Chapter 17, Problem 86AE

Consider the following reaction:

${\text{H}}_{\text{2}}\text{O}\left(g\right)+{\text{CI}}_{\text{2}}\text{O}\left(g\right)\rightleftharpoons 2\text{HOCI}\left(g\right)K_{298}=0.090$

For Cl₂O(g),

$\begin{array}{l}\Delta G_{\text{f}}^{\text{o}}=97.9\text{KJ/mol}\\ \Delta H_{\text{f}}^{\text{o}}=80.3\text{KJ/mol}\\ S^{\circ }=266.1J/K\cdot \text{mol}\end{array}$

a. Calculate ∆G° for the reaction using the equation ∆G°=−RT In(K).

b. Use bond energy values (Table 3-3) to estimate ∆H° for the reaction.

c. Use the results from parts a and b to estimate ∆S° for the reaction.

d. Estimate $\Delta H_{\text{f}}^{\text{o}}$ and S° for HOCI(g).

e. Estimate the value of K at 500. K.

f. Calculate ∆G at 25°C when $P_{{\text{H}}_{\text{2}}\text{O}}=18\text{torr},P_{{\text{CI}}_{\text{2}}\text{O}}=2.0\text{torr},and{P}_{\text{HOCI}}=0.10\text{torr}$.

Interpretation Introduction

Interpretation: The reaction corresponding to the formation of $\text{HOCl}\left(g\right)$, its equilibrium constant at $\text{298}\text{K}$ and values of $\Delta G_{\text{f}}°,\Delta H_{\text{f}}°,S°$ are given. The questions are to be answered on the basis of the given data.

Concept introduction: Equilibrium constant, $K$, is defined as the ratio of the concentration of products to that of the reactants at equilibrium. If a given reaction is at equilibrium, the free energy change is,

$\begin{array}{c}\Delta G=\text{0}\\ Q=K\end{array}$

The expression for free energy change is,

$\Delta G°=-RT\ln \left(K\right)$

The formula of $\Delta S°$ is,

$\Delta S°=\sum n_{\text{p}}\Delta S{°}_{\left(\text{product}\right)}-\sum n_{\text{f}}\Delta S{°}_{\left(\text{reactant}\right)}$

To determine: The value of $\Delta G°$ for the given reaction, using the equation $\Delta G°=-RT\ln \left(K\right)$.

Explanation of Solution

Explanation

Given

Temperature is $298\text{K}$.

The value of equilibrium constant is $0.090$.

The stated reaction is,

${\text{H}}_{\text{2}}\text{O}\left(g\right)+{\text{Cl}}_{\text{2}}\text{O}\left(g\right)\stackrel{}{\rightleftharpoons }2\text{HOCl}\left(g\right)$

Formula

The expression for free energy change is,

$\Delta G°=-RT\ln \left(K\right)$

Where,

• $\Delta G°$ is the standard Gibbs free energy change.
• $R$ is the gas law constant $(\text{8}\text{.3145}\text{J/K}\cdot \text{mol})$.
• $T$ is the absolute temperature.
• $K$ is the equilibrium constant.

Substitute the values of $R,T$ and $K$ in the above expression.

$\begin{array}{c}\Delta G°=-RT\ln \left(K\right)\\ =-(\text{8}\text{.3145}\text{J/K}\cdot \text{mol})\left(298\text{K}\right)\ln \left(0.090\right)\\ =6.0\times {10}^3\text{J/mol}\end{array}$

The conversion of joule per mole $\left(\text{J/mol}\right)$ into kilo-joule per mole $\left(\text{kJ/mol}\right)$ is done as,

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

Hence,

The conversion of $6.0\times {10}^3\text{J/mol}$ into kilo-joule per mole is,

$\begin{array}{c}\text{6}\text{.0}\times {\text{10}}^{\text{3}}\text{J/mol}=\left(\text{6}\text{.0}\times {\text{10}}^{\text{3}}\times {\text{10}}^{-\text{3}}\right)\text{kJ/mol}\\ =\underset{\_}{6.0kJ/mol}\end{array}$

Interpretation Introduction

Interpretation: The reaction corresponding to the formation of $\text{HOCl}\left(g\right)$, its equilibrium constant at $\text{298}\text{K}$ and values of $\Delta G_{\text{f}}°,\Delta H_{\text{f}}°,S°$ are given. The questions are to be answered on the basis of the given data.

