Chapter 2, Problem 2C.8AE

Interpretation Introduction

Interpretation: The value of ${\Delta }_{\text{r}}H°$ at $500\text{K}$ 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.

Answer to Problem 2C.8AE

The value of ${\Delta }_{\text{r}}H°$ at $500\text{K}$ is $-\underset{\_}{389.31kJmo{l}^{-1}}$.

Explanation of Solution

The given reaction is,

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

The standard enthalpy of the reaction is, ${\Delta }_{\text{r}}H°$ is calculated by the expression,

    ${\Delta }_{\text{r}}{H}^{\text{ο}}={\displaystyle \sum m\cdot \Delta H_{\text{f}}^{\text{ο}}}\left(\text{products}\right)-{\displaystyle \sum n\cdot \Delta H_{\text{f}}^{\text{ο}}}\left(\text{reactants}\right)$                                (1)

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.

The temperature dependent heat capacity is given by the expression,

    ${\Delta }_{\text{r}}{C}_{\text{p}}^{\text{ο}}=\Delta a+\Delta bT+\Delta c/T^2$                                                                             (2)

Where,

• • $\Delta a,\Delta b,\Delta c$ are the temperature dependent variables.
• • $T$ is the temperature.

For the reaction, $\Delta a$ is calculated by the expression,

    $\Delta a={\displaystyle \sum va}\left(\text{products}\right)-{\displaystyle \sum va}\left(\text{reactants}\right)$                                                      (3)

Where,

• • ${\displaystyle \sum va}\left(\text{products}\right)$ is the sum of $a$ of products.
• • ${\displaystyle \sum va}\left(\text{reactants}\right)$ is the sum of $a$ of reactants.
• • $v$ is the stoichiometry.

For the reaction, $\Delta b$ is calculated by the expression,

    $\Delta b={\displaystyle \sum vb}\left(\text{products}\right)-{\displaystyle \sum vb}\left(\text{reactants}\right)$                                                      (4)

Where,

• • ${\displaystyle \sum vb}\left(\text{products}\right)$ is the sum of $b$ of products.
• • ${\displaystyle \sum vb}\left(\text{reactants}\right)$ is the sum of $b$ of reactants.
• • $v$ is the stoichiometry.

For the reaction, $\Delta c$ is calculated by the expression,

    $\Delta c={\displaystyle \sum vc}\left(\text{products}\right)-{\displaystyle \sum vc}\left(\text{reactants}\right)$                                                      (5)

Where,

• • ${\displaystyle \sum vc}\left(\text{products}\right)$ is the sum of $c$ of products.
• • ${\displaystyle \sum vc}\left(\text{reactants}\right)$ is the sum of $c$ of reactants.
• • $v$ is the stoichiometry.

Enthalpy at a temperature $T_2=500\text{K}$ is calculated by the expression,

    $\begin{array}{c}{\Delta }_{\text{r}}H°(T_2)={\Delta }_{\text{r}}H°(T_1)+{\displaystyle \underset{T_1}{\overset{T_2}{\int }}{\Delta }_{\text{r}}{C}_{\text{p}}}°dT\\ ={\Delta }_{\text{r}}H°(T_1)+{\displaystyle \underset{T_1}{\overset{T_2}{\int }}\left(a+bT+c/T^2\right)}dT\end{array}$

    ${\Delta }_{\text{r}}H°(T_2)={\Delta }_{\text{r}}H°(T_1)+\Delta a\left(T_2-T_1\right)+\frac{1}{2}\Delta b\left(T_2^2-T_1^2\right)-\Delta c\left(\frac{1}{T_2}-\frac{1}{T_1}\right)$             (6)

The value of $\Delta a$ is calculated for the given reaction by equation (3)

    $\Delta a={\displaystyle \sum a}\left({\text{CO}}_{\text{2}}\text{,}\text{g}\right)-{\displaystyle \sum a}\left({\text{O}}_{\text{2}}\text{,g}\right)$

The value of $a$ for ${\text{CO}}_{\text{2}}$ and ${\text{O}}_{\text{2}}$ is $44.22\text{J}{\text{K}}^{-1}{\text{mol}}^{-1}$ and $29.96\text{J}{\text{K}}^{-1}{\text{mol}}^{-1}$, respectively.  Substitute the values in the above equation to calculate the value of $\Delta a$.

    $\begin{array}{c}\Delta a=44.22\text{J}{\text{K}}^{-1}{\text{mol}}^{-1}-29.96\text{J}{\text{K}}^{-1}{\text{mol}}^{-1}\\ =14.26\text{J}{\text{K}}^{-1}{\text{mol}}^{-1}\\ =14.26\times {10}^{-3}\text{kJ}{\text{K}}^{-1}{\text{mol}}^{-1}\end{array}$

The value of $\Delta b$ is calculated for the given reaction by equation (4)

    $\Delta b={\displaystyle \sum b}\left({\text{CO}}_{\text{2}}\text{,}\text{g}\right)-{\displaystyle \sum b}\left({\text{O}}_{\text{2}}\text{,g}\right)$

The value of $b$ for ${\text{CO}}_{\text{2}}$ and ${\text{O}}_{\text{2}}$ is $8.79\times {10}^{-3}\text{J}{\text{K}}^{-2}{\text{mol}}^{-1}$ and $4.18\times {10}^{-3}\text{J}{\text{K}}^{-2}{\text{mol}}^{-1}$, respectively.  Substitute the values in the above equation to calculate the value of $\Delta b$.

