Chapter 10, Problem 10.142QP

(a)

Interpretation Introduction

Interpretation:

The final temperature of the product and the standard enthalpy of formation of ${\text{Ca(OH)}}_{\text{2}}$ has to be calculated.

Concept Introduction:

Specific heat can be defined as quantity of heat required to raise the temperature of $\text{1}\text{g}$ substance by $\text{1°C}$.  The relationship between heat and change in temperature can be expressed by the equation given below.

$\text{q=smΔT}$

Where,

$\text{q=}$ Heat added

s = Specific heat

$\text{m=}$ Mass

$\text{ΔT=}$ Change in temperature.

The unit of specific heat is $\text{J}{\text{g}}^{\text{-1}}\text{.°C}$.

Answer to Problem 10.142QP

The final temperature is $\text{758°C}$ .

Explanation of Solution

To calculate the moles of ${\text{H}}_{\text{2}}\text{O}$

Mass of Water = $\text{500}\text{g}$

Moles of Water = $\text{500}\text{g×}\frac{\text{1}\text{mol}{\text{H}}_{\text{2}}\text{O}}{\text{18}\text{.02}\text{g}{\text{H}}_{\text{2}}\text{O}}$

= $\text{27}\text{.75}\text{mol}$

Moles of Water = $\text{27}\text{.75}\text{mol}$

To calculate the heat generated by the reaction

Moles of Water = $\text{27}\text{.75}\text{mol}$

Enthalpy of production of ${\text{Ca(OH)}}_{\text{2}}$= $\text{-65}\text{.2}\text{kJ}$

Heat generated = $\text{27}\text{.75}\text{mol}{\text{Ca(OH)}}_{\text{2}}\text{×}\frac{\text{-65}\text{.2}\text{kJ}}{\text{1}\text{mol}{\text{Ca(OH)}}_{\text{2}}}$

= $\text{-1}{\text{.809×10}}^{\text{3}}\text{kJ}$

Heat generated by the reaction= $\text{-1}{\text{.809×10}}^{\text{3}}\text{kJ}$

To calculate the mass of ${\text{Ca(OH)}}_{\text{2}}$ in $\text{27}\text{.75}\text{mol}{\text{Ca(OH)}}_{\text{2}}$

Mass of ${\text{Ca(OH)}}_{\text{2}}$= $\text{27}\text{.75}{\text{mol Ca(OH)}}_{\text{2}}\text{×}\frac{\text{74}\text{.10}\text{g}{\text{Ca(OH)}}_{\text{2}}}{\text{1}\text{mol}{\text{Ca(OH)}}_{\text{2}}}$

= $\text{2}{\text{.056×10}}^{\text{3}}\text{g}$

Mass of ${\text{Ca(OH)}}_{\text{2}}$= $\text{2}{\text{.056×10}}^{\text{3}}\text{g}$

To calculate the final temperature

$\text{q=smΔT}$

Rearranging we get,

$\begin{array}{l}\text{ΔT=}\frac{\text{q}}{\text{ms}}\\ \\ \text{ΔT=}\frac{\text{1}{\text{.809×10}}^{\text{6}}\text{J}}{\text{(2}{\text{.056×10}}^{\text{3}}\text{g)(1}\text{.20}\text{J}{\text{g}}^{\text{-1}}\text{°C)}}\text{=733°C}\\ \\ {\text{ΔT=T}}_{\text{initial}}{\text{-T}}_{\text{final}}\\ \\ \text{ΔT=}\left(\text{733+25}\right)\text{°C}\\ \\ \text{ΔT=758°C}\end{array}$

Final temperature of the product = $\text{758°C}$

Conclusion

The final temperature was calculated by using the difference in values of initial temperature and change in temperature. The initial temperature was calculated from the equation of specific heat capacity. The final temperature of the product was found to be

$\text{758°C}$.

(b)

Interpretation Introduction

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 10.142QP

The standard enthalpy of formation is $\text{-986}\text{.6}\text{kJ}{\text{mol}}^{\text{-1}}$ .

Explanation of Solution

The reaction can be given as,

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

Enthalpy of reaction = $\text{-65}\text{.2}\text{kJ}{\text{mol}}^{\text{-1}}$

Standard enthalpy of formation of $\text{CaO=-635}\text{.6}\text{kJ}{\text{mol}}^{-1}$

Standard enthalpy of formation of ${\text{H}}_{\text{2}}\text{O=-285}\text{.8}\text{kJ}{\text{mol}}^{\text{-1}}$

Standard enthalpy of formation of ${\text{Ca(OH)}}_{\text{2}}$,

$\begin{array}{l}{\text{ΔH}}_{\text{reaction}}{\text{=ΔH°}}_{\text{f}}\left[{\text{Ca(OH)}}_{\text{2}}\right]\text{-}\left[{\text{ΔH°}}_{\text{f}}{\text{(CaO)+ΔH°}}_{\text{f}}{\text{(H}}_{\text{2}}\text{O)}\right]\\ \\ \text{-65}\text{.2}\text{kJ}{\text{mol}}^{\text{-1}}{\text{=ΔH°}}_{\text{f}}\left[{\text{Ca(OH)}}_{\text{2}}\right]\text{-}\left[\text{(1)(-635}\text{.6}\text{kJ}{\text{mol}}^{\text{-1}}\text{)+(1)(-285}\text{.8}\text{kJ}{\text{mol}}^{\text{-1}}\text{)}\right]\\ \\ {\text{ΔH°}}_{\text{f}}\left[{\text{Ca(OH)}}_{\text{2}}\right]\text{=-986}\text{.6}\text{kJ}{\text{mol}}^{\text{-1}}\end{array}$

Standard enthalpy of formation of ${\text{Ca(OH)}}_{\text{2}}$= $\text{-986}\text{.6}\text{kJ}{\text{mol}}^{\text{-1}}$