Chapter 5, Problem 92AP

At $\text{25}°\text{C}$, the standard enthalpy of formation of $\text{HF}\left(aq\right)\text{ is -320}{\text{.1 kJ/mol; of OH}}^{\text{-}}\left(aq\right)\text{, it is -229}{\text{.6 kJ/mol; of F}}^{\text{-}}\left(aq\right)\text{,it is -329}{\text{.1 kJ/mol; and of H}}_{\text{2}}\text{O}\left(l\right)\text{, it is -285}\text{.8 kJ/mol}\text{.}$

(a)

Calculate the standard enthalpy of neutralisation of $\text{HF}\left(aq\right)$:

$\text{HF}\left(aq\right)+{\text{OH}}^{\text{-}}\left(aq\right)\to {\text{F}}^{\text{-}}\left(aq\right)+{\text{H}}_{\text{2}}\text{O}\left(l\right)$

(b)

Using the value of -56.2 kJ as the standard enthalpy change for the reaction

$\text{H+}\left(aq\right)+{\text{OH}}^{\text{-}}\left(aq\right)\to {\text{H}}_{\text{2}}\text{O}\left(l\right)$

calculate the standard enthalpy change for the reaction

$\text{HF}\left(aq\right)\to {\text{H}}^{+}\left(aq\right)+{\text{F}}^{\text{-}}\left(aq\right)$

Interpretation Introduction

Interpretation:

The standard enthalpy of neutralization for the given compound and the standard enthalpy change for the given reaction are to be calculated.

Concept introduction:

The standard enthalpy for a reaction is the amount of enthalpy that occurs under standard conditions.

The standard enthalpy of a reaction is determined by using the equation given below:

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

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

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

Answer to Problem 92AP

Solution:

(a) $-65.2\text{ kJ}/\text{mol}$

(b) $-9\text{ kJ}/\text{mol}$

Explanation of Solution

Given information:

$\begin{array}{c}\Delta H{°}_{\text{f}}\left[\text{HF}\left(aq\right)\right]=-320.1\text{ kJ}/\text{mol}\\ \Delta H{°}_{\text{f}}\left[{\text{OH}}^{-}\left(aq\right)\right]=-229.6\text{ kJ}/\text{mol}\\ \Delta H{°}_{\text{f}}\left[{\text{F}}^{-}\left(aq\right)\right]=-329.1\text{ kJ}/\text{mol}\\ \Delta H{°}_{\text{f}}\left[{\text{H}}_{\text{2}}\text{O}\left(l\right)\right]=-285.8\text{ kJ}/\text{mol}\end{array}$

a) The standard enthalpy of neutralization of $\text{HF}(aq)$

The given reaction is as follows:

$\text{HF}\left(aq\right)+{\text{OH}}^{-}\left(aq\right)\to {\text{F}}^{-}\left(aq\right)+{\text{H}}_{\text{2}}\text{O}\left(l\right)$

Calculate the enthalpy of neutralization for the given reaction as follows:

$\Delta H°=\left\{\Delta H{°}_{\text{f}}\left[{\text{F}}^{-}\left(aq\right)\right]+\Delta H{°}_{\text{f}}\left[{\text{H}}_{\text{2}}\text{O}\left(l\right)\right]\right\}-\left\{\Delta H{°}_{\text{f}}\left[\text{HF}\left(aq\right)\right]+\Delta H{°}_{\text{f}}\left[{\text{OH}}^{-}\left(aq\right)\right]\right\}$

Substitute $-320.1\text{ kJ}/\text{mol}$ for $\Delta H{°}_{\text{f}}\left[\text{HF}\left(aq\right)\right]$, $-229.6\text{ kJ}/\text{mol}$ for $\Delta H{°}_{\text{f}}\left[{\text{OH}}^{-}\left(aq\right)\right]$, $-329.1\text{ kJ}/\text{mol}$ for $\Delta H{°}_{\text{f}}\left[{\text{F}}^{-}\left(aq\right)\right]$, and $-285.8\text{ kJ}/\text{mol}$ for $\Delta H{°}_{\text{f}}\left[{\text{H}}_{\text{2}}\text{O}\left(l\right)\right]$ in the above equation

$\begin{array}{c}\Delta H°=\left\{(-329.1\text{ kJ}/\text{mol)}+(-285.8\text{ kJ}/\text{mol)}\right\}-\left[(-320.1\text{ kJ}/\text{mol)}+(-229.6\text{ kJ}/\text{mol)}\right]\\ =\left(-614.9\text{ kJ}/\text{mol}\right)-\left(-549.7\text{ kJ}/\text{mol}\right)\\ =-65.2\text{ kJ}/\text{mol}\end{array}$

