Chapter 6, Problem 6.63QP

From the following heats of combustion,

$\begin{array}{l}{\text{CH}}_3\text{OH}(l)+\frac{3}{2}{\text{O}}_2(g)\stackrel{}{\to }{\text{CO}}_2(g)+2{\text{H}}_2\text{O}(l)\hfill \\ \Delta H_{rxn}^{\text{o}}=-726.4\text{kJ/mol}\hfill \\ \text{C}(\text{graphite})+{\text{O}}_2(g)\stackrel{}{\to }{\text{CO}}_2(g)\hfill \\ \Delta H_{\text{rxn}}^{\text{o}}=-393.5\text{kJ/mol}\hfill \\ {\text{H}}_2(g)+\frac{1}{2}{\text{O}}_2(g)\stackrel{}{\to }{\text{H}}_2\text{O}(l)\hfill \\ \Delta H_{rxn}^{\text{o}}=-285.8\text{kJ/mol}\hfill \end{array}$

calculate the enthalpy of formation of methanol (CH₃OH) from its elements:

$\text{C}(\text{graphite})+2{\text{H}}_2(g)+\frac{1}{2}{\text{O}}_2(g)\stackrel{}{\to }{\text{CH}}_3\text{OH}(l)$

Interpretation Introduction

Interpretation: The enthalpy of formation of Methanol from its elements 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 enthalpy of formation of Methanol

Answer to Problem 6.63QP

The standard heat for the formation of Methanol is $\text{-238}\text{.7}\text{kJ}{\text{mol}}^{\text{-1}}$

Explanation of Solution

Some changes in the reactions and their enthalpies are made to get the enthalpy of formation of Methanol. The changes are as follows,

$\begin{array}{l}\text{Reaction}\text{ΔH°(kJ}{\text{mol}}^{\text{-1}}\text{)}\\ \\ {\text{CO}}_{\text{2(g)}}{\text{+2H}}_{\text{2}}{\text{O}}_{\left(\text{l}\right)}\to {\text{CH}}_{\text{3}}{\text{OH}}_{\left(\text{l}\right)}\text{+}\frac{\text{3}}{\text{2}}{\text{O}}_{\text{2(g)}}\text{726}\text{.4}\\ \\ {\text{C}}_{\text{(Graphite)}}{\text{+O}}_{\text{2(g)}}\to {\text{CO}}_{\text{2(g)}}\text{-393}\text{.5}\\ \\ {\text{2H}}_{\text{2(g)}}{\text{+O}}_{\text{2(g)}}\to {\text{2H}}_{\text{2}}{\text{O}}_{\text{(l)}}\text{2(-285}\text{.8)}\end{array}$

${\text{C}}_{\text{(Graphite)}}{\text{+2H}}_{\text{2(g)}}\text{+}\frac{\text{1}}{\text{2}}{\text{O}}_{\text{2(g)}}\to {\text{CH}}_{\text{3}}{\text{OH}}_{{}_{\left(\text{l}\right)}}{\text{ΔH°}}_{\text{reaction}}\text{=-238}\text{.7}\text{kJ}{\text{mol}}^{\text{-1}}$

Standard heat of formation of Methanol = $\text{-238}\text{.7}\text{kJ}{\text{mol}}^{\text{-1}}$ since the reaction involves in the formation of one mole of Methanol from its elements in their standard states.

Conclusion

The standard enthalpy of the reaction was calculated using the standard enthalpies of formation of the products formed.  Since the reaction involves in the formation of one mole of Methanol from its elements in their standard states.  The standard enthalpy of the reaction and the standard heat of formation of Methanol are found to be the same.  The standard heat of formation of Methanol was found to be $\text{-238}\text{.7}\text{kJ}{\text{mol}}^{\text{-1}}$.