Chapter 4, Problem 4D.21E

(a)

Interpretation Introduction

Interpretation:

The standard enthalpy of the final stage of the production of nitric acid has to be determined.

Concept Introduction:

The standard enthalpy of the reaction is the difference of the sum of standard enthalpy of formation of all the products and the sum of standard enthalpy of formation of all the reactants.  The mathematical equation for the calculation of standard enthalpy of the reaction is shown below.

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

Answer to Problem 4D.21E

The standard enthalpy of the final stage of the production of nitric acid is $\underset{\_}{-138.18kJ}$.

Explanation of Solution

The balanced chemical equation that represents the final stage of the production of nitric acid is shown below.

    $3N{O}_2\left(g\right)+H_{\text{2}}O\left(l\right)\to 2HN{O}_{\text{3}}\left(aq\right)+NO\left(g\right)$

The standard enthalpy of formation of products that are $HN{O}_{\text{3}}\left(aq\right)$ and $NO\left(g\right)$ is $-\text{207}\text{.36}\text{kJ}\cdot {\text{mol}}^{-1}$ and $+90.25\text{kJ}\cdot {\text{mol}}^{-1}$ respectively.

The total enthalpy of formation of products is calculated by the expression shown below.

  ${\displaystyle \sum \text{n}\Delta {\text{H}}_f^{\circ }\left(products\right)}={\text{n}}_{HN{O}_{\text{3}}}\Delta {\text{H}}_f^{\circ }\left(HN{O}_{\text{3}},aq\right)+{\text{n}}_{NO}\Delta {\text{H}}_f^{\circ }\left(NO,g\right)$        (1)

Where,

• ${\displaystyle \sum \text{n}\Delta {\text{H}}_f^{\circ }\left(products\right)}$ is total enthalpy of formation of products.
• ${\text{n}}_{HN{O}_{\text{3}}}$ is the number of moles of $HN{O}_{\text{3}}\left(aq\right)$.
• $\Delta {\text{H}}_f^{\circ }\left(HN{O}_{\text{3}},aq\right)$ is the standard enthalpy of formation of $HN{O}_{\text{3}}\left(aq\right)$.
• ${\text{n}}_{NO}$ is the number of moles of $NO\left(g\right)$.
• $\Delta {\text{H}}_f^{\circ }\left(NO,g\right)$ is the standard enthalpy of formation of $NO\left(g\right)$.

The value of ${\text{n}}_{HN{O}_{\text{3}}}$ is $2mol$.

The value of $\Delta {\text{H}}_f^{\circ }\left(HN{O}_{\text{3}},aq\right)$ is $-\text{207}\text{.36}\text{kJ}\cdot {\text{mol}}^{-1}$.

The value of ${\text{n}}_{NO}$ is $1mol$.

The value of $\Delta {\text{H}}_f^{\circ }\left(NO,g\right)$ is $+90.25\text{kJ}\cdot {\text{mol}}^{-1}$.

Substitute the value of ${\text{n}}_{HN{O}_{\text{3}}}$, $\Delta {\text{H}}_f^{\circ }\left(HN{O}_{\text{3}},aq\right)$, ${\text{n}}_{NO}$ and $\Delta {\text{H}}_f^{\circ }\left(NO,g\right)$ in equation (1).

  $\begin{array}{c}{\displaystyle \sum \text{n}\Delta {\text{H}}_f^{\circ }\left(products\right)}=\left(2mol\right)\times \left(-\text{207}\text{.36}\text{kJ}\cdot {\text{mol}}^{-1}\right)+\left(1mol\right)\times \left(+90.25\text{kJ}\cdot {\text{mol}}^{-1}\right)\\ =-414.72\text{kJ}+90.25\text{kJ}\\ =-324.47\text{kJ}\end{array}$

The standard enthalpy of formation of reactant that is $N{O}_2\left(g\right)$ and $H_{\text{2}}O\left(l\right)$ is $33.18\text{kJ}\cdot {\text{mol}}^{-1}$ and $-285.83\text{kJ}\cdot {\text{mol}}^{-1}$ respectively.

The total enthalpy of formation of reactants is calculated by the expression shown below.

