Chapter 2, Problem 2C.7AE

(i)

Interpretation Introduction

Interpretation: The values of ${\Delta }_{\text{r}}H°$ and ${\Delta }_{\text{r}}U°$ at $298\text{K}$ has to be calculated for the given reaction.

Concept introduction: Enthalpy is the internal heat content of the system.  Under constant pressure, a system can do expansion work under the effect of energy supplied as heat known as enthalpy change of the system.  Therefore, the measure of amount of heat supplied to the system gives the enthalpy change of the system.  The total amount of energy possessed within the given system is known as internal energy of the system. It is denoted by $U$.

Answer to Problem 2C.7AE

The value of ${\Delta }_{\text{r}}H°$ for the reaction is ${\underset{\_}{131.29kJmol}}^{-1}$.  The value of ${\Delta }_{\text{r}}U°$ for the reaction is $\underset{\_}{128.81kJmo{l}^{-1}}$.

Explanation of Solution

The given reaction is:

    $\text{C}\left(\text{graphite}\right)\text{+}{\text{H}}_{\text{2}}\text{O}\left(\text{g}\right)\to \text{CO(g)}\text{+}{\text{H}}_2\left(\text{g}\right)$                                                    (1)

The standard enthalpy for the reaction is calculated by the formula,

    ${\Delta }_{\text{r}}{H}^{\text{ο}}={\displaystyle \sum m\cdot \Delta H_{\text{f}}^{\text{ο}}}\left(\text{products}\right)-{\displaystyle \sum n\cdot \Delta H_{\text{f}}^{\text{ο}}}\left(\text{reactants}\right)$                                (2)

Where,

• • $\Delta H_{\text{f}}^{\text{ο}}\left(\text{products}\right)$ is the standard enthalpy of formation of products.
• • $\Delta H_{\text{f}}^{\text{ο}}\left(\text{reactants}\right)$ is the standard enthalpy of formation of reactants.
• • $m$ is the stoichiometric coefficient of products.
• • $n$ is the stoichiometric coefficient of reactants

For standard enthalpy of the reaction, equation (2) is written as,

    ${\Delta }_{\text{r}}{H}^{\text{ο}}=\left(1\times \Delta H_{\text{f}}^{\text{ο}}\left(\text{CO}\right)+1\times \Delta H_{\text{f}}^{\text{ο}}\left({\text{H}}_{\text{2}}\right)-\left(1\times \Delta H_{\text{f}}^{\text{ο}}\left(\text{C}\left(\text{graphite}\right)\right)+1\times \Delta H_{\text{f}}^{\text{ο}}\left({\text{H}}_{\text{2}}\text{O}\right)\right)\right)$

The standard enthalpy of formation of $\text{CO}$ is $-110.53{\text{ kJ mol}}^{-1}$.

The standard enthalpy of formation of ${\text{H}}_{\text{2}}$ is ${\text{0 kJ mol}}^{-1}$.

The standard enthalpy of formation of ${\text{H}}_{\text{2}}\text{O}$ is $-\text{241}{\text{.82 kJ mol}}^{-1}$.

The standard enthalpy of formation of $\text{C}\left(\text{graphite}\right)$ is ${\text{0 kJ mol}}^{-1}$.

Substitute the values of standard enthalpy of formation of reactants and products in the above equation to calculate the standard enthalpy of the reaction.

    $\begin{array}{c}{\Delta }_{\text{r}}{H}^{\text{ο}}=\left(\begin{array}{l}1\times \left(-110.53{\text{ kJ mol}}^{-1}\right)+1\times \left({\text{0 kJ mol}}^{-1}\right)-\\ \left(\text{1}\times \left({\text{0 kJ mol}}^{-1}\right)+1\times \left(-\text{241}{\text{.82 kJ mol}}^{-1}\right)\right)\end{array}\right)\\ =-110.53\text{ kJ}{\text{mol}}^{-1}+241.82\text{ kJ}{\text{mol}}^{-1}\\ ={\underset{\_}{131.29kJmol}}^{-1}\end{array}$

Hence, the value of ${\Delta }_{\text{r}}H°$ for the reaction is ${\underset{\_}{131.29kJmol}}^{-1}$.

The change in number of moles of the reaction is calculated by the formula,

  $\text{Number}\text{of}\text{moles,}\Delta n_{\text{g}}=\text{Moles}\text{of}\text{gaseous}\text{product}-\text{Moles}\text{of}\text{gaseous}\text{reactant}$

Hence, the number of moles for the reaction is,

  $\begin{array}{c}\Delta n_{\text{g}}=\left(1+1\right)-\left(1\right)\\ =1\end{array}$

The expression to calculate the value of ${\Delta }_{\text{r}}U°$ for the reaction is,

    ${\Delta }_{\text{r}}U°={\Delta }_{\text{r}}H°-\Delta n_{\text{g}}RT$

Substitute the values of ${\Delta }_{\text{r}}H°$, $R$, $\Delta n_{\text{g}}$ and $T=298\text{K}$ in the above equation to calculate the value of ${\Delta }_{\text{r}}U°$.

  $\begin{array}{c}{\Delta }_{\text{r}}U°=131.29\text{kJ}{\text{mol}}^{-1}-\left(1\times 8.314\times {10}^{-3}\text{kJ}{\text{K}}^{-1}{\text{mol}}^{-1}\times 298\text{K}\right)\\ =131.29\text{kJ}{\text{mol}}^{-1}-\left(2477.572\times {10}^{-3}\text{kJ}{\text{mol}}^{-1}\right)\\ =\underset{\_}{128.81kJmo{l}^{-1}}\end{array}$

Hence, the value of ${\Delta }_{\text{r}}U°$ for the reaction is $\underset{\_}{128.81kJmo{l}^{-1}}$.

