Chapter 6, Problem 6.137QP

Lime is a term that includes calcium oxide (CaO, also called quicklime) and calcium hydroxide [Ca(OH)₂, also called slaked lime]. It is used in the steel industry to remove acidic impurities, in air-pollution control to remove acidic oxides such as SO₂, and in water treatment. Quicklime is made industrially by heating limestone (CaCO₃) above 2000°C:

$\begin{array}{l}{\text{CaCo}}_3(s)\stackrel{}{\to }\text{CaO}(s)+{\text{CO}}_2(g)\hfill \\ \Delta H^{\text{o}}=177.8\text{kJ/mol}\hfill \end{array}$

Slaked lime is produced by treating quicklime with water:

$\begin{array}{l}\text{CaO}(s)+{\text{H}}_2\text{O}(l)\stackrel{}{\to }\text{Ca}{(}_{\text{OH}}(s)\\ \Delta H^{\text{o}}=-65.2\text{kJ/mol}\end{array}$

The exothermic reaction of quicklime with water and the rather small specific heats of both quicklime (0.946 J/g · °C) and slaked lime (1.20 J/g · °C) make it hazardous to store and transport lime in vessels made of wood. Wooden sailing ships carrying lime would occasionally catch fire when water leaked into the hold. (a) If a 500-g sample of water reacts with an equimolar amount of CaO (both at an initial temperature of 25°C), what is the final temperature of the product, Ca(OH)₂? Assume that the product absorbs all of the heat released in the reaction. (b) Given that the standard enthalpies of formation of CaO and H₂O are −635.6 kJ/mol and −285.8 kJ/mol, respectively, calculate the standard enthalpy of formation of Ca(OH)₂.

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 6.137QP

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 = $\text{758°C}$

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 6.137QP

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