Chapter 13, Problem 13.109QP

Interpretation Introduction

Interpretation: The fuel values of formaldehyde and formic acid is to be calculated and the compound having higher fuel value is to be predicted.

Concept Introduction: The calorie content of substance is calculated by determining the energy released in the combustion of the one gram mass of substance.

To determine: The fuel values of formaldehyde and formic acid and the compound having higher fuel value out of it.

Answer to Problem 13.109QP

The fuel value of formaldehyde is $-19.00\text{kJ/g}$ and the fuel value of formic acid is $-13.064\text{kJ/g}$. The fuel value of formaldehyde is more than formic acid.

Explanation of Solution

Given

The enthalpy of formation of formaldehyde is $-108.6\text{kJ/mol}$.

The enthalpy of formation of formic acid is $-378.7\text{kJ/mol}$.

The balanced chemical equation for the combustion of formaldehyde is,

${\text{CH}}_2\text{O}+{\text{O}}_2\left(g\right)\to {\text{CO}}_2\left(g\right)+{\text{H}}_2\text{O}\left(l\right)$

The enthalpy change of the combustion reaction of formaldehyde is calculated by the formula,

$\begin{array}{c}\Delta H^{\circ }={\displaystyle \sum n\Delta H^{\circ }\left(\text{products}\right)}-{\displaystyle \sum m\Delta H^{\circ }\left(\text{reactants}\right)}\\ =\left(\Delta H_{{\text{CO}}_2\left(g\right)}^{\circ }+\Delta H_{{\text{H}}_2\text{O}\left(l\right)}^{\circ }\right)-\left(\Delta H_{{\text{CH}}_2\text{O}}^{\circ }+\Delta H_{{\text{O}}_2\left(g\right)}^{\circ }\right)\end{array}$

Substitute the values of $\Delta H_{{\text{CO}}_2\left(g\right)}^{\circ }$, $\Delta H_{{\text{H}}_2\text{O}\left(l\right)}^{\circ }$, $\Delta H_{{\text{CH}}_2\text{O}}^{\circ }$ and $\Delta H_{{\text{O}}_2\left(g\right)}^{\circ }$ from appendix C in equation.

$\begin{array}{c}\Delta H^{\circ }=\left(1\times -393.5\text{kJ/mol}+1\times -285.83\text{kJ/mol}\right)\\ -\left(1\times -108.6\text{kJ/mol}+1\times 0.0\text{kJ/mol}\right)\\ =-570.73\text{kJ/mol}\end{array}$

The chemical equation for the combustion of formic acid is,

${\text{2CH}}_2{\text{O}}_2+{\text{O}}_2\left(g\right)\to 2{\text{CO}}_2\left(g\right)+2{\text{H}}_2\text{O}\left(l\right)$

The enthalpy change of the combustion reaction of palmitic acid is calculated by the formula,

$\begin{array}{c}\Delta H^{\circ }={\displaystyle \sum n\Delta H^{\circ }\left(\text{products}\right)}-{\displaystyle \sum m\Delta H^{\circ }\left(\text{reactants}\right)}\\ =\left(2\Delta H_{{\text{CO}}_2\left(g\right)}^{\circ }+2\Delta H_{{\text{H}}_2\text{O}\left(l\right)}^{\circ }\right)-\left(2\Delta H_{{\text{CH}}_2{\text{O}}_2}^{\circ }+\Delta H_{{\text{O}}_2\left(g\right)}^{\circ }\right)\end{array}$

Substitute the values of $\Delta H_{{\text{CO}}_2\left(g\right)}^{\circ }$, $\Delta H_{{\text{H}}_2\text{O}\left(l\right)}^{\circ }$, $\Delta H_{{\text{CH}}_2{\text{O}}_2}^{\circ }$ and $\Delta H_{{\text{O}}_2\left(g\right)}^{\circ }$ from appendix C in equation.

$\begin{array}{c}\Delta H^{\circ }=\left(2\times -393.5\text{kJ/mol}+2\times -285.83\text{kJ/mol}\right)\\ -\left(2\times -378.7\text{kJ/mol}+1\times 0.0\text{kJ/mol}\right)\\ =-601.26\text{kJ/mol}\end{array}$

The heat produced per gram in the combustion of formaldehyde is calculated by dividing its enthalpy change of combustion by the molar mass.

$\begin{array}{c}\text{Fuel}\text{value}=\frac{-570.73\text{kJ/mol}}{30.031\text{g/mol}}\\ =-19.00\text{kJ/g}\end{array}$

The heat produced per gram in the combustion of formic acid is calculated by dividing its enthalpy change of combustion by the molar mass.

$\begin{array}{c}\text{Fuel}\text{value}=\frac{-601.26\text{kJ/mol}}{46.025\text{g/mol}}\\ =-13.064\text{kJ/g}\end{array}$

Conclusion

The fuel value of formaldehyde is $-19.00\text{kJ/g}$ and the fuel value of formic acid is $-13.064\text{kJ/g}$. The fuel value of formaldehyde is more than formic acid.