Chapter 5, Problem 1KSP

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

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

$\begin{array}{l}\left(\text{a}\right)\hfill \\ \text{-608}\text{.7 kJ/mol}\hfill \\ \left(\text{b}\right)\hfill \\ \text{-81}\text{.1 kJ/mol}\hfill \\ \left(\text{c}\right)\hfill \\ \text{-37}\text{.1 kJ/mol}\hfill \\ \left(\text{d}\right)\hfill \\ \text{+81}\text{.1 kJ/mol}\hfill \\ \left(\text{e}\right)\hfill \\ \text{+37}\text{.1 kJ/mol}\hfill \end{array}$

Interpretation Introduction

Interpretation:

The standard enthalpy of reaction is to be calculated.

Concept introduction:

The standard enthalpy for a reaction is the amount of enthalpy changethat occursunder standard conditions, that is, at 27°C and $1\text{atm}$.

The standard enthalpy of the reaction is to be determined using the equation given below:

$\Delta H{°}_{\text{rxn}}={\displaystyle \sum n}\Delta H_{\text{f}}^{\text{°}}\left(\text{products}\right)-{\displaystyle \sum m}\Delta H_{\text{f}}^{°}\left(\text{reactants}\right)$

Here, the stoichiometric coefficients are represented by m for reactants and n for products, and enthalpy of formation under standard conditions is represented by $\Delta H_{\text{f}}^{°}$.

The value of enthalpy of formation of an element is zero at its most stable state.

Answer to Problem 1KSP

Solution: Option (e).

Explanation of Solution

Reason for the correct option:

The given reaction is as follows:

$\text{Mg}{\left(\text{OH}\right)}_{\text{2}}\left(s\right)\to \text{MgO}\left(s\right)+{\text{H}}_{\text{2}}\text{O}\left(l\right)$

From appendix 2,

$\begin{array}{c}\Delta H{°}_{\text{f}}\left(\text{Mg}{\left(\text{OH}\right)}_2\right)=-924.54\text{ kJ}/\text{mol}\\ \Delta H{°}_{\text{f}}\left(\text{MgO}\right)=-601.7\text{ kJ}/\text{mol}\\ \Delta H{°}_{\text{f}}\left({\text{H}}_{\text{2}}\text{O}\right)=-285.8\text{ kJ}/\text{mol}\end{array}$

Calculate the standard enthalpy of the reactionas follows:

$\Delta H{°}_{\text{rxn}}=\left[2\Delta H{°}_{\text{f}}\left(\text{MgO}\right)+\Delta H{°}_{\text{f}}\left({\text{H}}_{\text{2}}\text{O}\right)\right]-\left[\Delta H{°}_{\text{f}}\left(\text{Mg}{\left(\text{OH}\right)}_{\text{2}}\right)\right]$

Substitute $-924.54\text{ kJ}/\text{mol}$ for $\Delta H{°}_{\text{f}}\left(\text{Mg}{\left(\text{OH}\right)}_2\right)$,$-601.7\text{ kJ}/\text{mol}$ for $\Delta H{°}_{\text{f}}\left(\text{MgO}\right)$, and $-285.8\text{ kJ}/\text{mol}$ for $\Delta H{°}_{\text{f}}\left({\text{H}}_{\text{2}}\text{O}\right)$ in the above equation

$\begin{array}{c}\Delta H{°}_{\text{rxn}}=\left[\left(-601.7\text{ kJ}/\text{mol}\right)+\left(-285.8\text{ kJ}/\text{mol}\right)\right]-\left[\left(-924.54\text{ kJ}/\text{mol}\right)\right]\\ =37.04\text{ kJ}/\text{mol}\end{array}$

Hence, the enthalpy change for the reaction is $37.04\text{ kJ}/\text{mol}$.

Hence, option (e) is correct.

Reason for the incorrect options:

Option (a) is incorrect as the calculated value of enthalpy change for the reactiondoes not match with the valuegiven in option (a).

Option (b)is incorrect as the calculated value of enthalpy change for the reactiondoes not match with the value given in option (b).

Option (c) is incorrect as the calculated value of enthalpy change for the reactiondoes not match with the valuegiven in option (c).

Option (d) is incorrect as the calculated value of enthalpy change for the reactiondoes not match with the value given in option (d).

Hence, option (a), (b), (c), and (d) are incorrect.