Chapter 5, Problem 5.129AP

Interpretation Introduction

Interpretation: The standard molar enthalpy of for the complete combustion of liquid ethanol with the use of Appendix $4$ is to be calculated.

Concept introduction: The standard molar enthalpy change for a balanced reaction $\left(\Delta H_{\text{balanced rxn}}^{\text{ο}}\right)$ is calculated by the formula,

$\Delta H_{\text{balanced rxn}}^{\text{ο}}={\displaystyle \sum n\times \Delta H_{\text{f}}^{\text{ο}}\left(\text{Products}\right)}-{\displaystyle \sum m\times \Delta H_{\text{f}}^{\text{ο}}\left(\text{Reactants}\right)}$

To determine: The standard molar enthalpy of for the complete combustion of liquid ethanol.

Answer to Problem 5.129AP

The standard molar enthalpy of for the complete combustion of liquid ethanol is $\underset{\_}{-1,234.7kJ}$.

Explanation of Solution

Explanation

Given

Refer Appendix $4$.

The standard enthalpy of formation of ${\text{CO}}_2\left(g\right)$, $\left(\Delta H_{{\text{f, CO}}_2\left(g\right)}^{\text{ο}}\right)$ is $-393.5\text{kJ}/\text{mol}$.

The standard enthalpy of formation of ${\text{C}}_{\text{2}}{\text{H}}_{\text{5}}\text{OH}\left(l\right)$, $\left(\Delta H_{{\text{f, C}}_{\text{2}}{\text{H}}_{\text{5}}\text{OH}\left(l\right)}^{\text{ο}}\right)$ is $-277.7\text{kJ}/\text{mol}$.

The standard enthalpy of formation of ${\text{H}}_2\text{O}\left(g\right)$, $\left(\Delta H_{{\text{f, H}}_2\text{O}\left(g\right)}^{\text{ο}}\right)$ is $-241.8\text{kJ}/\text{mol}$.

The standard enthalpy of formation of ${\text{O}}_2\left(g\right)$, $\left(\Delta H_{{\text{f, O}}_2\left(g\right)}^{\text{ο}}\right)$ is $0\text{kJ}/\text{mol}$.

The combustion reaction of ethanol is,

${\text{C}}_{\text{2}}{\text{H}}_{\text{5}}\text{OH}\left(l\right)+{\text{O}}_2\left(g\right)\to {\text{CO}}_2\left(g\right)+{\text{H}}_2\text{O}\left(g\right)$

The above reaction is unbalanced.

To balance the number of carbon atoms multiply ${\text{CO}}_2\left(g\right)$, with two.

${\text{C}}_{\text{2}}{\text{H}}_{\text{5}}\text{OH}\left(l\right)+{\text{O}}_2\left(g\right)\to 2{\text{CO}}_2\left(g\right)+{\text{H}}_2\text{O}\left(g\right)$

To balance the oxygen atoms multiply ${\text{O}}_2\left(g\right)$, with three.

${\text{C}}_{\text{2}}{\text{H}}_{\text{5}}\text{OH}\left(l\right)+3{\text{O}}_2\left(g\right)\to 2{\text{CO}}_2\left(g\right)+{\text{H}}_2\text{O}\left(g\right)$

To balance the hydrogen atoms multiply ${\text{H}}_2\text{O}\left(g\right)$, with three.

${\text{C}}_{\text{2}}{\text{H}}_{\text{5}}\text{OH}\left(l\right)+3{\text{O}}_2\left(g\right)\to 2{\text{CO}}_2\left(g\right)+3{\text{H}}_2\text{O}\left(g\right)$ (1)

The above reaction is balanced now.

