Chapter 2, Problem 2C.9P

Interpretation Introduction

Interpretation: The standard enthalpy of combustion of methane to carbon dioxide and water vapour has to be calculated.

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.9P

The value of standard enthalpy of formation at $500\text{K}$ is $\underset{\_}{60.65\times 10^{8}Jmo{l}^{-1}}$.

Explanation of Solution

The reaction for the combustion of methane is,

    ${\text{CH}}_{\text{4}}\left(\text{g}\right){\text{+2O}}_{\text{2}}\left(\text{g}\right)\to {\text{CO}}_{\text{2}}\left(\text{g}\right){\text{+2H}}_{\text{2}}\text{O}\left(\text{g}\right)$

The standard enthalpy of combustion of ${\text{CH}}_{\text{4}}\left(\text{g}\right)$ is calculated by the formula,

    $\Delta H_{\text{f}}^{\text{ο}}={\displaystyle \sum _{\text{Products}}v\Delta H_{\text{f}}^{\text{ο}}}-{\displaystyle \sum _{\text{Reactants}}v\Delta H_{\text{f}}^{\text{ο}}}$

Where,

• • ${\displaystyle \sum _{\text{Products}}v\Delta H_{\text{f}}^{\text{ο}}}$ is the summation of enthalpy of formation of products.
• • ${\displaystyle \sum _{\text{Reactants}}v\Delta H_{\text{f}}^{\text{ο}}}$ is the summation of enthalpy of formation of reactants.
• • $v$ is the stoichiometric coefficient.

The above expression can be re-written as,

    $\Delta H_{\text{f}}^{\text{ο}}=\left(\begin{array}{c}\left(\Delta H_{\text{f}}^{\text{ο}}\left({\text{CO}}_{\text{2}},\text{g}\right)+2\times \Delta H_{\text{f}}^{\text{ο}}\left({\text{H}}_{\text{2}}\text{O},\text{g}\right)\right)-\\ \left(\Delta H_{\text{f}}^{\text{ο}}\left({\text{CH}}_{\text{4}}\text{,}\text{g}\right)+2\times \Delta H_{\text{f}}^{\text{ο}}\left({\text{O}}_{\text{2}},\text{g}\right)\right)\end{array}\right)$                                            (1)

Where,

• • $\Delta H_{\text{f}}^{\text{ο}}\left({\text{CO}}_{\text{2}},\text{g}\right)$ is the standard enthalpy of formation of ${\text{CO}}_{\text{2}}\left(\text{g}\right)$.
• • $\Delta H_{\text{f}}^{\text{ο}}\left({\text{H}}_{\text{2}}\text{O},\text{g}\right)$ is the standard enthalpy of formation of ${\text{H}}_{\text{2}}\text{O}\left(\text{g}\right)$.
• • $\Delta H_{\text{f}}^{\text{ο}}\left({\text{CH}}_{\text{4}}\text{,}\text{g}\right)$ is standard enthalpy of formation of ${\text{CH}}_{\text{4}}\left(\text{g}\right)$.
• • $\Delta H_{\text{f}}^{\text{ο}}\left({\text{O}}_{\text{2}},\text{g}\right)$ is standard enthalpy of formation of ${\text{O}}_{\text{2}}\left(\text{g}\right)$

The standard enthalpy of formation of ${\text{O}}_{\text{2}}\left(\text{g}\right)$ is ${\text{0 kJ mol}}^{-1}$.

The standard enthalpy of formation of ${\text{CO}}_{\text{2}}\left(\text{g}\right)$ is $-\text{393}{\text{.51 kJ mol}}^{-1}$.

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

The standard enthalpy of formation of ${\text{CH}}_{\text{4}}\left(\text{g}\right)$ is $-\text{74}{\text{.81 kJ mol}}^{-1}$.

Substitute the values in equation (1) to calculate the standard enthalpy of formation.

    $\begin{array}{c}\Delta H_{\text{f}}^{\text{ο}}=\left(\begin{array}{c}\left(-\text{393}{\text{.51 kJ mol}}^{-1}+2\times \left(-\text{241}{\text{.82 kJ mol}}^{-1}\right)\right)-\\ \left(-\text{74}{\text{.81 kJ mol}}^{-1}-2\times {\text{0 kJ mol}}^{-1}\right)\end{array}\right)\\ =\underset{\_}{-802.34kJmo{l}^{-1}}\end{array}$

Hence, the enthalpy of formation of $\underset{\_}{-802.34kJmo{l}^{-1}}$.

The expression for heat capacity can be written as difference of the heat capacities of products and reactants.