    $\begin{array}{c}\Delta b=8.79\times {10}^{-3}\text{J}{\text{K}}^{-2}{\text{mol}}^{-1}-4.18\times {10}^{-3}\text{J}{\text{K}}^{-2}{\text{mol}}^{-1}\\ =4.61\times {10}^{-3}\text{J}{\text{K}}^{-2}{\text{mol}}^{-1}\\ =4.61\times {10}^{-6}\text{kJ}{\text{K}}^{-2}{\text{mol}}^{-1}\end{array}$

The value of $\Delta c$ is calculated for the given reaction by equation (5)

    $\Delta c={\displaystyle \sum c}\left({\text{CO}}_{\text{2}}\text{,}\text{g}\right)-{\displaystyle \sum c}\left({\text{O}}_{\text{2}}\text{,g}\right)$

The value of $c$ for ${\text{CO}}_{\text{2}}$ and ${\text{O}}_{\text{2}}$ is $-8.62\times {10}^5\text{J}\text{K}{\text{mol}}^{-1}$ and $-1.67\times {10}^5\text{J}\text{K}{\text{mol}}^{-1}$, respectively.  Substitute the values in the above equation to calculate the value of $\Delta c$.

    $\begin{array}{c}\Delta c=-8.62\times {10}^5\text{J}\text{K}{\text{mol}}^{-1}-\left(-1.67\times {10}^5\text{J}\text{K}{\text{mol}}^{-1}\right)\\ =6.95\times {10}^5\text{J}\text{K}{\text{mol}}^{-1}\\ =6.95\times {10}^2\text{kJ}\text{K}{\text{mol}}^{-1}\end{array}$

To calculate the value of ${\Delta }_{\text{r}}H°$ at $298\text{K}$, use equation (1).

    ${\Delta }_{\text{r}}{H}^{\text{ο}}={\displaystyle \sum 1\times \Delta H_{\text{f}}^{\text{ο}}}\left({\text{CO}}_{\text{2}}\text{,}\text{g}\right)-{\displaystyle \sum 1\times \Delta H_{\text{f}}^{\text{ο}}}\left({\text{O}}_{\text{2}}\text{,g}\right)$

Substitute the values of $\Delta H_{\text{f}}^{\text{ο}}\left({\text{CO}}_{\text{2}}\text{,}\text{g}\right)=-393.51\text{kJ}{\text{mol}}^{-1}$ and $\Delta H_{\text{f}}^{\text{ο}}\left({\text{O}}_{\text{2}}\text{,g}\right)=0\text{kJ}{\text{mol}}^{-1}$ in the above equation to calculate ${\Delta }_{\text{r}}H°$.

    $\begin{array}{c}{\Delta }_{\text{r}}{H}^{\text{ο}}=-393.51\text{kJ}{\text{mol}}^{-1}-0\text{kJ}{\text{mol}}^{-1}\\ =-393.51\text{kJ}{\text{mol}}^{-1}\end{array}$

To calculate the value of ${\Delta }_{\text{r}}H°$ at $500\text{K}$, use equation (6).

    $\begin{array}{c}{\Delta }_{\text{r}}H°(500\text{K})={\Delta }_{\text{r}}H°(298\text{K})+\Delta a\left(500\text{K}-298\text{K}\right)+\frac{1}{2}\Delta b\left({\left(500\text{K}\right)}^2-{\left(298\text{K}\right)}^2\right)\\ -\Delta c\left(\frac{1}{500\text{K}}-\frac{1}{298\text{K}}\right)\\ =-393.51\text{kJ}{\text{mol}}^{-1}+14.26\times {10}^{-3}\text{kJ}{\text{K}}^{-1}{\text{mol}}^{-1}\left(500\text{K}-298\text{K}\right)+\\ \frac{1}{2}\times 4.61\times {10}^{-6}\text{kJ}{\text{K}}^{-2}{\text{mol}}^{-1}\left({\left(500\text{K}\right)}^2-{\left(298\text{K}\right)}^2\right)-6.95\times {10}^2\text{kJ}\text{K}{\text{mol}}^{-1}\left(\frac{1}{500\text{K}}-\frac{1}{298\text{K}}\right)\\ =-\underset{\_}{389.31kJmo{l}^{-1}}\end{array}$

Hence, the value of ${\Delta }_{\text{r}}H°$ at $500\text{K}$ is $-\underset{\_}{389.31kJmo{l}^{-1}}$.