\text{g}\right)\to {\text{HgCl}}_{\text{2}}\left(\text{s}\right)$

The standard reaction enthalpy of the reaction is calculated by the following formula.

    $\begin{array}{l}\Delta H^{\Theta }{}_{\text{reaction}}={\Delta }_{\text{f}}{H}^{\Theta }{}_{\text{products}}-{\Delta }_{\text{f}}{H}^{\Theta }{}_{\text{reactants}}\\ \Delta H^{\Theta }{}_{\text{reaction}}=\left(\begin{array}{l}\left({\Delta }_{\text{f}}{H}^{\Theta }\left({\text{HgCl}}_{\text{2}}\left(\text{s}\right)\right)\right)\\ -\left({\Delta }_{\text{f}}{H}^{\Theta }\left(\text{Hg}\left(\text{l}\right)\right)+{\Delta }_{\text{f}}{H}^{\Theta }\left({\text{Cl}}_{\text{2}}\left(\text{g}\right)\right)\right)\end{array}\right)\end{array}$

The values of standard enthalpies of formations are:

• ${\Delta }_{\text{f}}{H}^{\Theta }\left({\text{HgCl}}_{\text{2}}\left(\text{s}\right)\right)=-224.3\text{kJ}{\text{mol}}^{\text{-1}}$
• ${\Delta }_{\text{f}}{H}^{\Theta }\left({\text{Cl}}_{\text{2}}\left(\text{g}\right)\right)=0\text{kJ}{\text{mol}}^{\text{-1}}$
• ${\Delta }_{\text{f}}{H}^{\Theta }\left(\text{Hg}\left(\text{l}\right)\right)=0\text{kJ}{\text{mol}}^{\text{-1}}$

Substitute the corresponding values in the above equation.    $\begin{array}{c}\Delta H^{\Theta }{}_{\text{reaction}}=\left(\left(-224.3\text{kJ}{\text{mol}}^{\text{-1}}\right)-\left(0+0\right)\right)\\ =\underset{\_}{-224.3kJmo{l}^{\text{-1}}}\end{array}$

Hence, the standard entropy of the reaction at $298\text{K}$ is $\underset{\_}{-224.3kJmo{l}^{\text{-1}}}$.

The reaction Gibbs energy $\left(\Delta G^{\Theta }{}_{\text{reaction}}\right)$ is given as,

    $\Delta G^{\Theta }{}_{\text{reaction}}=\Delta H^{\Theta }{}_{\text{reaction}}-T\Delta S^{\Theta }{}_{\text{reaction}}$                                                             (1)

Where,

• $T$ is the temperature of the reaction
• $\Delta S^{\Theta }{}_{\text{reaction}}$ is the standard reaction entropy of the reaction.

The standard reaction entropy of the given reaction, $\Delta S^{\Theta }{}_{\text{reaction}}$ is $\text{-153}\text{.09}\text{J}{\text{K}}^{\text{-1}}{\text{mol}}^{\text{-1}}$.

The temperature of the reaction, $T$ is $298\text{K}$.

Substitute the corresponding values in equation (1).

    $\begin{array}{c}\Delta G^{\Theta }{}_{\text{reaction}}=-\text{224}\text{.3}\times {10}^3\text{J}{\text{mol}}^{-1}-\left(298\text{K}\times \text{-153}\text{.09}\text{J}{\text{K}}^{\text{-1}}{\text{mol}}^{\text{-1}}\right)\\ =-178679.18\text{J}{\text{mol}}^{-1}\\ =-178679.18\text{J}{\text{mol}}^{-1}\times \frac{\text{1}\text{kJ}}{1000\text{J}}\\ =-\underset{\_}{178.67918kJmo{l}^{-1}}\end{array}$

Hence, the standard reaction Gibbs energy for the given reaction at $298\text{K}$ is $-\underset{\_}{178.67918kJmo{l}^{-1}}$.