Chapter 4, Problem 4D.22E

(a)

Interpretation Introduction

Interpretation:

The standard enthalpy of the removal of hydrogen sulfide from natural gas 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.22E

The standard enthalpy of the removal of hydrogen sulfide from natural gas is $\underset{\_}{-354.23kJ}$.

Explanation of Solution

The balanced chemical equation that represents the removal of hydrogen sulfide from natural gas is shown below.

    $2H_{2}S\left(g\right)+S{O}_{\text{2}}\left(g\right)\to 3S\left(s\right)+2H_{2}O\left(l\right)$

The standard enthalpy of formation of products that are $S\left(s\right)$ and $H_{\text{2}}O\left(l\right)$ is $0.0\text{kJ}\cdot {\text{mol}}^{-1}$ and $-285.83\text{kJ}\cdot {\text{mol}}^{-1}$ respectively.

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

  ${\displaystyle \sum \text{n}\Delta {\text{H}}_f^{\circ }\left(products\right)}={\text{n}}_S\Delta {\text{H}}_f^{\circ }\left(S,s\right)+{\text{n}}_{H_{\text{2}}O}\Delta {\text{H}}_f^{\circ }\left(H_{\text{2}}O,l\right)$        (1)

Where,

• ${\displaystyle \sum \text{n}\Delta {\text{H}}_f^{\circ }\left(products\right)}$ is total enthalpy of formation of products.
• ${\text{n}}_S$ is the number of moles of $S\left(s\right)$.
• $\Delta {\text{H}}_f^{\circ }\left(S,s\right)$ is the standard enthalpy of formation of $S\left(s\right)$.
• ${\text{n}}_{H_{\text{2}}O}$ 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 ${\text{n}}_S$ is $3mol$.

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

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

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 ${\text{n}}_S$, $\Delta {\text{H}}_f^{\circ }\left(S,s\right)$, ${\text{n}}_{H_{\text{2}}O}$ and $\Delta {\text{H}}_f^{\circ }\left(H_{\text{2}}O,l\right)$ in equation (1).

  $\begin{array}{c}{\displaystyle \sum \text{n}\Delta {\text{H}}_f^{\circ }\left(products\right)}=\left(3mol\right)\times \left(0.0\text{kJ}\cdot {\text{mol}}^{-1}\right)+\left(2mol\right)\times \left(-285.83\text{kJ}\cdot {\text{mol}}^{-1}\right)\\ =0.0\text{kJ}-571.66\text{kJ}\\ =-571.66\text{kJ}\end{array}$

The standard enthalpy of formation of reactant that is $H_{2}S\left(g\right)$ and $S{O}_{\text{2}}\left(g\right)$ is $-39.7\text{kJ}\cdot {\text{mol}}^{-1}$ and $-296.83\text{kJ}\cdot {\text{mol}}^{-1}$ respectively.

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

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

Where,

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

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

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 ${\text{n}}_{S{O}_{\text{2}}}$ is $1mol$.

The value of $\Delta {\text{H}}_f^{\circ }\left(S{O}_{\text{2}},g\right)$ is $-296.83\text{kJ}\cdot {\text{mol}}^{-1}$.

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

  $\begin{array}{c}{\displaystyle \sum \text{n}\Delta {\text{H}}_f^{\circ }\left(r\text{eactants}\right)}=\left(2mol\right)\times \left(-39.7\text{kJ}\cdot {\text{mol}}^{-1}\right)+\left(1mol\right)\times \left(-296.83\text{kJ}\cdot {\text{mol}}^{-1}\right)\\ =79.4\text{kJ}-296.83\text{kJ}\\ =-217.43\text{kJ}\end{array}$

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 $-571.66\text{kJ}$.

The value of ${\displaystyle \sum \text{n}\Delta {\text{H}}_f^{\circ }}\left(\text{reactants}\right)$ is $-217.43\text{kJ}$.

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).

