Chapter 13, Problem 13.110QP

Interpretation Introduction

Interpretation: The fuel values of formamide and methyl formate are to be calculated.

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 13.110QP

The fuel value of formamide is $-11.95\text{kJ/g}$ and the fuel value of methyl formate is $-16.11\text{kJ/g}$.

Explanation of Solution

Given

The enthalpy of formation of formamide is $-251\text{kJ/mol}$.

The enthalpy of formation of methyl formate is $-391\text{kJ/mol}$.

The balanced chemical equation for the combustion of formamide is,

${\text{CHONH}}_2+3{\text{O}}_2\left(g\right)\to {\text{CO}}_2\left(g\right)+{\text{NO}}_2\left(g\right)+\frac{3}{2}{\text{H}}_2\text{O}\left(l\right)$

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

$\begin{array}{c}\Delta H^{\circ }={\displaystyle \sum n\Delta H^{\circ }\left(\text{products}\right)}-{\displaystyle \sum m\Delta H^{\circ }\left(\text{reactants}\right)}\\ =\left(\Delta H_{{\text{CO}}_2\left(g\right)}^{\circ }+\Delta H_{{\text{NO}}_2\left(g\right)}^{\circ }+\frac{3}{2}\Delta H_{{\text{H}}_2\text{O}\left(l\right)}^{\circ }\right)-\left(\Delta H_{{\text{CHONH}}_2}^{\circ }+3\Delta H_{{\text{O}}_2\left(g\right)}^{\circ }\right)\end{array}$

Substitute the values of $\Delta H_{{\text{CO}}_2\left(g\right)}^{\circ }$, $\Delta H_{{\text{NO}}_2\left(g\right)}^{\circ }$, $\Delta H_{{\text{H}}_2\text{O}\left(l\right)}^{\circ }$, $\Delta H_{{\text{CHONH}}_2}^{\circ }$ and $\Delta H_{{\text{O}}_2\left(g\right)}^{\circ }$ from appendix 4 in equation.

$\begin{array}{c}\Delta H^{\circ }=\left(\begin{array}{c}\left(1\times -393.5\text{kJ/mol}+1\times 33.2\text{kJ/mol}+\frac{3}{2}\times -285.83\text{kJ/mol}\right)\\ -\left(1\times -251\text{kJ/mol}+3\times 0.0\text{kJ/mol}\right)\end{array}\right)\\ =-538.05\text{kJ/mol}\end{array}$

The chemical equation for the combustion of methyl formate is,

${\text{C}}_2{\text{H}}_4{\text{O}}_2+2{\text{O}}_2\left(g\right)\to 2{\text{CO}}_2\left(g\right)+2{\text{H}}_2\text{O}\left(l\right)$

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

$\begin{array}{c}\Delta H^{\circ }={\displaystyle \sum n\Delta H^{\circ }\left(\text{products}\right)}-{\displaystyle \sum m\Delta H^{\circ }\left(\text{reactants}\right)}\\ =\left(2\Delta H_{{\text{CO}}_2\left(g\right)}^{\circ }+2\Delta H_{{\text{H}}_2\text{O}\left(l\right)}^{\circ }\right)-\left(\Delta H_{{\text{C}}_2{\text{H}}_4{\text{O}}_2}^{\circ }+2\Delta H_{{\text{O}}_2\left(g\right)}^{\circ }\right)\end{array}$

Substitute the values of $\Delta H_{{\text{CO}}_2\left(g\right)}^{\circ }$, $\Delta H_{{\text{H}}_2\text{O}\left(l\right)}^{\circ }$, $\Delta H_{{\text{C}}_2{\text{H}}_4{\text{O}}_2}^{\circ }$ and $\Delta H_{{\text{O}}_2\left(g\right)}^{\circ }$ from appendix C in equation.

$\begin{array}{c}\Delta H^{\circ }=\left(\begin{array}{c}\left(2\times -393.5\text{kJ/mol}+2\times -285.83\text{kJ/mol}\right)\\ -\left(1\times -391\text{kJ/mol}+2\times 0.0\text{kJ/mol}\right)\end{array}\right)\\ =-967.66\text{kJ/mol}\end{array}$

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

$\begin{array}{c}\text{Fuel}\text{value}=\frac{-538.05\text{kJ/mol}}{45.04\text{g/mol}}\\ =-11.95\text{kJ/g}\end{array}$

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

$\begin{array}{c}\text{Fuel}\text{value}=\frac{-967.66\text{kJ/mol}}{60.05\text{g/mol}}\\ =-16.11\text{kJ/g}\end{array}$

Conclusion

The fuel value of formamide is $-11.95\text{kJ/g}$ and the fuel value of methyl formate is $-16.11\text{kJ/g}$.