  ${\displaystyle \sum \text{n}\Delta {\text{H}}_f^{\circ }}\left(\text{reactants}\right)={\text{n}}_{N{O}_2}\Delta {\text{H}}_f^{\circ }\left(N{O}_2,g\right)+{\text{n}}_{H_{\text{2}}O}\Delta {\text{H}}_f^{\circ }\left(H_{\text{2}}O,l\right)$        (2)

Where,

• ${\displaystyle \sum \text{n}\Delta {\text{H}}_f^{\circ }}\left(\text{reactants}\right)$ is total enthalpy of formation of reactants.
• ${\text{n}}_{N{O}_2}$ is the number of moles of $N{O}_2\left(g\right)$.
• $\Delta {\text{H}}_f^{\circ }\left(N{O}_2,g\right)$ is the standard enthalpy of formation of $N{O}_2\left(g\right)$.
• ${\text{n}}_{H_{\text{2}}O}$ is the number of moles of $H_{\text{2}}O\left(l\right)$.
• $\Delta {\text{H}}_f^{\circ }\left(H_{\text{2}}O,l\right)$ is the standard enthalpy of formation of $H_{\text{2}}O\left(l\right)$.

The value of ${\text{n}}_{N{O}_2}$ is $3mol$.

The value of $\Delta {\text{H}}_f^{\circ }\left(N{O}_2,g\right)$ is $33.18\text{kJ}\cdot {\text{mol}}^{-1}$.

The value of ${\text{n}}_{H_{\text{2}}O}$ is $1mol$.

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

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

  $\begin{array}{c}{\displaystyle \sum \text{n}\Delta {\text{H}}_f^{\circ }\left(reac\tan ts\right)}=\left(3mol\right)\times \left(33.18\text{kJ}\cdot {\text{mol}}^{-1}\right)+\left(1mol\right)\times \left(-285.83\text{kJ}\cdot {\text{mol}}^{-1}\right)\\ =99.54\text{kJ}-285.83\text{kJ}\\ =-\text{186}\text{.29}\text{kJ}\end{array}$

The standard enthalpy of formation of ethyne is calculated by the expression shown below.

  $\Delta \text{H}°={\displaystyle \sum \text{n}\Delta {\text{H}}_f^{\circ }\left(products\right)}-{\displaystyle \sum \text{n}\Delta {\text{H}}_f^{\circ }}\left(\text{reactants}\right)$        (3)

Where,

• $\Delta \text{H}°$ is standard enthalpy of reaction.
• ${\displaystyle \sum \text{n}\Delta {\text{H}}_f^{\circ }\left(products\right)}$ is total enthalpy of formation of products.
• ${\displaystyle \sum \text{n}\Delta {\text{H}}_f^{\circ }}\left(\text{reactants}\right)$ is total enthalpy of formation of reactants.

The value of ${\displaystyle \sum \text{n}\Delta {\text{H}}_f^{\circ }\left(products\right)}$ is $-324.47\text{kJ}$.

The value of ${\displaystyle \sum \text{n}\Delta {\text{H}}_f^{\circ }}\left(\text{reactants}\right)$ is $-\text{186}\text{.29}\text{kJ}$.

Substitute the value of ${\displaystyle \sum \text{n}\Delta {\text{H}}_f^{\circ }\left(products\right)}$ and ${\displaystyle \sum \text{n}\Delta {\text{H}}_f^{\circ }}\left(\text{reactants}\right)$ in equation (3).

  $\begin{array}{c}\Delta \text{H}°=-324.47\text{kJ}-\left(-\text{186}\text{.29}\text{kJ}\right)\\ =-324.47\text{kJ}+\text{186}\text{.29}\text{kJ}\\ =-138.18kJ\end{array}$

Thus, the standard enthalpy of the final stage of the production of nitric acid is $\underset{\_}{-138.18kJ}$.

(b)

Interpretation Introduction

Interpretation:

The standard enthalpy of the industrial synthesis of boron trifluoride has to be determined.

Concept Introduction:

Same as part (a).

Answer to Problem 4D.21E

The standard enthalpy of the industrial synthesis of boron trifluoride is $\underset{\_}{1593.91kJ}$.

Explanation of Solution

The balanced chemical equation that represents the industrial synthesis of boron trifluoride is shown below.

    $B_2{O}_3\left(s\right)+3Ca{F}_{\text{2}}\left(s\right)\to 2B{F}_{\text{3}}\left(g\right)+3CaO\left(s\right)$

The standard enthalpy of formation of products that are $B{F}_{\text{3}}\left(g\right)$ and $CaO\left(s\right)$ is $-\text{1136}\text{.8}\text{kJ}\cdot {\text{mol}}^{-1}$ and $-635.09\text{kJ}\cdot {\text{mol}}^{-1}$ respectively.

The total enthalpy of formation of products is calculated by the expression shown below.

