Chapter 4, Problem 4.29E

(a)

Interpretation Introduction

Interpretation:

The thermochemical equation for the overall reaction taking place in the reactor 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 4.29E

The thermochemical equation for the production of sulfur is shown below.

    $H_{2}S\left(g\right)+\frac{1}{2}{O}_{\text{2}}\left(g\right)\to S\left(s\right)+H_2\text{O}\left(l\right),\Delta {\text{H}}^{\circ }=-265.2\text{kJ}$

Explanation of Solution

The chemical equations for the steps takes place in Claus process are shown below.

    $\begin{array}{c}\left(a\right)2H_{2}S\left(g\right)+3O_{\text{2}}\left(g\right)\to 2S{O}_2\left(g\right)+2H_2\text{O}\left(l\right)\\ \left(b\right)2H_{2}S\left(g\right)+S{O}_2\left(g\right)\to 3S\left(s\right)+2H_2\text{O}\left(l\right)\end{array}$

The $S{O}_2\left(g\right)$ molecules should not be the part of overall reaction.  Therefore, chemical equation (b) is multiplied by $2$.  the modified equation (b) is shown below.

    $4H_{2}S\left(g\right)+2S{O}_2\left(g\right)\to 6S\left(s\right)+4H_2\text{O}\left(l\right)$

Add the chemical equation (a) and modified chemical equation (b) to get the desired overall reaction.  The addition result is shown below.

    $\begin{array}{l}\underset{\_}{\begin{array}{c}2H_{2}S\left(g\right)+3O_{\text{2}}\left(g\right)\to 2S{O}_2\left(g\right)+2H_2\text{O}\left(l\right)\\ 4H_{2}S\left(g\right)+2S{O}_2\left(g\right)\to 6S\left(s\right)+4H_2\text{O}\left(l\right)\end{array}}\\ 6H_{2}S\left(g\right)+3O_{\text{2}}\left(g\right)\to 6S\left(s\right)+6H_2\text{O}\left(l\right)\end{array}$

The chemical equation obtained above is divided by $6$.  The overall reaction that takes place in reactor is shown below.

    $H_{2}S\left(g\right)+\frac{1}{2}{O}_{\text{2}}\left(g\right)\to S\left(s\right)+H_2\text{O}\left(l\right)$

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

  ${\displaystyle \sum \text{n}\Delta {\text{H}}_f^{\circ }\left(products\right)}={\text{n}}_S\Delta {\text{H}}_f^{\circ }\left(S,s\right)+{\text{n}}_{H_{2}O}\Delta {\text{H}}_f^{\circ }\left(H_{2}O,l\right)$        (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 $S\left(s\right)$.
• ${\text{n}}_{H_{2}O}$ is the number of moles of $H_2\text{O}\left(l\right)$.
• $\Delta {\text{H}}_f^{\circ }\left(H_{2}O,l\right)$ is the standard enthalpy of formation of $H_2\text{O}\left(l\right)$.

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

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

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

The value of $\Delta {\text{H}}_f^{\circ }\left(H_{2}O,l\right)$ is $-\text{285}\text{.83}\text{kJ}\cdot {\text{mol}}^{-1}$.

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

  $\begin{array}{c}{\displaystyle \sum \text{n}\Delta {\text{H}}_f^{\circ }\left(products\right)}=\left(1mol\right)\times \left(0.0\text{kJ}\cdot {\text{mol}}^{-1}\right)+\left(1mol\right)\times \left(-\text{285}\text{.83}\text{kJ}\cdot {\text{mol}}^{-1}\right)\\ =-\text{285}\text{.83}\text{kJ}\end{array}$

The standard enthalpy of formation of reactant that is $H_{2}S\left(g\right)$ and $O_{\text{2}}\left(g\right)$ is $-20.63\text{kJ}\cdot {\text{mol}}^{-1}$ and $0.0\text{kJ}\cdot {\text{mol}}^{-1}$ respectively.

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

  ${\displaystyle \sum \text{n}\Delta {\text{H}}_f^{\circ }\left(\text{reactants}\right)}={\text{n}}_{H_{2}S}\Delta {\text{H}}_f^{\circ }\left(H_{2}S,g\right)+{\text{n}}_{O_{\text{2}}}\Delta {\text{H}}_f^{\circ }\left(O_{\text{2}},g\right)$        (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 $H_{2}S\left(g\right)$.
• $\Delta {\text{H}}_f^{\circ }\left(H_{2}S,g\right)$ is the standard enthalpy of formation of $H_{2}S\left(g\right)$.
• ${\text{n}}_{O_{\text{2}}}$ 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 ${\text{n}}_{H_{2}S}$ is $1mol$.

