Chapter 15, Problem 111A

Interpretation Introduction

Interpretation:

The molar enthalpy of combustion of the two compounds methane gas ( ${\text{CH}}_{\text{4}}$ ) or methanal vapor ( ${\text{CH}}_{\text{2}}\text{O}$ ) needs to be determined.

Concept introduction:

The enthalpy of combustion ${\text{(ΔH}}_{\text{comb}})$ of a substance is the enthalpy change for the complete burning of one mole of the substance.

Answer to Problem 111A

Molar enthalpy of combustion of methane ( ${\text{CH}}_{\text{4}}$ ) is greater than the molar enthalpy of combustion of methanol ( ${\text{CH}}_{\text{2}}\text{O}$ )

Explanation of Solution

The balanced chemical equation for methane gas ( ${\text{CH}}_{\text{4}}$ ) or methanal vapor ( ${\text{CH}}_{\text{2}}\text{O}$ ) is

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

The molar enthalpy of combustion of methane gas ( ${\text{CH}}_{\text{4}}$ ) and methanal vapor ( ${\text{CH}}_{\text{2}}\text{O}$ ) are

$\begin{array}{c}{\text{ΔH}}_{{\text{CH}}_{\text{4}}}=\text{570}\text{.22kJ/mol}\\ {\text{ΔH}}_{{\text{CH}}_{\text{2}}\text{O}}=\text{890}\text{.3kJ/mol}\end{array}$

Therefore,

$\begin{array}{l}{\text{ΔH}}_{{\text{CH}}_{\text{4}}}>{\text{ΔH}}_{{\text{CH}}_{\text{2}}\text{O}}\\ \text{890kJ/mol}>\text{570kJ/mol}\end{array}$

Hence, molar enthalpy of combustion of methane ( ${\text{CH}}_{\text{4}}$ ) is greater than the molar enthalpy of combustion of methanol ( ${\text{CH}}_{\text{2}}\text{O}$ ).

Conclusion

In order to calculate Gibbs free energy( $\text{ΔG}$ ), it is usually necessary to convert units because $\text{ΔS}$ is usually expressed in $\text{j/K}$, whereas $\text{ΔH}$ is expressed in $\text{kJ}$.