Chapter 4, Problem 4.31E

(a)

Interpretation Introduction

Interpretation:

The production of water gas is exothermic or endothermic has to be predicted.

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 4.31E

The production of water gas is endothermic.

Explanation of Solution

The chemical equation for the production of water gas is shown below.

    $C\left(s\right)+H_2\text{O}\left(g\right)\to CO\left(g\right)+H_2\left(g\right)$

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

  ${\displaystyle \sum \text{n}\Delta {\text{H}}_f^{\circ }\left(products\right)}={\text{n}}_{CO}\Delta {\text{H}}_f^{\circ }\left(CO,g\right)+{\text{n}}_{H_2}\Delta {\text{H}}_f^{\circ }\left(H_2,g\right)$        (1)

Where,

• ${\displaystyle \sum \text{n}\Delta {\text{H}}_f^{\circ }\left(products\right)}$ is total enthalpy of formation of products.
• ${\text{n}}_{CO}$ is the number of moles of $CO\left(g\right)$.
• $\Delta {\text{H}}_f^{\circ }\left(CO,g\right)$ is the standard enthalpy of formation of $CO\left(g\right)$.
• ${\text{n}}_{H_2}$ is the number of moles of $H_2\left(g\right)$.
• $\Delta {\text{H}}_f^{\circ }\left(H_2,g\right)$ is the standard enthalpy of formation of $H_2\left(g\right)$.

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

The value of $\Delta {\text{H}}_f^{\circ }\left(CO,g\right)$ is $-110.53\text{kJ}\cdot {\text{mol}}^{-1}$.

The value of ${\text{n}}_{H_2}$ is $1mol$.

The value of $\Delta {\text{H}}_f^{\circ }\left(H_2,g\right)$ is $0.0\text{kJ}\cdot {\text{mol}}^{-1}$.

Substitute the value of ${\text{n}}_{CO}$, $\Delta {\text{H}}_f^{\circ }\left(CO,g\right)$, ${\text{n}}_{H_2}$ and $\Delta {\text{H}}_f^{\circ }\left(H_2,g\right)$ in equation (1).

  $\begin{array}{c}{\displaystyle \sum \text{n}\Delta {\text{H}}_f^{\circ }\left(products\right)}=\left(1mol\right)\times \left(-110.53\text{kJ}\cdot {\text{mol}}^{-1}\right)+\left(1mol\right)\times \left(0.0\text{kJ}\cdot {\text{mol}}^{-1}\right)\\ =-110.53\text{kJ}\end{array}$

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

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

Where,

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

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

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

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

  $\begin{array}{c}{\displaystyle \sum \text{n}\Delta {\text{H}}_f^{\circ }\left(\text{reactants}\right)}=\left(1mol\right)\times \left(0.0\text{kJ}\cdot {\text{mol}}^{-1}\right)+\left(1mol\right)\times \left(-241.82\text{kJ}\cdot {\text{mol}}^{-1}\right)\\ =-241.82\text{kJ}\end{array}$

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 $-110.53\text{kJ}$.

The value of ${\displaystyle \sum \text{n}\Delta {\text{H}}_f^{\circ }\left(\text{reactants}\right)}$ is $-241.82\text{kJ}$.

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).

  $\begin{array}{c}\Delta \text{H}°=-110.53\text{kJ}-\left(-241.82\text{kJ}\right)\\ =-110.53\text{kJ}+241.82\text{kJ}\\ =+131.29kJ\end{array}$

Therefore, the change in the enthalpy for the production of water is $+131.29kJ$.  As the value of for the given reaction is positive, it indicates that the reaction is endothermic in nature.

Thus, the production of water gas is endothermic.

(b)

Interpretation Introduction

Interpretation:

The change in the enthalpy for the production of given volume of hydrogen gas has to be determined.

Concept Introduction:

Same as part (a).

Answer to Problem 4.31E

The change in the enthalpy of production of ${\text{H}}_2\left(\text{g}\right)$ is $\underset{\_}{+622.3kJ}$.

Explanation of Solution

The volume of ${\text{H}}_2\left(\text{g}\right)$ produced at $500torr$ pressure and $65°C$ temperature is $200L$.

The unit conversion of temperature from $°C$ to $K$ is shown below.

    $\begin{array}{c}T=65+273\text{K}\\ =338\text{K}\end{array}$

The number of moles of ${\text{H}}_2\left(\text{g}\right)$ regenerated can be calculated by the relation shown below.

  $n=\frac{PV}{RT}$        (4)

Where,

• $n$ is the number of moles.
• $P$ is the pressure.
• $V$ is the volume.
• $R$ is the gas constant.
• $T$ is the temperature.

The value of $P$ is $500torr$.

The value of $V$ is $200L$.

The value of $R$ is $62.3635L\cdot torr\cdot mo{l}^{-}{}^1\cdot K^{-}{}^1$.

The value of $T$ is $338\text{K}$.

Substitute the value of $P$, $V$, $R$ and $T$ in equation (4).

    $\begin{array}{c}n=\frac{\left(500\text{torr}\right)\times \left(200\text{L}\right)}{\left(62.3635\text{ L}\cdot \text{torr}\cdot {\text{mol}}^{-}{}^1\cdot {\text{K}}^{-}{}^1\right)\times \left(338\text{K}\right)}\\ =\frac{100000\text{L}\cdot \text{torr}}{\text{21078}\text{.863 L}\cdot \text{torr}\cdot {\text{mol}}^{-}{}^1}\\ =4.74\text{mol}\end{array}$

Therefore, the number of moles of ${\text{H}}_2\left(\text{g}\right)$ produced at $338\text{K}$ is $4.74\text{mol}$.

The enthalpy of production of ${\text{H}}_2\left(\text{g}\right)$ is $+131.29\text{kJ}\cdot {\text{mol}}^{-1}$.  Therefore, the enthalpy of production of $1871.19\text{mol}\text{S}\left(\text{s}\right)$ is calculated as shown below.

    $\begin{array}{c}\Delta H°=+131.29\text{kJ}\cdot {\text{mol}}^{-1}\times 4.74\text{mol}\\ =+622.3\text{kJ}\end{array}$

Thus, the change in the enthalpy of production of ${\text{H}}_2\left(\text{g}\right)$ is $\underset{\_}{+622.3kJ}$.