Chapter 6, Problem 6.124QP

Determine the standard enthalpy of formation of ethanol (C₂H₅OH) from its standard enthalpy of combustion (−1367.4 kJ/mol).

Interpretation Introduction

Interpretation:

The standard enthalpy of formation of Ethanol has to be calculated.

Concept Introduction:

The change in enthalpy that is associated with the formation of one mole of a substance from its related elements being in standard state is called standard enthalpy of formation (${\text{ΔH}}_{\text{f}}\text{°}$).  The standard enthalpy of formation is used to determine the standard enthalpies of compound and element.

The standard enthalpy of reaction is the enthalpy of reaction that takes place under standard conditions.

 The equation for determining the standard enthalpies of compound and element can be given by,

${\text{ΔH°}}_{\text{reaction}}\text{=}{\displaystyle \sum {\text{nΔH°}}_{\text{f}}\text{(products)}}\text{-}{\displaystyle \sum {\text{mΔH°}}_{\text{f}}\text{(reactants)}}$

To calculate: The standard enthalpy of formation of Ethanol

Answer to Problem 6.124QP

The standard enthalpy of formation of Ethanol is $\text{-277}\text{.0}\text{kJ}{\text{mol}}^{\text{-1}}$

Explanation of Solution

Standard enthalpy of formation of ${\text{CO}}_{\text{2}}$= $\text{-393}\text{.5}\text{kJ}{\text{mol}}^{\text{-1}}$

Standard enthalpy of formation of Water = $\text{-285}\text{.8}\text{kJ}{\text{mol}}^{\text{-1}}$

Standard enthalpy of formation of Oxygen = $\text{0}\text{kJ}{\text{mol}}^{\text{-1}}$

The equation of formation of Ethanol from its elements can be given as,

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

The reverse reaction of the above equation gives the combustion of the Ethanol that can be given as,

${\text{2CO}}_{\text{2(g)}}{\text{+3H}}_{\text{2}}{\text{O}}_{\left(\text{l}\right)}\to {\text{C}}_{\text{2}}{\text{H}}_{\text{5}}{\text{OH}}_{\left(\text{l}\right)}{\text{+3O}}_{\text{2(g)}}\text{ΔH°=+1367}\text{.4}\text{kJ}{\text{mol}}^{\text{-1}}$

Water is removed from the reactant side and ${\text{C}}_{\left(\text{graphite}\right)}$ is added. The equation can be given as,

$\begin{array}{l}{\text{2CO}}_{\text{2(g)}}{\text{+3H}}_{\text{2}}{\text{O}}_{\left(\text{l}\right)}\to {\text{C}}_{\text{2}}{\text{H}}_{\text{5}}{\text{OH}}_{\left(\text{l}\right)}{\text{+3O}}_{\text{2(g)}}\text{ΔH°=+1367}\text{.4}\text{kJ}{\text{mol}}^{\text{-1}}\\ \\ {\text{2C}}_{\left(\text{graphite}\right)}{\text{+2O}}_{\text{2(g)}}\to {\text{2CO}}_{\left(\text{g}\right)}\text{ΔH°=2(-393}\text{.5}\text{kJ}{\text{mol}}^{\text{-1}}\text{)}\\ \\ {\text{3H}}_{\text{2(g)}}\text{+}\frac{\text{3}}{\text{2}}{\text{O}}_{\text{2(g)}}\to {\text{3H}}_{\text{2}}{\text{O}}_{\left(\text{l}\right)}\text{ΔH°=3(-285}\text{.8}\text{kJ}{\text{mol}}^{\text{-1}}\text{)}\end{array}$

$\begin{array}{l}{\text{2C}}_{\left(\text{Graphite}\right)}\text{+}\frac{\text{1}}{\text{2}}{\text{O}}_{\text{2(g)}}{\text{+3H}}_{\text{2(g)}}\to {\text{C}}_{\text{2}}{\text{H}}_{\text{5}}{\text{OH}}_{\left(\text{l}\right)}\text{ΔH°=-277}\text{.0}\text{kJ}{\text{mol}}^{\text{-1}}\\ \end{array}$

Standard enthalpy of formation of Ethanol=$\text{-277}\text{.0}\text{kJ}{\text{mol}}^{\text{-1}}$

Conclusion

The standard enthalpy of formation of Ethanol was calculated by using the values of standard enthalpies of reactants and the products.  The standard enthalpy of formation of Ethanol is found to be $\text{-277}\text{.0}\text{kJ}{\text{mol}}^{\text{-1}}$.