The value of $\Delta {\text{H}}_f^{\circ }\left(H_{2}S,g\right)$ is $-20.63\text{kJ}\cdot {\text{mol}}^{-1}$.

The value of ${\text{n}}_{O_{\text{2}}}$ is $\frac{1}{2}mol$.

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 ${\text{n}}_{H_{2}S}$, $\Delta {\text{H}}_f^{\circ }\left(H_{2}S,g\right)$, ${\text{n}}_{O_{\text{2}}}$ and $\Delta {\text{H}}_f^{\circ }\left(O_{\text{2}},g\right)$ in equation (2).

  $\begin{array}{c}{\displaystyle \sum \text{n}\Delta {\text{H}}_f^{\circ }\left(\text{reactants}\right)}=\left(1mol\right)\times \left(-20.63\text{kJ}\cdot {\text{mol}}^{-1}\right)+\left(\frac{1}{2}mol\right)\times \left(0.0\text{kJ}\cdot {\text{mol}}^{-1}\right)\\ =-20.63\text{kJ}\end{array}$

The standard enthalpy of formation of urea 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 $-\text{285}\text{.83}\text{kJ}$.

The value of ${\displaystyle \sum \text{n}\Delta {\text{H}}_f^{\circ }\left(\text{reactants}\right)}$ is $-20.63\text{kJ}$.

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).

  $\begin{array}{c}\Delta \text{H}°=-\text{285}\text{.83}\text{kJ}-\left(-20.63\text{kJ}\right)\\ =-\text{285}\text{.83}\text{kJ}+20.63\text{kJ}\\ =-265.2\text{kJ}\end{array}$

Thus, the desired thermochemical equation is shown below.

    $H_{2}S\left(g\right)+\frac{1}{2}{O}_{\text{2}}\left(g\right)\to S\left(s\right)+H_2\text{O}\left(l\right),\Delta {\text{H}}^{\circ }=-265.2\text{kJ}$

(b)

Interpretation Introduction

Interpretation:

The change in the enthalpy of production of $\text{S}\left(\text{s}\right)$ has to be determined.

Concept Introduction:

Same as part (a).

Answer to Problem 4.29E

The change in the enthalpy of production of $\text{S}\left(\text{s}\right)$ is $\underset{\_}{-4.96\times 10^{5}kJ}$.

Explanation of Solution

The mass of $\text{S}\left(\text{s}\right)$ produced is $60.0kg$.

The number of moles of $\text{S}\left(\text{s}\right)$ produced is calculated by the relation shown below.

  $n=\frac{m}{M_{\text{w}}}$        (4)

Where,

• $n$ is the number of moles.
• $m$ is the mass.
• $M_{\text{w}}$ is the molar mass.

The value of $m$ is $60.0kg$.

The value of $M_{\text{w}}$ is $32.065g\cdot mo{l}^{-1}$.

Substitute the value of $m$ and $M_{\text{w}}$ in equation (4).

    $\begin{array}{c}n=\frac{60.0\text{kg}\times \frac{1000\text{g}}{1\text{kg}}}{32.065\text{g}\cdot {\text{mol}}^{-1}}\\ =1871.19\text{mol}\end{array}$

Therefore, the number of moles of $\text{S}\left(\text{s}\right)$ produced is $1871.19\text{mol}$.

The enthalpy of production of $\text{S}\left(\text{s}\right)$ is $-265.2\text{kJ}\cdot {\text{mol}}^{-1}$.  Therefore, the enthalpy of production of $1871.19\text{mol}\text{S}\left(\text{s}\right)$ is calculated as shown below.

    $\begin{array}{c}\Delta H°=-265.2\text{kJ}\cdot {\text{mol}}^{-1}\times 1871.19\text{mol}\\ =-496239.58\text{kJ}\\ =-4.96\times {10}^5\text{kJ}\end{array}$

Thus, the change in the enthalpy of production of $\text{S}\left(\text{s}\right)$ is $\underset{\_}{-4.96\times 10^{5}kJ}$.

(c)

Interpretation Introduction

Interpretation:

The reactor need to be cooled or heat to maintain the constant temperature has to be explained.

Concept Introduction:

Same as part (a).

Answer to Problem 4.29E

The reactor would be cooled to maintain a constant temperature.

Explanation of Solution

The change in the enthalpy of the reaction that takes place in the reactor is calculated in part (b) is $-4.96\times {10}^5\text{kJ}$.  Since the value for the enthalpy change is negative, it indicates that the reaction is exothermic in nature.  Therefore, the temperature of the reactor will remain constant on cooling.

Thus, the reactor would be cooled to maintain a constant temperature.