  ${\displaystyle \sum \text{n}\Delta {\text{H}}_f^{\circ }\left(products\right)}={\text{n}}_{B{F}_{\text{3}}}\Delta {\text{H}}_f^{\circ }\left(B{F}_{\text{3}},g\right)+{\text{n}}_{CaO}\Delta {\text{H}}_f^{\circ }\left(CaO,s\right)$        (4)

Where,

• ${\displaystyle \sum \text{n}\Delta {\text{H}}_f^{\circ }\left(products\right)}$ is total enthalpy of formation of products.
• ${\text{n}}_{B{F}_{\text{3}}}$ is the number of moles of $B{F}_{\text{3}}\left(g\right)$.
• $\Delta {\text{H}}_f^{\circ }\left(B{F}_{\text{3}},g\right)$ is the standard enthalpy of formation of $B{F}_{\text{3}}\left(g\right)$.
• ${\text{n}}_{CaO}$ is the number of moles of $CaO\left(s\right)$.
• $\Delta {\text{H}}_f^{\circ }\left(CaO,s\right)$ is the standard enthalpy of formation of $CaO\left(s\right)$.

The value of ${\text{n}}_{B{F}_{\text{3}}}$ is $2mol$.

The value of $\Delta {\text{H}}_f^{\circ }\left(B{F}_{\text{3}},g\right)$ is $-\text{1136}\text{.8}\text{kJ}\cdot {\text{mol}}^{-1}$.

The value of ${\text{n}}_{CaO}$ is $1mol$.

The value of $\Delta {\text{H}}_f^{\circ }\left(CaO,g\right)$ is $-635.09\text{kJ}\cdot {\text{mol}}^{-1}$.

Substitute the value of ${\text{n}}_{B{F}_{\text{3}}}$, $\Delta {\text{H}}_f^{\circ }\left(B{F}_{\text{3}},g\right)$, ${\text{n}}_{CaO}$ and $\Delta {\text{H}}_f^{\circ }\left(CaO,s\right)$ in equation (4).

  $\begin{array}{c}{\displaystyle \sum \text{n}\Delta {\text{H}}_f^{\circ }\left(products\right)}=\left(2mol\right)\times \left(-\text{1136}\text{.8}\text{kJ}\cdot {\text{mol}}^{-1}\right)+\left(1mol\right)\times \left(-635.09\text{kJ}\cdot {\text{mol}}^{-1}\right)\\ =-2273.6\text{kJ}-635.09\text{kJ}\cdot {\text{mol}}^{-1}\\ =-2908.69\text{kJ}\end{array}$

The standard enthalpy of formation of reactant that is $B_2{O}_3\left(s\right)$ and $Ca{F}_{\text{2}}\left(s\right)$ is $-843.8\text{kJ}\cdot {\text{mol}}^{-1}$ and $-1219.6\text{kJ}\cdot {\text{mol}}^{-1}$ respectively.

The total enthalpy of formation of reactants is calculated by the expression shown below.

  ${\displaystyle \sum \text{n}\Delta {\text{H}}_f^{\circ }\left(\text{reactants}\right)}={\text{n}}_{B_2{O}_3}\Delta {\text{H}}_f^{\circ }\left(B_2{O}_3,s\right)+{\text{n}}_{Ca{F}_{\text{2}}}\Delta {\text{H}}_f^{\circ }\left(Ca{F}_{\text{2}},s\right)$        (5)

Where,

• ${\displaystyle \sum \text{n}\Delta {\text{H}}_f^{\circ }\left(\text{reactants}\right)}$ is total enthalpy of formation of reactants.
• ${\text{n}}_{B_2{O}_3}$ is the number of moles of $B_2{O}_3\left(s\right)$.
• $\Delta {\text{H}}_f^{\circ }\left(B_2{O}_3,s\right)$ is the standard enthalpy of formation of $B_2{O}_3\left(s\right)$.
• ${\text{n}}_{Ca{F}_{\text{2}}}$ is the number of moles of $Ca{F}_{\text{2}}\left(s\right)$.
• $\Delta {\text{H}}_f^{\circ }\left(Ca{F}_{\text{2}},s\right)$ is the standard enthalpy of formation of $Ca{F}_{\text{2}}\left(s\right)$.

The value of ${\text{n}}_{B_2{O}_3}$ is $1mol$.

The value of $\Delta {\text{H}}_f^{\circ }\left(B_2{O}_3,s\right)$ is $-843.8\text{kJ}\cdot {\text{mol}}^{-1}$.

The value of ${\text{n}}_{Ca{F}_{\text{2}}}$ is $3mol$.

The value of $\Delta {\text{H}}_f^{\circ }\left(Ca{F}_{\text{2}},s\right)$ is $-1219.6\text{kJ}\cdot {\text{mol}}^{-1}$.