  $\begin{array}{c}\Delta \text{H}°=-571.66\text{kJ}-\left(-217.43\text{kJ}\right)\\ =-571.66\text{kJ}+217.43\text{kJ}\\ =-354.23kJ\end{array}$

Thus, the standard enthalpy of the removal of hydrogen sulfide from natural gas is $\underset{\_}{-354.23kJ}$.

(b)

Interpretation Introduction

Interpretation:

The standard enthalpy of the oxidation of ammonia has to be determined.

Concept Introduction:

Same as part (a).

Answer to Problem 4D.22E

The standard enthalpy of the oxidation of ammonia is $\underset{\_}{-905.48kJ}$.

Explanation of Solution

The balanced chemical equation that represents the oxidation of ammonia is shown below.

    $4N{H}_3\left(g\right)+5O_{\text{2}}\left(g\right)\to 4NO\left(g\right)+6H_{2}O\left(g\right)$

The standard enthalpy of formation of products that are $NO\left(g\right)$ and $H_{2}O\left(g\right)$ is $+90.25\text{kJ}\cdot {\text{mol}}^{-1}$ and $-241.82\text{kJ}\cdot {\text{mol}}^{-1}$ respectively.

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

  ${\displaystyle \sum \text{n}\Delta {\text{H}}_f^{\circ }\left(products\right)}={\text{n}}_{NO}\Delta {\text{H}}_f^{\circ }\left(NO,g\right)+{\text{n}}_{H_{2}O}\Delta {\text{H}}_f^{\circ }\left(H_{2}O,g\right)$        (4)

Where,

• ${\displaystyle \sum \text{n}\Delta {\text{H}}_f^{\circ }\left(products\right)}$ is total enthalpy of formation of products.
• ${\text{n}}_{NO}$ 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 $NO\left(g\right)$.
• ${\text{n}}_{H_{2}O}$ is the number of moles of $H_{2}O\left(g\right)$.
• $\Delta {\text{H}}_f^{\circ }\left(H_{2}O,g\right)$ is the standard enthalpy of formation of $H_{2}O\left(g\right)$.

The value of ${\text{n}}_{NO}$ is $4mol$.

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

The value of ${\text{n}}_{H_{2}O}$ is $6mol$.

The value of $\Delta {\text{H}}_f^{\circ }\left(H_{2}O,g\right)$ is $-241.82\text{kJ}\cdot {\text{mol}}^{-1}$.

Substitute the value of ${\text{n}}_{NO}$, $\Delta {\text{H}}_f^{\circ }\left(NO,g\right)$, ${\text{n}}_{H_{2}O}$ and $\Delta {\text{H}}_f^{\circ }\left(H_{2}O,g\right)$ in equation (4).

  $\begin{array}{c}{\displaystyle \sum \text{n}\Delta {\text{H}}_f^{\circ }\left(products\right)}=\left(4mol\right)\times \left(90.25\text{kJ}\cdot {\text{mol}}^{-1}\right)+\left(6mol\right)\times \left(-241.82\text{kJ}\cdot {\text{mol}}^{-1}\right)\\ =361\text{kJ}-1450.92\text{kJ}\cdot {\text{mol}}^{-1}\\ =-1089.92\text{kJ}\end{array}$

The standard enthalpy of formation of reactant that is $N{H}_3\left(g\right)$ and $O_{\text{2}}\left(g\right)$ is $-46.11\text{kJ}\cdot {\text{mol}}^{-1}$ and $0.0\text{kJ}\cdot {\text{mol}}^{-1}$ respectively.

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

  ${\displaystyle \sum \text{n}\Delta {\text{H}}_f^{\circ }\left(\text{reactants}\right)}={\text{n}}_{N{H}_3}\Delta {\text{H}}_f^{\circ }\left(N{H}_3,g\right)+{\text{n}}_{O_{\text{2}}}\Delta {\text{H}}_f^{\circ }\left(O_{\text{2}},g\right)$        (5)

Where,

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

The value of ${\text{n}}_{N{H}_3}$ is $4mol$.