Concept introduction: Equilibrium constant, $K$, is defined as the ratio of the concentration of products to that of the reactants at equilibrium. If a given reaction is at equilibrium, the free energy change is,

$\begin{array}{c}\Delta G=\text{0}\\ Q=K\end{array}$

The expression for free energy change is,

$\Delta G°=-RT\ln \left(K\right)$

The formula of $\Delta S°$ is,

$\Delta S°=\sum n_{\text{p}}\Delta S{°}_{\left(\text{product}\right)}-\sum n_{\text{f}}\Delta S{°}_{\left(\text{reactant}\right)}$

To determine: The value of $\Delta G°$ for the given reaction, using the equation $\Delta G°=-RT\ln \left(K\right)$.

Explanation of Solution

Explanation

The reaction that takes place is,

${\text{H}}_{\text{2}}\text{O}\left(g\right)+{\text{Cl}}_{\text{2}}\text{O}\left(g\right)\stackrel{}{\rightleftharpoons }2\text{HOCl}\left(g\right)$

Refer to Table $\text{3-3}$.

The value of bond energy, $D°\left(\text{kJ/mol}\right)$, for the given reactant and product is,

Bonds	$D°\left(\text{kJ/mol}\right)$
$\text{O}-\text{H}$	$\text{467}$
$\text{Cl}-\text{O}$	$\text{203}$
$\text{O}-\text{Cl}$	$\text{203}$

The formula of $\Delta H°$ is,

$\Delta H°=\sum n_{\text{p}}D{°}_{\left(\text{bonds}\text{broken}\right)}-\sum n_{\text{f}}D{°}_{\left(\text{bonds}\text{formed}\right)}$

Where,

• $n_{\text{p}}$ is the number of moles of each product.
• $n_{\text{r}}$ is the number of moles each reactant.
• $D{°}_{\left(\text{bonds broken}\right)}$ is the bond energy of broken bonds.
• $D{°}_{\left(\text{bonds}\text{formed}\right)}$ is the bond energy of bonds formed.

Substitute all values from the table in the above equation.

$\begin{array}{c}\Delta H°=\sum n_{\text{p}}D{°}_{\left(\text{bonds}\text{broken}\right)}-\sum n_{\text{f}}D{°}_{\left(\text{bonds}\text{formed}\right)}\\ =\left(\text{2}\times D_{\text{O}-\text{H}}+\text{2}\times D_{\text{Cl}-\text{O}}\right)-\left(\text{2}\times D_{\text{O}-\text{H}}+\text{2}\times D_{\text{O}-\text{Cl}}\right)\\ =\left[\left(\text{2}\times \text{467}\right)+\left(\text{2}\times \text{203}\right)-\left\{\left(\text{2}\times \text{467}\right)+\left(\text{2}\times \text{203}\right)\right\}\right]\text{kJ/mol}\end{array}$

Simplify the above equation.

$\begin{array}{c}\Delta H°=\left[\left(\text{2}\times \text{467}\right)+\left(\text{2}\times \text{203}\right)-\left\{\left(\text{2}\times \text{467}\right)+\left(\text{2}\times \text{203}\right)\right\}\right]\text{kJ/mol}\\ =\underset{\_}{0kJ/mol}\end{array}$

Interpretation Introduction

Interpretation: The reaction corresponding to the formation of $\text{HOCl}\left(g\right)$, its equilibrium constant at $\text{298}\text{K}$ and values of $\Delta G_{\text{f}}°,\Delta H_{\text{f}}°,S°$ are given. The questions are to be answered on the basis of the given data.

Concept introduction: Equilibrium constant, $K$, is defined as the ratio of the concentration of products to that of the reactants at equilibrium. If a given reaction is at equilibrium, the free energy change is,

$\begin{array}{c}\Delta G=\text{0}\\ Q=K\end{array}$

The expression for free energy change is,

$\Delta G°=-RT\ln \left(K\right)$

The formula of $\Delta S°$ is,

$\Delta S°=\sum n_{\text{p}}\Delta S{°}_{\left(\text{product}\right)}-\sum n_{\text{f}}\Delta S{°}_{\left(\text{reactant}\right)}$

To determine: The value of $\Delta G°$ for the given reaction, using the equation $\Delta G°=-RT\ln \left(K\right)$.

Explanation of Solution

Explanation

The value of $\Delta G°$ is $6.0\times {10}^3\text{J/mol}$.