Substitute the value of ${\text{n}}_{B_2{O}_3}$, $\Delta {\text{H}}_f^{\circ }\left(B_2{O}_3,s\right)$, ${\text{n}}_{Ca{F}_{\text{2}}}$ and $\Delta {\text{H}}_f^{\circ }\left(Ca{F}_{\text{2}},s\right)$ in equation (5).

  $\begin{array}{c}{\displaystyle \sum \text{n}\Delta {\text{H}}_f^{\circ }\left(\text{reactants}\right)}=\left(1mol\right)\times \left(-843.8\text{kJ}\cdot {\text{mol}}^{-1}\right)+\left(3mol\right)\times \left(-1219.6\text{kJ}\cdot {\text{mol}}^{-1}\right)\\ =-843.8\text{kJ}-3658.8\text{kJ}\\ =-\text{4502}\text{.6}\text{kJ}\end{array}$

The value of ${\displaystyle \sum \text{n}\Delta {\text{H}}_f^{\circ }\left(products\right)}$ is $-2908.69\text{kJ}$.

The value of ${\displaystyle \sum \text{n}\Delta {\text{H}}_f^{\circ }\left(\text{reactants}\right)}$ is $-\text{4502}\text{.6}\text{kJ}$.

Substitute the value of ${\displaystyle \sum \text{n}\Delta {\text{H}}_f^{\circ }\left(products\right)}$ and ${\displaystyle \sum \text{n}\Delta {\text{H}}_f^{\circ }\left(\text{reactants}\right)}$ in equation (3).

  $\begin{array}{c}\Delta \text{H}°=-2908.69\text{kJ}-\left(-\text{4502}\text{.6}\text{kJ}\right)\\ =-2908.69\text{kJ}+\text{4502}\text{.6}\text{kJ}\\ =\text{1593}\text{.91}\text{kJ}\end{array}$

Thus, the standard enthalpy of the industrial synthesis of boron trifluoride is $\underset{\_}{1593.91kJ}$.

(c)

Interpretation Introduction

Interpretation:

The standard enthalpy of the formation of sulfide by the action of hydrogen sulfide on a aqueous solution of a base has to be determined.

Concept Introduction:

Same as part (a).

Answer to Problem 4D.21E

The standard enthalpy of the formation of sulfide by the action of hydrogen sulfide on a aqueous solution of a base is $\underset{\_}{15.28kJ}$.

Explanation of Solution

The balanced chemical equation that represents the formation of sulfide by the action of hydrogen sulfide on a aqueous solution of a baseis shown below.

    $H_{2}S\left(aq\right)+2KOH\left(aq\right)\to K_{\text{2}}S\left(aq\right)+2H_{2}O\left(l\right)$

The standard enthalpy of formation of products that are $K_{\text{2}}S\left(aq\right)$ and $H_{2}O\left(l\right)$ is $-\text{417}\text{.5}\text{kJ}\cdot {\text{mol}}^{-1}$ and $-285.83\text{kJ}\cdot {\text{mol}}^{-1}$ respectively.

The total enthalpy of formation of products is calculated by the expression shown below.

  ${\displaystyle \sum \text{n}\Delta {\text{H}}_f^{\circ }\left(products\right)}={\text{n}}_{K_{\text{2}}S}\Delta {\text{H}}_f^{\circ }\left(K_{\text{2}}S,aq\right)+{\text{n}}_{H_{2}O}\Delta {\text{H}}_f^{\circ }\left(H_{2}O,l\right)$        (6)

Where,

• ${\displaystyle \sum \text{n}\Delta {\text{H}}_f^{\circ }\left(products\right)}$ is total enthalpy of formation of products.
• ${\text{n}}_{K_{\text{2}}S}$ is the number of moles of $K_{\text{2}}S\left(aq\right)$.
• $\Delta {\text{H}}_f^{\circ }\left(K_{\text{2}}S,aq\right)$ is the standard enthalpy of formation of $K_{\text{2}}S\left(aq\right)$.
• ${\text{n}}_{H_{2}O}$ is the number of moles of $H_{2}O\left(l\right)$.
• $\Delta {\text{H}}_f^{\circ }\left(H_{2}O,l\right)$ is the standard enthalpy of formation of $H_{2}O\left(l\right)$.

The value of ${\text{n}}_{K_{\text{2}}S}$ is $1mol$.

The value of $\Delta {\text{H}}_f^{\circ }\left(K_{\text{2}}S,aq\right)$ is $-\text{417}\text{.5}\text{kJ}\cdot {\text{mol}}^{-1}$.

The value of ${\text{n}}_{H_{2}O}$ is $2mol$.

