Chapter 6, Problem 6.61QP

Interpretation Introduction

Interpretation:

The enthalpy change for the given transformation 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 6.61QP

The enthalpy change for the given transformation is $\text{0}\text{.30}\text{kJ}{\text{mol}}^{\text{-1}}$ .

Explanation of Solution

The standard enthalpy change for the given transformation is given as,

From the given equation, the second equation is reversed and the sign of ${\text{ΔH°}}_{\text{reaction}}$ is changed.

$\begin{array}{l}\text{Reaction}\text{ΔH°(kJ}{\text{mol}}^{\text{-1}}\text{)}\\ \\ {\text{S}}_{\text{(rhombic)}}{\text{+O}}_{\text{2(g)}}\to {\text{SO}}_{\text{2(g)}}\text{-296}\text{.06}\text{kJ}{\text{mol}}^{\text{-1}}\\ \\ {\text{SO}}_{\text{2(g)}}\to {\text{S}}_{\text{(monoclinic)}}{\text{+O}}_{\text{2(g)}}\text{296}\text{kJ}{\text{mol}}^{\text{-1}}\end{array}$

${\text{S}}_{\left(\text{rhombic}\right)}\to {\text{S}}_{\left(\text{monoclinic}\right)}{\text{ΔH°}}_{\text{reaction}}\text{=0}\text{.30}\text{kJ}{\text{mol}}^{\text{-1}}$

Standard enthalpy change for the given transformation = $\text{0}\text{.30}\text{kJ}{\text{mol}}^{\text{-1}}$

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 $\text{0}\text{.30}\text{kJ}{\text{mol}}^{\text{-1}}$