The value of $\Delta H°$ is $0\text{J/mol}$.

Temperature is $298\text{K}$.

The formula of $\Delta G°$ is,

$\Delta G°=\Delta H°-T\Delta S°$

Where,

• $\Delta H°$ is the standard enthalpy of reaction.
• $\Delta G°$ is the free energy change.
• $T$ is the given temperature.
• $\Delta S°$ is the standard entropy of the reaction.

Substitute the values of $\Delta G°,\Delta H°$ and $T$ in equation (1).

$\begin{array}{c}\Delta G°=\Delta H°-T\Delta S°\\ \text{6}\text{.0}\times {\text{10}}^{\text{3}}\text{J/mol}=\text{0}-\text{298}\left(\Delta S°\right)\\ \Delta S°=\underset{\_}{-20.13J/K\cdot mol}\end{array}$

Interpretation Introduction

Interpretation: The reaction corresponding to the formation of $\text{HOCl}\left(g\right)$, its equilibrium constant at $\text{298}\text{K}$ and values of $\Delta G_{\text{f}}°,\Delta H_{\text{f}}°,S°$ are given. The questions are to be answered on the basis of the given data.

Concept introduction: Equilibrium constant, $K$, is defined as the ratio of the concentration of products to that of the reactants at equilibrium. If a given reaction is at equilibrium, the free energy change is,

$\begin{array}{c}\Delta G=\text{0}\\ Q=K\end{array}$

The expression for free energy change is,

$\Delta G°=-RT\ln \left(K\right)$

The formula of $\Delta S°$ is,

$\Delta S°=\sum n_{\text{p}}\Delta S{°}_{\left(\text{product}\right)}-\sum n_{\text{f}}\Delta S{°}_{\left(\text{reactant}\right)}$

To determine: The value of $\Delta G°$ for the given reaction, using the equation $\Delta G°=-RT\ln \left(K\right)$.

Explanation of Solution

Explanation

Refer Appendix $4$.

The value of standard enthalpy of ${\text{H}}_{\text{2}}\text{O}$ is $-241.82\text{kJ/mol}$.

The value of standard enthalpy of ${\text{Cl}}_{\text{2}}\text{O}$ is $80.3\text{kJ/mol}$.

The formula of $\Delta H°$ is,

$\Delta H°=\sum n_{\text{p}}\Delta H{°}_{\left(\text{product}\right)}-\sum n_{\text{f}}\Delta H{°}_{\left(\text{reactant}\right)}$

Where,

• $\Delta H°$ is the standard enthalpy of reaction.
• $n_{\text{p}}$ is the number of moles of each product.
• $n_{\text{r}}$ is the number of moles each reactant.
• $\Delta H{°}_{\left(\text{product}\right)}$ is the standard enthalpy of product at a pressure of $\text{1}\text{atm}$.
• $\Delta H{°}_{\left(\text{reactant}\right)}$ is the standard enthalpy of reactant at a pressure of $\text{1}\text{atm}$.

Substitute all values from the table in the above equation.

$\begin{array}{c}\Delta H°=\sum n_{\text{p}}\Delta H{°}_{\left(\text{product}\right)}-\sum n_{\text{f}}\Delta H{°}_{\left(\text{reactant}\right)}\\ =\left[\text{2}\left(\Delta H{°}_{\text{HOCl}}\right)-\left(-\text{241}\text{.82}\right)-\left(\text{80}\text{.3}\right)\right]\text{kJ/mol}\\ =\underset{\_}{-80.76kJ/mol}\end{array}$

Refer Appendix $4$.

The value of standard entropy of ${\text{H}}_{\text{2}}\text{O}$ is $188.83\text{J/K}\cdot \text{mol}$.

The value of standard entropy of ${\text{Cl}}_{\text{2}}\text{O}$ is $266.1\text{J/K}\cdot \text{mol}$.

The formula of $\Delta S°$ is,

$\Delta S°=\sum n_{\text{p}}\Delta S{°}_{\left(\text{product}\right)}-\sum n_{\text{f}}\Delta S{°}_{\left(\text{reactant}\right)}$

Where,

• $\Delta S°$ is the standard enthalpy of reaction.
• $n_{\text{p}}$ is the number of moles of each product.
• $n_{\text{r}}$ is the number of moles each reactant.
• $\Delta S{°}_{\left(\text{product}\right)}$ is the standard entropy of the product at a pressure of $\text{1}\text{atm}$.
• $\Delta S{°}_{\left(\text{reactant}\right)}$ is the standard entropy of the reactant at a pressure of $\text{1}\text{atm}$.

