Chapter 4, Problem 4J.7E

(a)

Interpretation Introduction

Interpretation:

The standard change in the enthalpy, entropy and Gibbs free energy for the given reaction has to be determined.

Concept Introduction:

The importance of Gibbs energy is that it gives information about the spontaneity of the reaction or a process at constant temperature and pressure.  A compound is considered thermodynamically stable when it has negative Gibbs free energy of formation.  The relation to determine Gibbs free energy of a reaction is shown below.

$\Delta \text{G}°={\displaystyle \sum \text{n}\Delta {\text{G}}_f^{\circ }\left(products\right)}-{\displaystyle \sum \text{n}\Delta {\text{G}}_f^{\circ }}\left(\text{reactants}\right)$

Answer to Problem 4J.7E

The standard change in the enthalpy, entropy and Gibbs free energy for the given reaction $\underset{\_}{-196.1kJ}$, $\underset{\_}{125.758J\cdot K^{-1}}$ and $\underset{\_}{-233.56kJ}$ respectively.

Explanation of Solution

The given chemical equation for the decomposition of hydrogen peroxide is shown below.

    $2H_2{O}_2\left(l\right)\to 2H_{2}O\left(l\right)+O_2\left(g\right)$

The relation for the calculation of standard change in the enthalpy is shown below.

    $\Delta \text{H}°={\displaystyle \sum \text{n}\Delta {\text{H}}_f^{\circ }\left(products\right)}-{\displaystyle \sum \text{n}\Delta {\text{H}}_f^{\circ }}\left(\text{reactants}\right)$

The standard change in the enthalpy of the given reaction is calculated by the expression is shown below.

    $\Delta \text{H}°=\left\{\begin{array}{l}\left(2mol\right)\times \Delta {\text{H}}_f^{\circ }\left(H_{2}O,l\right)\\ +\left(1mol\right)\times \Delta {\text{H}}_f^{\circ }\left(O_2,g\right)\end{array}\right\}-\left\{\left(2mol\right)\times \Delta {\text{H}}_f^{\circ }\left(H_2{O}_2,l\right)\right\}$        (1)

Where,

• $\Delta {\text{H}}_f^{\circ }\left(H_2{O}_2,l\right)$ is the standard enthalpy of formation of $H_2{O}_2\left(l\right)$.
• $\Delta {\text{H}}_f^{\circ }\left(H_{2}O,l\right)$ is the standard enthalpy of formation of $H_{2}O\left(l\right)$.
• $\Delta {\text{H}}_f^{\circ }\left(O_2,g\right)$ is the standard enthalpy of formation of $O_2\left(g\right)$.

The value of $\Delta {\text{H}}_f^{\circ }\left(H_2{O}_2,l\right)$ is $-187.78\text{kJ}\cdot {\text{mol}}^{-1}$.

The value of $\Delta {\text{H}}_f^{\circ }\left(H_{2}O,l\right)$ is $-285.83\text{kJ}\cdot {\text{mol}}^{-1}$.

The value of $\Delta {\text{H}}_f^{\circ }\left(O_2,g\right)$ is $0.0\text{kJ}\cdot {\text{mol}}^{-1}$.

Substitute the value of $\Delta {\text{H}}_f^{\circ }\left(H_2{O}_2,l\right)$, $\Delta {\text{H}}_f^{\circ }\left(H_{2}O,l\right)$ and $\Delta {\text{H}}_f^{\circ }\left(O_2,g\right)$ in equation (1).

    $\begin{array}{c}\Delta \text{H}°=\left\{\begin{array}{l}\left(2mol\right)\times \left(-285.83\text{kJ}\cdot {\text{mol}}^{-1}\right)\\ +\left(1mol\right)\times \left(0.0\text{kJ}\cdot {\text{mol}}^{-1}\right)\end{array}\right\}-\left\{\left(2mol\right)\times \left(-187.78\text{kJ}\cdot {\text{mol}}^{-1}\right)\right\}\\ =\left\{-571.66\text{kJ}\right\}-\left\{-375.56\text{kJ}\right\}\\ =-196.1\text{kJ}\end{array}$

The relation for the calculation of standard change in the entropy is shown below.

