Chapter 6, Problem 6.118QP

(a)

Interpretation Introduction

Interpretation:

The standard enthalpy of neutralization of ${\text{HF}}_{\left(\text{aq}\right)}$ 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 6.118QP

The standard enthalpy of neutralization of ${\text{HF}}_{\left(\text{aq}\right)}$ is $\text{-65}\text{.2}\text{kJ/mole}\text{.}$

Explanation of Solution

The standard enthalpy of neutralization of ${\text{HF}}_{\left(\text{aq}\right)}$ is calculated as,

$\begin{array}{l}\text{ΔH°}\text{=}{\text{ΔH°}}_{\text{f}}\left({\text{F}}^{\text{-}}\right){\text{+ΔH°}}_{\text{f}}\left({\text{H}}_{\text{2}}\text{O}\right)\text{-}\left[{\text{ΔH°}}_{\text{f}}\left(\text{HF}\right){\text{+ΔH°}}_{\text{f}}\left({\text{OH}}^{\text{-}}\right)\right]\\ \\ \text{ΔH°}\text{=}\left[\left(\text{1}\right)\left(\text{-329}\text{.1kJ/mole}\right)\text{+}\left(\text{1}\right)\left(\text{-285}\text{.8kJ/mole}\right)\right]\text{-}\left[\left(\text{1}\right)\left(\text{-320}\text{.1}\text{kJ/mole}\right)\text{+}\left(\text{1}\right)\left(\text{-229}\text{.6}\text{kJ/mole}\right)\right]\\ \\ \text{ΔH°}\text{=}\text{-65}\text{.2}\text{kJ/mole}\end{array}$

The standard enthalpy of neutralization of ${\text{HF}}_{\left(\text{aq}\right)}$ is $\text{-65}\text{.2}\text{kJ/mole}\text{.}$

(b)

Interpretation Introduction

Interpretation:

The standard enthalpy change for the 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 6.118QP

The standard enthalpy change for the reaction is $\text{-9}\text{.0}\text{kJ/mole}\text{.}$

Explanation of Solution

The standard enthalpy of neutralization of ${\text{HF}}_{\left(\text{aq}\right)}$ is calculated as,

$\begin{array}{l}\text{ΔH°}\text{=}{\text{ΔH°}}_{\text{f}}\left({\text{F}}^{\text{-}}\right){\text{+ΔH°}}_{\text{f}}\left({\text{H}}_{\text{2}}\text{O}\right)\text{-}\left[{\text{ΔH°}}_{\text{f}}\left(\text{HF}\right){\text{+ΔH°}}_{\text{f}}\left({\text{OH}}^{\text{-}}\right)\right]\\ \\ \text{ΔH°}\text{=}\left[\left(\text{1}\right)\left(\text{-329}\text{.1kJ/mole}\right)\text{+}\left(\text{1}\right)\left(\text{-285}\text{.8kJ/mole}\right)\right]\text{-}\left[\left(\text{1}\right)\left(\text{-320}\text{.1}\text{kJ/mole}\right)\text{+}\left(\text{1}\right)\left(\text{-229}\text{.6}\text{kJ/mole}\right)\right]\\ \\ \text{ΔH°}\text{=}\text{-65}\text{.2}\text{kJ/mole}\end{array}$

The standard enthalpy of neutralization of ${\text{HF}}_{\left(\text{aq}\right)}$ is $\text{-65}\text{.2}\text{kJ/mole}\text{.}$

The given equations are added to get the final question.

The equations to be added are,

$\begin{array}{l}\underset{\_}{\begin{array}{l}{\text{HF}}_{\left(\text{aq}\right)}{\text{+OH}}^{\text{-}}{}_{\left(\text{aq}\right)}\stackrel{}{\to }{\text{F}}^{\text{-}}{}_{\left(\text{aq}\right)}{\text{+H}}_{\text{2}}{\text{O}}_{\left(\text{l}\right)}\text{ΔH°=-65}\text{.2}\text{kJ/mole}\\ {\text{H}}_{\text{2}}{\text{O}}_{\left(\text{l}\right)}\stackrel{}{\to }{\text{H}}^{\text{+}}{}_{\left(\text{aq}\right)}{\text{+OH}}^{\text{-}}{}_{\left(\text{aq}\right)}\text{ΔH°=+56}\text{.2}\text{kJ/mole}\end{array}}\\ {\text{HF}}_{\left(\text{aq}\right)}\stackrel{}{\to }{\text{H}}^{\text{+}}{}_{\left(\text{aq}\right)}{\text{+F}}^{\text{-}}{}_{\left(\text{aq}\right)}\text{ΔH°=-9}\text{.0}\text{kJ/mole}\end{array}$

The standard enthalpy change for the reaction is $\text{-9}\text{.0}\text{kJ/mole}\text{.}$