Chapter 10, Problem 10.1KSP

Using data from Appendix 2, calculate the standard enthalpy of the following reaction.

$Mg{(OH)}_2(\text{s})\to MgO(\text{s})+H_{2}O(\text{l})$

(a) −608.7 kJ/mol (b) −81.1 kJ/mol (c) −37.1 kJ/mol (d) +81.1 kJ/mol (e) +37.1 kJ/mol

Interpretation Introduction

Interpretation:

The standard enthalpy of reaction has to be calculated.

Concept Introduction:

Standard enthalpy of reaction:

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 given reaction

Answer to Problem 10.1KSP

The enthalpy change of the reaction is $+37.1\text{kJ}{\text{mol}}^{\text{-1}}$

Explanation of Solution

Reasons for true option:

Standard enthalpy of formation of ${\text{Mg(OH)}}_{2(s)}\text{=-924}\text{.66 kJ}{\text{mol}}^{\text{-1}}$

Standard enthalpy of formation of ${\text{MgO}}_{\left(s\right)}\text{=-601}\text{.8 kJ}{\text{mol}}^{\text{-1}}$

Standard enthalpy of formation of ${\text{H}}_2{\text{O}}_{\left(l\right)}\text{=-285}\text{.8 kJ}{\text{mol}}^{\text{-1}}$

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

$\begin{array}{l}{\text{ΔH°}}_{\text{reaction}}\text{= }\left[{\text{ΔH°}}_{\text{f}}{\text{(MgO}}_{\left(\text{s}\right)}{\text{)+ΔH°}}_{\text{f}}{\text{(H}}_{\text{2}}{\text{O}}_{\left(\text{l}\right)}\text{)}\right]\text{-}\left[{\text{ΔH°}}_{\text{f}}{\text{(MgOH}}_{\left(\text{2}\right)\text{s}}\text{)}\right]\\ \\ {\text{ΔH°}}_{\text{reaction}}\text{= }\left[\text{(-601}\text{.8}\text{kJ}{\text{mol}}^{\text{-1}}\text{-285}\text{.8}\text{kJ}{\text{mol}}^{\text{-1}}\right]\text{-}\left[\text{-924}\text{.66}\text{kJ}{\text{mol}}^{\text{-1}}\right]\\ \\ {\text{ΔH°}}_{\text{reaction}}\text{= +37}\text{.1}\text{kJ}{\text{mol}}^{\text{-1}}\end{array}$

Standard enthalpy of formation of $\text{reaction=+37}\text{.1 kJ}{\text{mol}}^{\text{-1}}$

Reasons for false option:

The enthalpy of reaction from the given data shows that the options (a), (b), (c) and (d) are incorrect.

Conclusion

The standard enthalpy of given reaction was calculated.