(ii)

Interpretation Introduction

Interpretation: The values of ${\Delta }_{\text{r}}H°$ and ${\Delta }_{\text{r}}U°$ has to be calculated for the given reaction at $478\text{K}$.

Concept introduction: Enthalpy is the internal heat content of the system.  Under constant pressure, a system can do expansion work under the effect of energy supplied as heat known as enthalpy change of the system.  Therefore, the measure of amount of heat supplied to the system gives the enthalpy change of the system.  The total amount of energy possessed within the given system is known as internal energy of the system. It is denoted by $U$.

Answer to Problem 2C.7AE

The value of ${\Delta }_{\text{r}}H°$ for the reaction is ${\underset{\_}{131.29kJmol}}^{-1}$.  The value of ${\Delta }_{\text{r}}U°$ for the reaction is $\underset{\_}{127.31kJmo{l}^{-1}}$.

Explanation of Solution

The given reaction is:

    $\text{C(graphite)}\text{+}{\text{H}}_{\text{2}}\text{O(g)}\to \text{CO(g)}\text{+}{\text{H}}_2\left(\text{g}\right)$                                                     (1)

The standard enthalpy for the reaction is calculated by the formula,

    ${\Delta }_{\text{r}}{H}^{\text{ο}}={\displaystyle \sum m\cdot \Delta H_{\text{f}}^{\text{ο}}}\left(\text{products}\right)-{\displaystyle \sum n\cdot \Delta H_{\text{f}}^{\text{ο}}}\left(\text{reactants}\right)$                                (2)

Where,

• • $\Delta H_{\text{f}}^{\text{ο}}\left(\text{products}\right)$ is the standard enthalpy of formation of products.
• • $\Delta H_{\text{f}}^{\text{ο}}\left(\text{reactants}\right)$ is the standard enthalpy of formation of reactants.
• • $m$ is the stoichiometric coefficient of products.
• • $n$ is the stoichiometric coefficient of reactants

For standard enthalpy of the reaction, equation (2) is written as,

    ${\Delta }_{\text{r}}{H}^{\text{ο}}=\left(1\times \Delta H_{\text{f}}^{\text{ο}}\left(\text{CO}\right)+1\times \Delta H_{\text{f}}^{\text{ο}}\left({\text{H}}_{\text{2}}\right)-\left(1\times \Delta H_{\text{f}}^{\text{ο}}\left(\text{C}\left(\text{graphite}\right)\right)+1\times \Delta H_{\text{f}}^{\text{ο}}\left({\text{H}}_{\text{2}}\text{O}\right)\right)\right)$

The standard enthalpy of formation of $\text{CO}$ is $-110.53{\text{ kJ mol}}^{-1}$.

The standard enthalpy of formation of ${\text{H}}_{\text{2}}$ is ${\text{0 kJ mol}}^{-1}$.

The standard enthalpy of formation of ${\text{H}}_{\text{2}}\text{O}$ is $-\text{241}{\text{.82 kJ mol}}^{-1}$.

The standard enthalpy of formation of $\text{C}\left(\text{graphite}\right)$ is ${\text{0 kJ mol}}^{-1}$.

Substitute the values of standard enthalpy of formation of reactants and products in the above equation to calculate the standard enthalpy of the reaction.

    $\begin{array}{c}{\Delta }_{\text{r}}{H}^{\text{ο}}=\left(\begin{array}{l}1\times \left(-110.53{\text{ kJ mol}}^{-1}\right)+1\times \left({\text{0 kJ mol}}^{-1}\right)-\\ \left(\text{1}\times \left({\text{0 kJ mol}}^{-1}\right)+1\times \left(-\text{241}{\text{.82 kJ mol}}^{-1}\right)\right)\end{array}\right)\\ =-110.53\text{ kJ}{\text{mol}}^{-1}+241.82\text{ kJ}{\text{mol}}^{-1}\\ ={\underset{\_}{131.29kJmol}}^{-1}\end{array}$

Hence, the value of ${\Delta }_{\text{r}}H°$ for the reaction is ${\underset{\_}{131.29kJmol}}^{-1}$.

The change in number of moles of the reaction is calculated by the formula,

  $\text{Number}\text{of}\text{moles,}\Delta n_{\text{g}}=\text{Moles}\text{of}\text{gaseous}\text{product}-\text{Moles}\text{of}\text{gaseous}\text{reactant}$

Hence, the number of moles for the reaction is,

  $\begin{array}{c}\Delta n_{\text{g}}=\left(1+1\right)-\left(+1\right)\\ =1\end{array}$

The expression to calculate the value of ${\Delta }_{\text{r}}U°$ for the reaction is,

    ${\Delta }_{\text{r}}U°={\Delta }_{\text{r}}H°-\Delta n_{\text{g}}RT$

Substitute the values of ${\Delta }_{\text{r}}H°$, $R$, $\Delta n_{\text{g}}$ and $T=478\text{K}$ in the above equation to calculate the value of ${\Delta }_{\text{r}}U°$.

  $\begin{array}{c}{\Delta }_{\text{r}}U°=131.29\text{kJ}{\text{mol}}^{-1}-\left(1\times 8.314\times {10}^{-3}\text{kJ}{\text{K}}^{-1}{\text{mol}}^{-1}\times 478\text{K}\right)\\ =131.29\text{kJ}{\text{mol}}^{-1}-\left(\text{3}\text{.97}\text{kJ}{\text{mol}}^{-1}\right)\\ =\underset{\_}{127.31kJmo{l}^{-1}}\end{array}$

Hence, the value of ${\Delta }_{\text{r}}U°$ for the reaction is $\underset{\_}{127.31kJmo{l}^{-1}}$.