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 ${\text{n}}_{K_{\text{2}}S}$, $\Delta {\text{H}}_f^{\circ }\left(K_{\text{2}}S,aq\right)$, ${\text{n}}_{H_{2}O}$ and $\Delta {\text{H}}_f^{\circ }\left(H_{2}O,l\right)$ in equation (6).

  $\begin{array}{c}{\displaystyle \sum \text{n}\Delta {\text{H}}_f^{\circ }\left(products\right)}=\left(1mol\right)\times \left(-417.5\text{kJ}\cdot {\text{mol}}^{-1}\right)+\left(2mol\right)\times \left(-285.83\text{kJ}\cdot {\text{mol}}^{-1}\right)\\ =-417.5\text{kJ}-571.66\text{kJ}\\ =-989.16\text{kJ}\end{array}$

The standard enthalpy of formation of reactant that is $H_{2}S\left(aq\right)$ and $KOH\left(aq\right)$ is $-39.7\text{kJ}\cdot {\text{mol}}^{-1}$ and $-482.37\text{kJ}\cdot {\text{mol}}^{-1}$ respectively.

The total enthalpy of formation of reactants is calculated by the expression shown below.

  ${\displaystyle \sum \text{n}\Delta {\text{H}}_f^{\circ }\left(\text{reactants}\right)}={\text{n}}_{H_{2}S}\Delta {\text{H}}_f^{\circ }\left(H_{2}S,aq\right)+{\text{n}}_{KOH}\Delta {\text{H}}_f^{\circ }\left(KOH,aq\right)$        (7)

Where,

• ${\displaystyle \sum \text{n}\Delta {\text{H}}_f^{\circ }\left(\text{reactants}\right)}$ is total enthalpy of formation of reactants.
• ${\text{n}}_{H_{2}S}$ is the number of moles of $H_{2}S\left(aq\right)$.
• $\Delta {\text{H}}_f^{\circ }\left(H_{2}S,aq\right)$ is the standard enthalpy of formation of $H_{2}S\left(aq\right)$.
• ${\text{n}}_{KOH}$ is the number of moles of $KOH\left(aq\right)$.
• $\Delta {\text{H}}_f^{\circ }\left(KOH,aq\right)$ is the standard enthalpy of formation of $KOH\left(aq\right)$.

The value of ${\text{n}}_{H_{2}S}$ is $1mol$.

The value of $\Delta {\text{H}}_f^{\circ }\left(H_{2}S,aq\right)$ is $-39.7\text{kJ}\cdot {\text{mol}}^{-1}$.

The value of ${\text{n}}_{KOH}$ is $2mol$.

The value of $\Delta {\text{H}}_f^{\circ }\left(KOH,aq\right)$ is $-482.37\text{kJ}\cdot {\text{mol}}^{-1}$.

Substitute the value of ${\text{n}}_{H_{2}S}$, $\Delta {\text{H}}_f^{\circ }\left(H_{2}S,aq\right)$, ${\text{n}}_{KOH}$ and $\Delta {\text{H}}_f^{\circ }\left(KOH,aq\right)$ in equation (7).

  $\begin{array}{c}{\displaystyle \sum \text{n}\Delta {\text{H}}_f^{\circ }\left(\text{reactants}\right)}=\left(1mol\right)\times \left(-39.7\text{kJ}\cdot {\text{mol}}^{-1}\right)+\left(2mol\right)\times \left(-482.37\text{kJ}\cdot {\text{mol}}^{-1}\right)\\ =-39.7\text{kJ}-964.74\text{kJ}\\ =-\text{1004}\text{.44}\text{kJ}\end{array}$

The value of ${\displaystyle \sum \text{n}\Delta {\text{H}}_f^{\circ }\left(products\right)}$ is $-989.16\text{kJ}$.

The value of ${\displaystyle \sum \text{n}\Delta {\text{H}}_f^{\circ }\left(\text{reactants}\right)}$ is $-\text{1004}\text{.44}\text{kJ}$.

Substitute the value of ${\displaystyle \sum \text{n}\Delta {\text{H}}_f^{\circ }\left(products\right)}$ and ${\displaystyle \sum \text{n}\Delta {\text{H}}_f^{\circ }\left(\text{reactants}\right)}$ in equation (3).

  $\begin{array}{c}\Delta \text{H}°=-989.16\text{kJ}-\left(-\text{1004}\text{.44}\text{kJ}\right)\\ =-989.16\text{kJ}+\text{1004}\text{.44}\text{kJ}\\ =\text{15}\text{.28}\text{kJ}\end{array}$

Thus, the standard enthalpy of the formation of sulfide by the action of hydrogen sulfide on a aqueous solution of a base is $\underset{\_}{15.28kJ}$.