The value of  $\Delta {\text{H}}_f^{\circ }\left(N{H}_3,g\right)$ is $-46.11\text{kJ}\cdot {\text{mol}}^{-1}$.

The value of ${\text{n}}_{O_{\text{2}}}$ is $5mol$.

The value of $\Delta {\text{H}}_f^{\circ }\left(O_{\text{2}},g\right)$ is $0.0\text{kJ}\cdot {\text{mol}}^{-1}$.

Substitute the value of ${\text{n}}_{N{H}_3}$, $\Delta {\text{H}}_f^{\circ }\left(N{H}_3,g\right)$, ${\text{n}}_{O_{\text{2}}}$ and $\Delta {\text{H}}_f^{\circ }\left(O_{\text{2}},g\right)$ in equation (5).

  $\begin{array}{c}{\displaystyle \sum \text{n}\Delta {\text{H}}_f^{\circ }\left(\text{reactants}\right)}=\left(4mol\right)\times \left(-46.11\text{kJ}\cdot {\text{mol}}^{-1}\right)+\left(5mol\right)\times \left(0.0\text{kJ}\cdot {\text{mol}}^{-1}\right)\\ =-184.44\text{kJ}\end{array}$

The value of ${\displaystyle \sum \text{n}\Delta {\text{H}}_f^{\circ }\left(products\right)}$ is $-1089.92\text{kJ}$.

The value of ${\displaystyle \sum \text{n}\Delta {\text{H}}_f^{\circ }\left(\text{reactants}\right)}$ is $-184.44\text{kJ}$.

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).

  $\begin{array}{c}\Delta \text{H}°=-1089.92\text{kJ}-\left(-184.44\text{kJ}\right)\\ =-1089.92\text{kJ}+184.44\text{kJ}\\ =-905.48kJ\end{array}$

Thus, the standard enthalpy of the oxidation of ammonia is $\underset{\_}{-905.48kJ}$.

(c)

Interpretation Introduction

Interpretation:

The standard enthalpy of the formation of phosphorous acid has to be determined.

Concept Introduction:

Same as part (a).

Answer to Problem 4D.22E

The standard enthalpy of the formation of phosphorous acid is $\underset{\_}{-504.12kJ}$.

Explanation of Solution

The balanced chemical equation that represents the formation of phosphorous acid is shown below.

    $P_4{O}_6\left(s\right)+6H_{2}O\left(l\right)\to 4H_{\text{3}}P{O}_3\left(aq\right)$

The standard enthalpy of formation of product that is $H_{\text{3}}P{O}_3\left(aq\right)$ is $-\text{964}\text{.8}\text{kJ}\cdot {\text{mol}}^{-1}$.

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

  ${\displaystyle \sum \text{n}\Delta {\text{H}}_f^{\circ }\left(products\right)}={\text{n}}_{H_{\text{3}}P{O}_3}\Delta {\text{H}}_f^{\circ }\left(H_{\text{3}}P{O}_3,aq\right)$        (6)

Where,

• ${\displaystyle \sum \text{n}\Delta {\text{H}}_f^{\circ }\left(products\right)}$ is total enthalpy of formation of products.
• ${\text{n}}_{H_{\text{3}}P{O}_3}$ is the number of moles of $H_{\text{3}}P{O}_3\left(aq\right)$.
• $\Delta {\text{H}}_f^{\circ }\left(H_{\text{3}}P{O}_3,aq\right)$ is the standard enthalpy of formation of $H_{\text{3}}P{O}_3\left(aq\right)$.

The value of ${\text{n}}_{H_{\text{3}}P{O}_3}$ is $4mol$.

The value of $\Delta {\text{H}}_f^{\circ }\left(H_{\text{3}}P{O}_3,aq\right)$ is $-\text{964}\text{.8}\text{kJ}\cdot {\text{mol}}^{-1}$.