    ${\Delta }_{\text{r}}{C}_{\text{p}}°={\displaystyle \sum _{\text{Products}}v{C}_{\text{p,m}}°}-{\displaystyle \sum _{\text{Reactants}}v{C}_{\text{p,m}}°}$                                                              (2)

Where,

• • $v$ is the stoiciometric coefficient.
• • ${\displaystyle \sum _{\text{Products}}v{C}_{\text{p,m}}°}$ is the summation of molar heat capacities of the products.
• • ${\displaystyle \sum _{\text{Reactants}}v{C}_{\text{p,m}}°}$ is the summation of molar heat capacities of the reactants.

The molar heat capacity can also be written as,

    $C_{\text{p,m}}=\alpha +\beta T+\gamma T^2$

Hence, equation (2) can be written as,

    $\begin{array}{c}{\Delta }_{\text{r}}{C}_{\text{p}}°={\displaystyle \sum _{\text{Products}}v\left(\alpha +\beta T+\gamma T^2\right)}-{\displaystyle \sum _{\text{Reactants}}v\left(\alpha +\beta T+\gamma T^2\right)}\\ =\left(\begin{array}{l}\left({\displaystyle \sum _{\text{Products}}v\left(\alpha \right)}-{\displaystyle \sum _{\text{Reactants}}v\left(\alpha \right)}\right)+\left({\displaystyle \sum _{\text{Products}}v\left(\beta \right)}-{\displaystyle \sum _{\text{Reactants}}v\left(\beta \right)}\right)\times T+\\ \left({\displaystyle \sum _{\text{Products}}v\left(\gamma \right)}-{\displaystyle \sum _{\text{Reactants}}v\left(\gamma \right)}\right)\times T^2\end{array}\right)\\ =\Delta \alpha +\Delta \beta T+\Delta \gamma T^2\end{array}$

The value of $\Delta \alpha$ is calculated as shown below.

    $\Delta \alpha =\left(\alpha \left({\text{CO}}_{\text{2}},\text{g}\right)+2\times \alpha \left({\text{H}}_{\text{2}}\text{O},\text{g}\right)\right)-\left(\alpha \left({\text{CH}}_{\text{4}}\text{,}\text{g}\right)+2\times \alpha \left({\text{O}}_{\text{2}},\text{g}\right)\right)$

It is given that $\alpha \left({\text{CO}}_{\text{2}},\text{g}\right)$ is $26.86\text{J}{\text{K}}^{-1}{\text{mol}}^{-1}$, $\alpha \left({\text{H}}_{\text{2}}\text{O},\text{g}\right)$ is $30.36\text{J}{\text{K}}^{-1}{\text{mol}}^{-1}$, $\alpha \left({\text{CH}}_{\text{4}}\text{,}\text{g}\right)$ is $14.46\text{J}{\text{K}}^{-1}{\text{mol}}^{-1}$, $\alpha \left({\text{O}}_{\text{2}},\text{g}\right)$ is $25.72\text{J}{\text{K}}^{-1}{\text{mol}}^{-1}$.  Hence, substitute these values in the above equation to calculate the value of $\Delta \alpha$.

  $\begin{array}{c}\Delta \alpha =\left(\begin{array}{c}\left(26.86\text{J}{\text{K}}^{-1}{\text{mol}}^{-1}+2\times 30.36\text{J}{\text{K}}^{-1}{\text{mol}}^{-1}\right)-\\ \left(14.46\text{J}{\text{K}}^{-1}{\text{mol}}^{-1}+2\times 25.72\text{J}{\text{K}}^{-1}{\text{mol}}^{-1}\right)\end{array}\right)\\ =87.58-65.9\\ =21.68\text{J}{\text{K}}^{-1}{\text{mol}}^{-1}\end{array}$

The value of $\Delta \beta$ is calculated as shown below.

    $\Delta \beta =\left(\beta \left({\text{CO}}_{\text{2}},\text{g}\right)+2\times \beta \left({\text{H}}_{\text{2}}\text{O},\text{g}\right)\right)-\left(\beta \left({\text{CH}}_{\text{4}}\text{,}\text{g}\right)+2\times \beta \left({\text{O}}_{\text{2}},\text{g}\right)\right)$

It is given that $\beta \left({\text{CO}}_{\text{2}},\text{g}\right)$ is $6.97\text{mJ}{\text{K}}^{-2}{\text{mol}}^{-1}$, $\beta \left({\text{H}}_{\text{2}}\text{O},\text{g}\right)$ is $9.61\text{mJ}{\text{K}}^{-2}{\text{mol}}^{-1}$, $\beta \left({\text{CH}}_{\text{4}}\text{,}\text{g}\right)$ is $75.5\text{mJ}{\text{K}}^{-2}{\text{mol}}^{-1}$, $\beta \left({\text{O}}_{\text{2}},\text{g}\right)$ is $12.98\text{mJ}{\text{K}}^{-2}{\text{mol}}^{-1}$.

Hence, substitute these values in the above equation to calculate the value of $\Delta \beta$.