Substitute all values from the table in the above equation.

$\begin{array}{c}\Delta S°=\sum n_{\text{p}}\Delta S{°}_{\left(\text{product}\right)}-\sum n_{\text{f}}\Delta S{°}_{\left(\text{reactant}\right)}\\ -\text{20}\text{.13}\text{J/K}\cdot \text{mol}=\left[\text{2}\left(\Delta S{°}_{\text{HOCl}}\right)-\left(\text{188}\text{.83}\right)-\left(\text{266}\text{.1}\right)\right]\text{J/K}\cdot \text{mol}\\ \Delta S{°}_{\text{HOCl}}=\underset{\_}{-217.4J/K\cdot mol}\end{array}$

Interpretation Introduction

Interpretation: The reaction corresponding to the formation of $\text{HOCl}\left(g\right)$, its equilibrium constant at $\text{298}\text{K}$ and values of $\Delta G_{\text{f}}°,\Delta H_{\text{f}}°,S°$ are given. The questions are to be answered on the basis of the given data.

Concept introduction: Equilibrium constant, $K$, is defined as the ratio of the concentration of products to that of the reactants at equilibrium. If a given reaction is at equilibrium, the free energy change is,

$\begin{array}{c}\Delta G=\text{0}\\ Q=K\end{array}$

The expression for free energy change is,

$\Delta G°=-RT\ln \left(K\right)$

The formula of $\Delta S°$ is,

$\Delta S°=\sum n_{\text{p}}\Delta S{°}_{\left(\text{product}\right)}-\sum n_{\text{f}}\Delta S{°}_{\left(\text{reactant}\right)}$

To determine: The value of $\Delta G°$ for the given reaction, using the equation $\Delta G°=-RT\ln \left(K\right)$.

Explanation of Solution

Explanation

The value of $\Delta S°$ is $-20.3\text{J/K}\cdot \text{mol}$.

The value of $\Delta H°$ is $0\text{J/mol}$.

Temperature is $500\text{K}$.

It is assumed that $\Delta H°$ and $\Delta S°$ are independent of $T$.

The formula of $\Delta G°$ is,

$\Delta G°=\Delta H°-T\Delta S°$

Where,

• $\Delta H°$ is the standard enthalpy of reaction.
• $\Delta G°$ is the free energy change.
• $T$ is the given temperature.
• $\Delta S°$ is the standard entropy of the reaction.

Substitute the values of $\Delta G°,\Delta H°$ and $T$ in equation (1).

$\begin{array}{c}\Delta G°=\Delta H°-T\Delta S°\\ =\text{0}-\text{500}\left(-\text{20}\text{.13}\text{J/K}\cdot \text{mol}\right)\\ =\underset{\_}{1.01\times {10}^{4}J/mol}\end{array}$

Formula

The expression for free energy change is,

$\begin{array}{c}\Delta G°=-RT\ln \left(K\right)\\ \ln \left(K\right)=-\frac{\Delta G°}{RT}\end{array}$

Where,

• $\Delta G$ is the free energy change for a reaction at specified pressure.
• $R$ is the gas law constant $(\text{8}\text{.3145}\text{J/K}\cdot \text{mol})$.
• $K$ is the equilibrium constant.
•   $T$ is the absolute temperature.

Substitute the values of $R,\Delta G°$ and $T$ in the above expression.

$\begin{array}{c}\ln \left(K\right)=-\frac{\Delta G°}{RT}\\ =-\frac{1.01\times {\text{10}}^{\text{4}}\text{J/mol}}{(\text{8}\text{.3145}\text{J/K}\cdot \text{mol})\left(500\text{K}\right)}\\ K=\underset{\_}{0.088}\end{array}$

Interpretation Introduction

Interpretation: The reaction corresponding to the formation of $\text{HOCl}\left(g\right)$, its equilibrium constant at $\text{298}\text{K}$ and values of $\Delta G_{\text{f}}°,\Delta H_{\text{f}}°,S°$ are given. The questions are to be answered on the basis of the given data.