Substitute the value of ${\text{n}}_{H_{\text{3}}P{O}_3}$ and $\Delta {\text{H}}_f^{\circ }\left(H_{\text{3}}P{O}_3,aq\right)$ in equation (6).

  $\begin{array}{c}{\displaystyle \sum \text{n}\Delta {\text{H}}_f^{\circ }\left(products\right)}=\left(4mol\right)\times \left(-\text{964}\text{.8}\text{kJ}\cdot {\text{mol}}^{-1}\right)\\ =-3859.2\text{kJ}\end{array}$

The standard enthalpy of formation of reactant that is $P_4{O}_6\left(s\right)$ and $H_{2}O\left(l\right)$ is $-1640.1\text{kJ}\cdot {\text{mol}}^{-1}$ and $-285.83\text{kJ}\cdot {\text{mol}}^{-1}$ respectively.

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

  ${\displaystyle \sum \text{n}\Delta {\text{H}}_f^{\circ }\left(\text{reactants}\right)}={\text{n}}_{P_4{O}_6}\Delta {\text{H}}_f^{\circ }\left(P_4{O}_6,s\right)+{\text{n}}_{H_{2}O}\Delta {\text{H}}_f^{\circ }\left(H_{2}O,l\right)$        (7)

Where,

• ${\displaystyle \sum \text{n}\Delta {\text{H}}_f^{\circ }\left(\text{reactants}\right)}$ is total enthalpy of formation of reactants.
• ${\text{n}}_{P_4{O}_6}$ is the number of moles of $P_4{O}_6\left(s\right)$.
• $\Delta {\text{H}}_f^{\circ }\left(P_4{O}_6,s\right)$ is the standard enthalpy of formation of $P_4{O}_6\left(s\right)$.
• ${\text{n}}_{H_{2}O}$ 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 ${\text{n}}_{P_4{O}_6}$ is $1mol$.

The value of  $\Delta {\text{H}}_f^{\circ }\left(P_4{O}_6,s\right)$ is $-1640.1\text{kJ}\cdot {\text{mol}}^{-1}$.

The value of ${\text{n}}_{H_{2}O}$ is $6mol$.

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 ${\text{n}}_{P_4{O}_6}$, $\Delta {\text{H}}_f^{\circ }\left(P_4{O}_6,s\right)$, ${\text{n}}_{H_{2}O}$ and $\Delta {\text{H}}_f^{\circ }\left(H_{2}O,l\right)$ in equation (7).

  $\begin{array}{c}{\displaystyle \sum \text{n}\Delta {\text{H}}_f^{\circ }\left(\text{reactants}\right)}=\left(1mol\right)\times \left(-1640.1\text{kJ}\cdot {\text{mol}}^{-1}\right)+\left(6mol\right)\times \left(-285.83\text{kJ}\cdot {\text{mol}}^{-1}\right)\\ =-1640.1\text{kJ}-1714.98\text{kJ}\\ =-\text{3355}\text{.08}\text{kJ}\end{array}$

The value of ${\displaystyle \sum \text{n}\Delta {\text{H}}_f^{\circ }\left(products\right)}$ is $-3859.2\text{kJ}$.

The value of ${\displaystyle \sum \text{n}\Delta {\text{H}}_f^{\circ }\left(\text{reactants}\right)}$ is $-\text{3355}\text{.08}\text{kJ}$.

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).

  $\begin{array}{c}\Delta \text{H}°=-3859.2\text{kJ}-\left(-\text{3355}\text{.08}\text{kJ}\right)\\ =-3859.2\text{kJ}+\text{3355}\text{.08}\text{kJ}\\ =-\text{504}\text{.12}\text{kJ}\end{array}$

Thus, the standard enthalpy of the formation of phosphorous acid is $\underset{\_}{-504.12kJ}$.