Chapter 10, Problem 10.48QP

Interpretation Introduction

Interpretation: The standard enthalpy change for the given reaction 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,

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

Answer to Problem 10.48QP

The standard enthalpy change for the given transformation is $-84.6\text{kJ}{\text{mol}}^{\text{-1}}$ .

Explanation of Solution

To calculate the ${\text{ΔH°}}_{\text{reaction}}$ for the first equation

Two moles of Graphite are needed; hence the first equation is multiplied by two and the ${\text{ΔH°}}_{\text{reaction}}$ value becomes,

$\begin{array}{l}{\text{2C}}_{\left(\text{graphite}\right)}{\text{+2O}}_{\text{2(g)}}\to {\text{2CO}}_{\text{2(g)}}\\ \\ {\text{ΔH°}}_{\text{reaction}}\text{=-787}\text{.0}\text{kJ}{\text{mol}}^{\text{-1}}\end{array}$

To calculate the ${\text{ΔH°}}_{\text{reaction}}$ for the second equation

Three moles of Hydrogen are needed as reactant; hence the second equation is multiplied by three and the ${\text{ΔH°}}_{\text{reaction}}$ value becomes,

$\begin{array}{l}{\text{3H}}_{\text{2(g)}}\text{+}\frac{\text{3}}{\text{2}}{\text{O}}_{\text{2(g)}}\to {\text{3H}}_{\text{2}}{\text{O}}_{\text{(l)}}\\ \\ {\text{ΔH}}^{\text{°}}{}_{\text{reaction}}\text{=-857}\text{.4}\text{kJ}{\text{mol}}^{\text{-1}}\end{array}$

To calculate the ${\text{ΔH°}}_{\text{reaction}}$ for the third equation

One mole of Ethane is obtained as product. The third equation has two moles of Ethane as reactants hence, the equation is reversed and divided by two and the ${\text{ΔH°}}_{\text{reaction}}$ value becomes,

$\begin{array}{l}{\text{2CO}}_{\text{2(g)}}{\text{+3H}}_{\text{2}}{\text{O}}_{{}_{\left(\text{l}\right)}}\to {\text{C}}_{\text{2}}{\text{H}}_{{\text{6}}_{\left(\text{g}\right)}}\text{+}\frac{\text{7}}{\text{2}}{\text{O}}_{\text{2(g)}}\\ \\ {\text{ΔH°}}_{\text{reaction}}\text{=+1559}\text{.8}\text{kJ}{\text{mol}}^{\text{-1}}\end{array}$

All the three equations from the above steps are summed up

$\begin{array}{l}\text{Reaction}\text{ΔH°(kJ}{\text{mol}}^{\text{-1}}\text{)}\\ \\ {\text{2C}}_{\left(\text{graphite}\right)}{\text{+2O}}_{\text{2(g)}}\to {\text{2CO}}_{\text{2(g)}}\text{-787}\text{.0}\text{kJ}{\text{mol}}^{\text{-1}}\\ \\ {\text{3H}}_{\text{2(g)}}\text{+}\frac{\text{3}}{\text{2}}{\text{O}}_{\text{2(g)}}\to {\text{3H}}_{\text{2}}{\text{O}}_{\text{(l)}}\text{-857}\text{.4}\text{kJ}{\text{mol}}^{\text{-1}}\\ \\ {\text{2CO}}_{\text{2(g)}}{\text{+3H}}_{\text{2}}{\text{O}}_{{}_{\left(\text{l}\right)}}\to {\text{C}}_{\text{2}}{\text{H}}_{{\text{6}}_{\left(\text{g}\right)}}\text{+}\frac{\text{7}}{\text{2}}{\text{O}}_{\text{2(g)}}\text{+1559}\text{.8}\text{kJ}{\text{mol}}^{\text{-1}}\end{array}$

${\text{2C}}_{\text{(graphite)}}{\text{+3H}}_{\text{2(g)}}\to {\text{C}}_{\text{2}}{\text{H}}_{\text{6(g)}}{\text{ΔH°}}_{\text{reaction}}\text{=-84}\text{.6}\text{kJ}{\text{mol}}^{\text{-1}}$

Standard enthalpy change for the given transformation = $-84.6\text{kJ}{\text{mol}}^{\text{-1}}$

Conclusion

The standard enthalpy of the reaction was calculated using the standard enthalpies of formation of the products formed.  The standard enthalpy change for the given reaction was found to be $-84.6\text{kJ}{\text{mol}}^{\text{-1}}$.