Hence, the enthalpy of neutralization for $\text{HF}$ is $-65.2\text{ kJ}/\text{mol}$.

b) The standard enthalpy change for the reaction

$\text{HF}\left(aq\right)\to {\text{H}}^{+}\left(aq\right)+{\text{F}}^{-}\left(aq\right)$

The given reaction with its enthalpy of reaction is as follows:

${\text{H}}^{+}\left(aq\right)+{\text{OH}}^{-}\left(aq\right)\to {\text{H}}_{\text{2}}\text{O}\left(l\right)\Delta H°=-56.2\text{ KJ}/\text{mol}$

Calculate the standard enthalpy of reaction as follows:

$\Delta H°=\left\{\Delta H{°}_f\left[{\text{H}}_{\text{2}}\text{O}\left(l\right)\right]\right\}-\left\{\Delta H{°}_{\text{f}}\left[{\text{H}}^{+}\left(aq\right)\right]+\Delta H{°}_{\text{f}}\left[{\text{OH}}^{-}\left(aq\right)\right]\right\}$

Substitute $-229.6\text{ kJ}/\text{mol}$ for $\Delta H{°}_{\text{f}}\left[{\text{OH}}^{-}\left(aq\right)\right]$, $-56.2\text{ kJ}/\text{mol}$ for $\Delta H°$, and $-285.8\text{ kJ}/\text{mol}$ for $\Delta H{°}_{\text{f}}\left[{\text{H}}_{\text{2}}\text{O}\left(l\right)\right]$ in the above equation

$\begin{array}{c}-56.2\text{ kJ}/\text{mol}=\left\{-285.8\text{ kJ}/\text{mol}\right\}-\left[\Delta H{°}_{\text{f}}\left[{\text{H}}^{+}\left(aq\right)\right]+\left(-229.6\text{ kJ}/\text{mol}\right)\right]\\ =-285.8\text{ kJ}/\text{mol}-\Delta H{°}_{\text{f}}\left[{\text{H}}^{+}\left(aq\right)\right]+229.6\text{ kJ}/\text{mol}\end{array}$

Rearrange the above equation for $\Delta H{°}_{\text{f}}\left[{\text{H}}^{+}\left(aq\right)\right]$

as follows:

$\begin{array}{c}\Delta H{°}_{\text{f}}\left[{\text{H}}^{+}\left(aq\right)\right]=-285.8\text{ kJ}/\text{mol}+56.2\text{ kJ}/\text{mol}+229.6\text{ kJ}/\text{mol}\\ =-285.8\text{ kJ}/\text{mol}+285.8\text{ kJ}/\text{mol}\\ =0\text{ kJ}/\text{mol}\end{array}$

The reaction of hydrogen fluoride is as follows:

$\text{HF}\left(aq\right)\to {\text{H}}^{+}\left(aq\right)+{\text{F}}^{-}\left(aq\right)$

Calculate the standard enthalpy of the reaction as follows:

$\Delta H°=\left\{\Delta H{°}_{\text{f}}\left[{\text{H}}^{+}\left(aq\right)\right]+\Delta H{°}_{\text{f}}\left[{\text{F}}^{-}\left(aq\right)\right]\right\}-\left\{\Delta H{°}_{\text{f}}\left[\text{HF}\left(aq\right)\right]\right\}$

Substitute $-320.1\text{ kJ}/\text{mol}$ for $\Delta H{°}_{\text{f}}\left[\text{HF}\left(aq\right)\right]$, $\text{0 kJ}/\text{mol}$ for $\Delta H{°}_{\text{f}}\left[{\text{H}}^{+}\left(aq\right)\right]$, and $-329.1\text{ kJ}/\text{mol}$ for $\Delta H{°}_{\text{f}}\left[{\text{F}}^{-}\left(aq\right)\right]$ in the above equation

$\begin{array}{c}\Delta H°=\left[\left(0\text{ kJ}/\text{mol}\right)+\left(-329.1\text{ kJ}/\text{mol}\right)\right]-\left\{-320.1\text{ kJ}/\text{mol}\right\}\\ =-329.1\text{ kJ}/\text{mol}+320.1\text{ kJ}/\text{mol}\\ =-9\text{ kJ}/\text{mol}\end{array}$

Hence, the required enthalpy of reaction is $-9\text{ kJ}/\text{mol}$.