Concept introduction: Equilibrium constant, $K$, is defined as the ratio of the concentration of products to that of the reactants at equilibrium. If a given reaction is at equilibrium, the free energy change is,

$\begin{array}{c}\Delta G=\text{0}\\ Q=K\end{array}$

The expression for free energy change is,

$\Delta G°=-RT\ln \left(K\right)$

The formula of $\Delta S°$ is,

$\Delta S°=\sum n_{\text{p}}\Delta S{°}_{\left(\text{product}\right)}-\sum n_{\text{f}}\Delta S{°}_{\left(\text{reactant}\right)}$

To determine: The value of $\Delta G°$ for the given reaction, using the equation $\Delta G°=-RT\ln \left(K\right)$.

Explanation of Solution

Explanation

Given

Partial pressure of ${\text{H}}_{\text{2}}\text{O}$, $P_{{\text{H}}_{\text{2}}\text{O}}$, is $18\text{torr}$.

Partial pressure of ${\text{Cl}}_{\text{2}}\text{O}$, $P_{{\text{Cl}}_{\text{2}}\text{O}}$, is $2.0\text{torr}$.

Partial pressure of $\text{HOCl}$, $P_{\text{HOCl}}$, is $0.10\text{torr}$.

The stated reaction is,

${\text{H}}_{\text{2}}\text{O}\left(g\right)+{\text{Cl}}_{\text{2}}\text{O}\left(g\right)\stackrel{}{\rightleftharpoons }2\text{HOCl}\left(g\right)$

If the partial pressure of reactant and product is given, then equilibrium pressure is expressed as $K_{\text{p}}$ and its expression is written as,

$K_{\text{p}}=\frac{{\left(\text{Partial}\text{pressure of product}\right)}^a}{{\left(\text{Partial}\text{pressure of reactant}\right)}^b}$

Where,

• $a$ is the number of moles of product.
• $b$ is the number of moles of reactant.

The equilibrium constant expression for the given reaction is,

$K_{\text{p}}=\frac{{\left(P_{\text{HOCl}}\right)}^2}{\left(P_{{\text{H}}_{\text{2}}\text{O}}\right)\left(P_{{\text{Cl}}_{\text{2}}\text{O}}\right)}$

Substitute the values of $P_{{\text{Cl}}_{\text{2}}\text{O}},P_{{\text{H}}_{\text{2}}\text{O}}$ and $P_{\text{HOCl}}$ in the above expression.

$\begin{array}{c}{K}_{\text{p}}=\frac{{\left(P_{\text{HOCl}}\right)}^2}{\left(P_{{\text{H}}_{\text{2}}\text{O}}\right)\left(P_{{\text{Cl}}_{\text{2}}\text{O}}\right)}\\ =\frac{{\left(0.10\text{torr}\right)}^2}{\left(18\text{torr}\right)\left(2.0\text{torr}\right)}\\ =\underset{\_}{2.7\times 10^{-4}}\end{array}$

The value of $K_{\text{p}}$ is $2.7\times {10}^{-4}$.

Temperature is $\text{298}\text{K}$.

The expression for free energy change is,

$\Delta G=\Delta G°+RT\ln \left(K_{\text{p}}\right)$

Where,

• $\Delta G$ is the free energy change for a reaction at specified pressure.
• $R$ is the gas law constant $(\text{8}\text{.3145}\text{J/K}\cdot \text{mol})$.
• $T$ is the absolute temperature.
• $K_{\text{p}}$ is the equilibrium constant.

Substitute the values of $R,K_{\text{p}}$ and $T$ in the equation.

$\begin{array}{c}\Delta G=\Delta G°+RT\ln \left(K_{\text{p}}\right)\\ =\left[6.0\times {10}^3+\left(8.3145\right)\left(298\right)\ln \left(2.7\times {10}^{-4}\right)\right]\text{J/mol}\\ =-14.3\times {10}^3\text{J/mol}\end{array}$

The conversion of joule-per mole $\left(\text{J/mol}\right)$ into kilo-joule per mole $\left(\text{kJ/mol}\right)$ is done as,

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

Hence,

The conversion of $-14.3\times {10}^3\text{J/mol}$ into kilo- joule per mole is,

$\begin{array}{c}-\text{14}\text{.3}\times {\text{10}}^{\text{3}}\text{J/mol}=\left(-\text{14}\text{.3}\times {\text{10}}^{\text{3}}\times {\text{10}}^{-\text{3}}\right)\text{kJ/mol}\\ =\underset{\_}{-14.3kJ/mol}\end{array}$