The standard molar enthalpy change for a balanced reaction $\left(\Delta H_{\text{balanced rxn}}^{\text{ο}}\right)$ is calculated by the formula,

$\Delta H_{\text{balanced rxn}}^{\text{ο}}={\displaystyle \sum n\times \Delta H_{\text{f}}^{\text{ο}}\left(\text{Products}\right)}-{\displaystyle \sum m\times \Delta H_{\text{f}}^{\text{ο}}\left(\text{Reactants}\right)}$ (2)

Where,

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

In the balanced chemical equation (1) the,

• Number of moles of product ${\text{H}}_2\text{O}\left(g\right)$ is $3$.
• Number of moles of product ${\text{CO}}_2\left(g\right)$ is $2$.
• Number of moles of reactant ${\text{C}}_{\text{2}}{\text{H}}_{\text{5}}\text{OH}\left(l\right)$ is $1$.
• Number of moles of reactant ${\text{O}}_2\left(g\right)$ is $3$.

The ${\displaystyle \sum n\times \Delta H_{\text{f}}^{\text{ο}}\left(\text{Products}\right)}$ for the balanced chemical reaction is calculated by the formula,

${\displaystyle \sum n\times \Delta H_{\text{f}}^{\text{ο}}\left(\text{Products}\right)}=3\text{mol}\times \left(\Delta H_{{\text{f, H}}_2\text{O}\left(g\right)}^{\text{ο}}\right)+2\text{mol}\times \left(\Delta H_{{\text{f, CO}}_2\left(g\right)}^{\text{ο}}\right)$ (3)

Substitute the value of $\Delta H_{{\text{f, H}}_2\text{O}\left(g\right)}^{\text{ο}}$ and $\Delta H_{{\text{f, CO}}_2\left(g\right)}^{\text{ο}}$ in equation (3).

$\begin{array}{c}{\displaystyle \sum n\times \Delta H_{\text{f}}^{\text{ο}}\left(\text{Products}\right)}=3\text{mol}\times -241.8\text{kJ}/\text{mol}+2\text{mol}\times -393.5\text{kJ}/\text{mol}\\ =-725.4\text{kJ}-787.\text{kJ}\\ =-1,512.4\text{kJ}\end{array}$

The ${\displaystyle \sum m\times \Delta H_{\text{f}}^{\text{ο}}\left(\text{Reactants}\right)}$ for the balanced chemical reaction (1) is calculated by the formula,

${\displaystyle \sum m\times \Delta H_{\text{f}}^{\text{ο}}\left(\text{Reactants}\right)}=1\text{mol}\times \left(\Delta H_{{\text{f, C}}_{\text{2}}{\text{H}}_{\text{5}}\text{OH}\left(l\right)}^{\text{ο}}\right)+1\text{mol}\times \left(\Delta H_{{\text{f, O}}_2\left(g\right)}^{\text{ο}}\right)$ (4)

Substitute the values of $\Delta H_{{\text{f, C}}_{\text{2}}{\text{H}}_{\text{5}}\text{OH}\left(l\right)}^{\text{ο}}$ and $\Delta H_{{\text{f, O}}_2\left(g\right)}^{\text{ο}}$ in equation (4).

$\begin{array}{c}{\displaystyle \sum m\times \Delta H_{\text{f}}^{\text{ο}}\left(\text{Reactants}\right)}=1\text{mol}\times -277.7\text{kJ}/\text{mol}+1\text{mol}\times 0\text{kJ}/\text{mol}\\ =-277.7\text{kJ}\end{array}$

Substitute the values of ${\displaystyle \sum n\times \Delta H_{\text{f}}^{\text{ο}}\left(\text{Products}\right)}$ and ${\displaystyle \sum m\times \Delta H_{\text{f}}^{\text{ο}}\left(\text{Reactants}\right)}$ in equation (2).

$\begin{array}{c}\Delta H_{\text{balanced rxn}}^{\text{ο}}=-1,512.4\text{kJ}-\left(-277.7\text{kJ}\right)\\ =-1,512.4\text{kJ}+277.7\text{kJ}\\ =-1,234.7\text{kJ}\end{array}$

Hence, standard molar enthalpy of for the complete combustion of liquid ethanol is $\underset{\_}{-1,234.7kJ}$.

Conclusion

Conclusion

The standard molar enthalpy of for the complete combustion of liquid ethanol is $\underset{\_}{-1,234.7kJ}$.