    $\Delta \text{S}°={\displaystyle \sum \text{n}\Delta {\text{S}}_f^{\circ }\left(products\right)}-{\displaystyle \sum \text{n}\Delta {\text{S}}_f^{\circ }}\left(\text{reactants}\right)$

The standard change in the entropy of the given reaction is calculated by the expression is shown below.

    $\Delta \text{S}°=\left\{\begin{array}{l}\left(2mol\right)\times \Delta {\text{S}}_f^{\circ }\left(H_{2}O,l\right)\\ +\left(1mol\right)\times \Delta {\text{S}}_f^{\circ }\left(O_2,g\right)\end{array}\right\}-\left\{\left(2mol\right)\times \Delta {\text{S}}_f^{\circ }\left(H_2{O}_2,l\right)\right\}$        (2)

Where,

• $\Delta {\text{S}}_f^{\circ }\left(H_2{O}_2,l\right)$ is the standard entropy of formation of $H_2{O}_2\left(l\right)$.
• $\Delta {\text{S}}_f^{\circ }\left(H_{2}O,l\right)$ is the standard entropy of formation of $H_{2}O\left(l\right)$.
• $\Delta {\text{S}}_f^{\circ }\left(O_2,g\right)$ is the standard entropy of formation of $O_2\left(g\right)$.

The value of $\Delta {\text{S}}_f^{\circ }\left(H_2{O}_2,l\right)$ is $+109.6\text{J}\cdot {\text{K}}^{-1}\cdot {\text{mol}}^{-1}$.

The value of $\Delta {\text{S}}_f^{\circ }\left(H_{2}O,l\right)$ is $+69.91\text{J}\cdot {\text{K}}^{-1}\cdot {\text{mol}}^{-1}$.

The value of $\Delta {\text{S}}_f^{\circ }\left(O_2,g\right)$ is $+205.138\text{J}\cdot {\text{K}}^{-1}\cdot {\text{mol}}^{-1}$.

Substitute the value of $\Delta {\text{S}}_f^{\circ }\left(H_2{O}_2,l\right)$, $\Delta {\text{S}}_f^{\circ }\left(H_{2}O,l\right)$ and $\Delta {\text{S}}_f^{\circ }\left(O_2,g\right)$ in equation (2).

    $\begin{array}{c}\Delta \text{S}°=\left\{\begin{array}{l}\left(2mol\right)\times \left(69.91\text{J}\cdot {\text{K}}^{-1}\cdot {\text{mol}}^{-1}\right)\\ +\left(1mol\right)\times \left(205.138\text{J}\cdot {\text{K}}^{-1}\cdot {\text{mol}}^{-1}\right)\end{array}\right\}-\left\{\left(2mol\right)\times \left(\text{109}\text{.6}\text{J}\cdot {\text{K}}^{-1}\cdot {\text{mol}}^{-1}\right)\right\}\\ =\left\{139.82\text{J}\cdot {\text{K}}^{-1}+205.138\text{J}\cdot {\text{K}}^{-1}\right\}-\left\{219.2\text{J}\cdot {\text{K}}^{-1}\right\}\\ =125.758\text{J}\cdot {\text{K}}^{-1}\end{array}$

The relation for the calculation of standard Gibbs free energy for the given reaction is shown below.

    $\Delta \text{G}°=\left\{\begin{array}{l}\left(2mol\right)\times \Delta {\text{G}}_f^{\circ }\left(H_{2}O,l\right)\\ +\left(1mol\right)\times \Delta {\text{G}}_f^{\circ }\left(O_2,g\right)\end{array}\right\}-\left\{\left(2mol\right)\times \Delta {\text{G}}_f^{\circ }\left(H_2{O}_2,l\right)\right\}$        (3)

Where,

• $\Delta {\text{G}}_f^{\circ }\left(H_2{O}_2,l\right)$ is the standard Gibbs free energy of formation of $H_2{O}_2\left(l\right)$.
• $\Delta {\text{G}}_f^{\circ }\left(H_{2}O,l\right)$ is the standard Gibbs free energy of formation of $H_{2}O\left(l\right)$.
• $\Delta {\text{G}}_f^{\circ }\left(O_2,g\right)$ is the standard Gibbs free energy of formation of $O_2\left(g\right)$.