  $\begin{array}{c}\Delta \beta =\left(\begin{array}{l}\left(6.97\text{mJ}{\text{K}}^{-2}{\text{mol}}^{-1}+2\times 9.61\text{mJ}{\text{K}}^{-2}{\text{mol}}^{-1}\right)-\\ \left(75.5\text{mJ}{\text{K}}^{-2}{\text{mol}}^{-1}+2\times 12.98\text{mJ}{\text{K}}^{-2}{\text{mol}}^{-1}\right)\end{array}\right)\\ =26.19-101.46\\ =-75.27\text{mJ}{\text{K}}^{-2}{\text{mol}}^{-1}\end{array}$

The value of $\Delta \gamma$ is calculated as shown below.

    $\Delta \gamma =\left(\gamma \left({\text{CO}}_{\text{2}},\text{g}\right)+2\times \gamma \left({\text{H}}_{\text{2}}\text{O},\text{g}\right)\right)-\left(\gamma \left({\text{CH}}_{\text{4}}\text{,}\text{g}\right)+2\times \gamma \left({\text{O}}_{\text{2}},\text{g}\right)\right)$

It is given that $\gamma \left({\text{CO}}_{\text{2}},\text{g}\right)$ is $-0.82\text{μJ}{\text{K}}^{-3}{\text{mol}}^{-1}$, $\gamma \left({\text{H}}_{\text{2}}\text{O},\text{g}\right)$ is $1.184\text{μJ}{\text{K}}^{-3}{\text{mol}}^{-1}$, $\gamma \left({\text{CH}}_{\text{4}}\text{,}\text{g}\right)$ is $-17.99\text{μJ}{\text{K}}^{-3}{\text{mol}}^{-1}$, $\gamma \left({\text{O}}_{\text{2}},\text{g}\right)$ is $-3.862\text{μJ}{\text{K}}^{-3}{\text{mol}}^{-1}$.

Hence, substitute these values in the above equation to calculate the value of $\Delta \gamma$.

  $\begin{array}{c}\Delta \gamma =\left(\begin{array}{c}\left(-0.82\text{μJ}{\text{K}}^{-3}{\text{mol}}^{-1}+2\times 1.184\text{μJ}{\text{K}}^{-3}{\text{mol}}^{-1}\right)-\\ \left(-17.99\text{μJ}{\text{K}}^{-3}{\text{mol}}^{-1}+2\times -3.862\text{μJ}{\text{K}}^{-3}{\text{mol}}^{-1}\right)\end{array}\right)\\ =27.2\text{62 μJ}{\text{K}}^{-3}{\text{mol}}^{-1}\end{array}$

The formula to calculate the standard enthalpy of combustion at a given temperature is,

  $\begin{array}{c}\Delta H_{\text{f}}^{\text{ο}}(T_2)=\Delta H_{\text{f}}^{\text{ο}}(T_1)+{\displaystyle \underset{T_1}{\overset{T_2}{\int }}{C}_{\text{p}}°}dT\\ =\Delta H_{\text{f}}^{\text{ο}}(T_1)+{\displaystyle \underset{T_1}{\overset{T_2}{\int }}\left(\Delta \alpha +\Delta \beta T+\Delta \gamma T^2\right)}dT\\ =\Delta H_{\text{f}}^{\text{ο}}(T_1)+\Delta \alpha \left(T_2-T_1\right)+\frac{1}{2}\Delta \beta \left(T_2^2-T_1^2\right)+\frac{1}{3}\Delta \gamma \left(T_2^3-T_1^3\right)\end{array}$

The value of $T_1$ is $298\text{K}$ and the value of $T_2$ is $500\text{K}$.

Substitute the values of $T_1$, $T_2$, $\Delta \alpha$, $\Delta \beta$, $\Delta \gamma$ in the above expression to calculate the standard enthalpy of formation at $500\text{K}$.

    $\begin{array}{c}\Delta H_{\text{f}}^{\text{ο}}(500\text{K})=\left(\begin{array}{c}-802.34\times {10}^3{\text{J mol}}^{-1}+21.68\text{J}{\text{K}}^{-1}{\text{mol}}^{-1}\left(500\text{K}-298\text{K}\right)\\ +\frac{1}{2}\times \left(-75.27\times {10}^3\text{J}{\text{K}}^{-2}{\text{mol}}^{-1}\right)\left({\left(500\text{K}\right)}^2-{\left(298\text{K}\right)}^2\right)\\ +\frac{1}{3}\times 27.262\times {10}^{-6}\text{J}{\text{K}}^{-3}{\text{mol}}^{-1}\left({\left(500\text{K}\right)}^3-{\left(298\text{K}\right)}^3\right)\end{array}\right)\\ =-797960.64+6066611460+895.433\\ =\underset{\_}{60.65\times 10^{8}Jmo{l}^{-1}}\end{array}$

Hence, the value of standard enthalpy of formation at $500\text{K}$ is $\underset{\_}{60.65\times 10^{8}Jmo{l}^{-1}}$.