The value of $\Delta {\text{G}}_f^{\circ }\left(H_2{O}_2,l\right)$ is $-120.35k\text{J}\cdot {\text{mol}}^{-1}$.

The value of $\Delta {\text{G}}_f^{\circ }\left(H_{2}O,l\right)$ is $-237.13k\text{J}\cdot {\text{mol}}^{-1}$.

The value of $\Delta {\text{G}}_f^{\circ }\left(O_2,g\right)$ is $0.0k\text{J}\cdot {\text{mol}}^{-1}$.

Substitute the value of $\Delta {\text{G}}_f^{\circ }\left(H_2{O}_2,l\right)$, $\Delta {\text{G}}_f^{\circ }\left(H_{2}O,l\right)$ and $\Delta {\text{G}}_f^{\circ }\left(O_2,g\right)$ in equation (3).

    $\begin{array}{c}\Delta \text{G}°=\left\{\begin{array}{l}\left(2mol\right)\times \left(-237.13\text{kJ}\cdot {\text{mol}}^{-1}\right)\\ +\left(1mol\right)\times \left(0.0\text{kJ}\cdot {\text{mol}}^{-1}\right)\end{array}\right\}-\left\{\left(2mol\right)\times \left(-120.35\text{kJ}\cdot {\text{mol}}^{-1}\right)\right\}\\ =\left\{-474.26\text{kJ}\right\}-\left\{-240.7\text{kJ}\right\}\\ =-233.56\text{kJ}\end{array}$

Thus, the standard change in the enthalpy, entropy and Gibbs free energy for the given reaction $\underset{\_}{-196.1kJ}$, $\underset{\_}{125.758J\cdot K^{-1}}$ and $\underset{\_}{-233.56kJ}$ respectively.

(b)

Interpretation Introduction

Interpretation:

The standard change in the enthalpy, entropy and Gibbs free energy for the given reaction has to be determined.

Concept Introduction:

Same as part (a).

Answer to Problem 4J.7E

The standard change in the enthalpy, entropy and Gibbs free energy for the given reaction $\underset{\_}{-758.86kJ}$, $\underset{\_}{-395.442J\cdot K^{-1}}$ and $\underset{\_}{-640.9kJ}$ respectively.

Explanation of Solution

The given chemical equation for the production of hydrofluoric acid is shown below.

    $2F_2\left(g\right)+2H_{2}O\left(l\right)\to 4HF\left(g\right)+O_2\left(g\right)$

The standard change in the enthalpy of the given reaction is calculated by the expression is shown below.

    $\Delta \text{H}°=\left\{\begin{array}{l}\left(4mol\right)\times \Delta {\text{H}}_f^{\circ }\left(HF,g\right)\\ +\left(1mol\right)\times \Delta {\text{H}}_f^{\circ }\left(O_2,g\right)\end{array}\right\}-\left\{\begin{array}{l}\left(2mol\right)\times \Delta {\text{H}}_f^{\circ }\left(F_2,g\right)\\ +\left(2mol\right)\times \Delta {\text{H}}_f^{\circ }\left(H_{2}O,l\right)\end{array}\right\}$        (4)

Where,

• $\Delta {\text{H}}_f^{\circ }\left(HF,g\right)$ is the standard enthalpy of formation of $HF\left(g\right)$.
• $\Delta {\text{H}}_f^{\circ }\left(O_2,g\right)$ is the standard enthalpy of formation of $O_2\left(g\right)$.
• $\Delta {\text{H}}_f^{\circ }\left(F_2,g\right)$ is the standard enthalpy of formation of $F_2\left(g\right)$.
• $\Delta {\text{H}}_f^{\circ }\left(H_{2}O,l\right)$ is the standard enthalpy of formation of $H_{2}O\left(l\right)$.

The value of $\Delta {\text{H}}_f^{\circ }\left(HF,g\right)$ is $-332.63\text{kJ}\cdot {\text{mol}}^{-1}$.

The value of $\Delta {\text{H}}_f^{\circ }\left(O_2,g\right)$ is $0.0\text{kJ}\cdot {\text{mol}}^{-1}$.

The value of $\Delta {\text{H}}_f^{\circ }\left(F_2,g\right)$ is $0.0\text{kJ}\cdot {\text{mol}}^{-1}$.

The value of $\Delta {\text{H}}_f^{\circ }\left(H_{2}O,l\right)$ is $-285.83\text{kJ}\cdot {\text{mol}}^{-1}$.

Substitute the value of $\Delta {\text{H}}_f^{\circ }\left(HF,g\right)$, $\Delta {\text{H}}_f^{\circ }\left(O_2,g\right)$, $\Delta {\text{H}}_f^{\circ }\left(F_2,g\right)$ and $\Delta {\text{H}}_f^{\circ }\left(H_{2}O,l\right)$ in equation (4).

    $\begin{array}{c}\Delta \text{H}°=\left\{\begin{array}{l}\left(4mol\right)\times \left(-332.63\text{kJ}\cdot {\text{mol}}^{-1}\right)\\ +\left(1mol\right)\times \left(0.0\text{kJ}\cdot {\text{mol}}^{-1}\right)\end{array}\right\}-\left\{\begin{array}{l}\left(2mol\right)\times \left(0.0\text{kJ}\cdot {\text{mol}}^{-1}\right)\\ +\left(2mol\right)\times \left(-285.83\text{kJ}\cdot {\text{mol}}^{-1}\right)\end{array}\right\}\\ =\left\{-1330.52\text{kJ}\right\}-\left\{-571.66\text{kJ}\right\}\\ =-758.86\text{kJ}\end{array}$

The standard change in the entropy of the given reaction is calculated by the expression is shown below.

    $\Delta \text{S}°=\left\{\begin{array}{l}\left(4mol\right)\times \Delta {\text{S}}_f^{\circ }\left(HF,g\right)\\ +\left(1mol\right)\times \Delta {\text{S}}_f^{\circ }\left(O_2,g\right)\end{array}\right\}-\left\{\begin{array}{l}\left(2mol\right)\times \Delta {\text{S}}_f^{\circ }\left(F_2,g\right)\\ +\left(2mol\right)\times \Delta {\text{S}}_f^{\circ }\left(H_{2}O,l\right)\end{array}\right\}$        (5)

Where,

• $\Delta {\text{S}}_f^{\circ }\left(HF,g\right)$ is the standard entropy of formation of $HF\left(g\right)$.
• $\Delta {\text{S}}_f^{\circ }\left(O_2,g\right)$ is the standard entropy of formation of $O_2\left(g\right)$.
• $\Delta {\text{S}}_f^{\circ }\left(F_2,g\right)$ is the standard entropy of formation of $F_2\left(g\right)$.
• $\Delta {\text{S}}_f^{\circ }\left(H_{2}O,l\right)$ is the standard entropy of formation of $H_{2}O\left(l\right)$.

The value of $\Delta {\text{S}}_f^{\circ }\left(HF,g\right)$ is $-13.8\text{J}\cdot {\text{K}}^{-1}\cdot {\text{mol}}^{-1}$.

The value of $\Delta {\text{S}}_f^{\circ }\left(O_2,g\right)$ is $+205.138\text{J}\cdot {\text{K}}^{-1}\cdot {\text{mol}}^{-1}$.

The value of $\Delta {\text{H}}_f^{\circ }\left(F_2,g\right)$ is $+202.78\text{J}\cdot {\text{K}}^{-1}\cdot {\text{mol}}^{-1}$.

The value of $\Delta {\text{S}}_f^{\circ }\left(H_{2}O,l\right)$ is $69.91\text{J}\cdot {\text{K}}^{-1}\cdot {\text{mol}}^{-1}$.

Substitute the value of $\Delta {\text{S}}_f^{\circ }\left(HF,g\right)$, $\Delta {\text{S}}_f^{\circ }\left(F_2,g\right)$, $\Delta {\text{S}}_f^{\circ }\left(O_2,g\right)$ and $\Delta {\text{S}}_f^{\circ }\left(H_{2}O,l\right)$ in equation (5).

    $\begin{array}{c}\Delta \text{S}°=\left\{\begin{array}{l}\left(4mol\right)\times \left(-13.8\text{J}\cdot {\text{K}}^{-1}\cdot {\text{mol}}^{-1}\right)\\ +\left(1mol\right)\times \left(205.138\text{J}\cdot {\text{K}}^{-1}\cdot {\text{mol}}^{-1}\right)\end{array}\right\}-\left\{\begin{array}{l}\left(2mol\right)\times \left(202.78\text{J}\cdot {\text{K}}^{-1}\cdot {\text{mol}}^{-1}\right)\\ +\left(2mol\right)\times \left(69.91\text{J}\cdot {\text{K}}^{-1}\cdot {\text{mol}}^{-1}\right)\end{array}\right\}\\ =\left\{-55.2\text{J}\cdot {\text{K}}^{-1}+205.138\text{J}\cdot {\text{K}}^{-1}\right\}-\left\{405.56\text{J}\cdot {\text{K}}^{-1}+139.82\text{J}\cdot {\text{K}}^{-1}\right\}\\ =-395.442\text{J}\cdot {\text{K}}^{-1}\end{array}$

The relation for the calculation of standard Gibbs free energy for the given reaction is shown below.

    $\Delta \text{G}°=\left\{\begin{array}{l}\left(4mol\right)\times \Delta {\text{G}}_f^{\circ }\left(HF,g\right)\\ +\left(1mol\right)\times \Delta {\text{G}}_f^{\circ }\left(O_2,g\right)\end{array}\right\}-\left\{\begin{array}{l}\left(2mol\right)\times \Delta {\text{G}}_f^{\circ }\left(F_2,g\right)\\ +\left(2mol\right)\times \Delta {\text{G}}_f^{\circ }\left(H_{2}O,l\right)\end{array}\right\}$        (6)

Where,

• $\Delta {\text{G}}_f^{\circ }\left(HF,g\right)$ is the standard Gibbs free energy of formation of $HF\left(g\right)$.
• $\Delta {\text{G}}_f^{\circ }\left(O_2,g\right)$ is the standard Gibbs free energy of formation of $O_2\left(g\right)$.
• $\Delta {\text{G}}_f^{\circ }\left(F_2,g\right)$ is the standard Gibbs free energy of formation of $F_2\left(g\right)$.
• $\Delta {\text{G}}_f^{\circ }\left(H_{2}O,l\right)$ is the standard Gibbs free energy of formation of $H_{2}O\left(l\right)$.

The value of $\Delta {\text{G}}_f^{\circ }\left(HF,g\right)$ is $-278.79k\text{J}\cdot {\text{mol}}^{-1}$.

The value of $\Delta {\text{G}}_f^{\circ }\left(O_2,g\right)$ is $0.0k\text{J}\cdot {\text{mol}}^{-1}$.

The value of $\Delta {\text{G}}_f^{\circ }\left(F_2,g\right)$ is $0.0k\text{J}\cdot {\text{mol}}^{-1}$.

The value of $\Delta {\text{G}}_f^{\circ }\left(H_{2}O,l\right)$ is $-237.13k\text{J}\cdot {\text{mol}}^{-1}$.

Substitute the value of $\Delta {\text{G}}_f^{\circ }\left(HF,g\right)$, $\Delta {\text{G}}_f^{\circ }\left(O_2,g\right)$, $\Delta {\text{G}}_f^{\circ }\left(H_{2}O,l\right)$ and $\Delta {\text{G}}_f^{\circ }\left(F_2,g\right)$ in equation (6).

    $\begin{array}{c}\Delta \text{G}°=\left\{\begin{array}{l}\left(4mol\right)\times \left(-278.79k\text{J}\cdot {\text{mol}}^{-1}\right)\\ +\left(1mol\right)\times \left(0.0\text{kJ}\cdot {\text{mol}}^{-1}\right)\end{array}\right\}-\left\{\begin{array}{l}\left(2mol\right)\times \left(0.0\text{kJ}\cdot {\text{mol}}^{-1}\right)\\ +\left(2mol\right)\times \left(-237.13\text{kJ}\cdot {\text{mol}}^{-1}\right)\end{array}\right\}\\ =\left\{-1115.16\text{kJ}\right\}-\left\{-474.26\text{kJ}\right\}\\ =-640.9\text{kJ}\end{array}$

Thus, the standard change in the enthalpy, entropy and Gibbs free energy for the given reaction $\underset{\_}{-758.86kJ}$, $\underset{\_}{-395.442J\cdot K^{-1}}$ and $\underset{\_}{-640.9kJ}$ respectively.