Chapter 22, Problem 22.86P

(a)

Interpretation Introduction

Interpretation:

The temperature at which each decomposition is spontaneous has to be calculated.

Concept Introduction: 

Free energy, ${\text{ΔG}}^{\text{ο}}$, is the measure of the ability of a reaction to proceed spontaneously. ${\text{ΔG}}^{\text{ο}}$ can be calculated with the relationship.

    ${\text{ΔG}}^{\text{ο}}{}_{\text{rxn}}\text{=}{\displaystyle \sum _{}^{}{\text{mΔG}}_{{}^{\text{f}}\text{(products)}}^{\text{o}}\text{-}}{\displaystyle \sum _{}^{}{\text{mΔG}}_{{}^{\text{f}}\text{(reactants)}}^{\text{o}}}$

    ${\text{ΔG}}^{\text{ο}}{}_{\text{rxn}}\text{=}{\text{ΔH}}_{\text{rxn}}^{\text{o}}\text{-}{\text{TΔS}}_{\text{rxn}}^{\text{o}}$

Explanation of Solution

The change in enthalpy and entropy of decomposition of calcium carbonate are calculated as,

$\begin{array}{l}{\text{CaCO}}_{\text{3(s)}}\underset{}{\overset{\text{0}}{\to }}{\text{CaO}}_{\text{(s)}}{\text{ + CO}}_{\text{2(g)}}\\ \\ {\text{ΔH}}^{\text{ο}}{}_{\text{rxn}}\text{=}{\displaystyle \sum _{}^{}{\text{mΔH}}_{{}^{\text{f}}\text{(products)}}^{\text{o}}\text{-}}{\displaystyle \sum _{}^{}{\text{nΔH}}_{{}^{\text{f}}\text{(reactants)}}^{\text{o}}}\\ \\ {\text{ΔH}}^{\text{ο}}{}_{\text{rxn}}{\text{=[(1mol)(ΔH}}_{{}^{\text{f}}}^{\text{o}}\text{CaO)}\text{+}\text{(1mol}{\text{)(ΔH}}_{{}^{\text{f}}}^{\text{o}}{\text{CO}}_{\text{2(g)}}\text{)]}\text{-}\text{[(1mol}{\text{)(ΔH}}_{{}^{\text{f}}}^{\text{o}}{\text{CaCO}}_{\text{3(s)}}\text{)]}\\ \\ {\text{ΔH}}^{\text{ο}}{}_{\text{rxn}}\text{=}\text{[(1 mol)(-635}\text{.1kJ/mol)}\text{+}\text{(1 mol)(-393}\text{.5kJ/mol)]}\text{-}\text{[(1 mol)(-1206}\text{.9kJ/mol)]}\\ \\ {\text{ΔH}}^{\text{ο}}{}_{\text{rxn}}\text{=}\text{178}\text{.3 kJ}\\ \\ {\text{ΔS}}^{\text{ο}}{}_{\text{rxn}}\text{=}{\displaystyle \sum _{}^{}{\text{mΔS}}_{{}^{\text{f}}\text{(products)}}^{\text{o}}\text{-}}{\displaystyle \sum _{}^{}{\text{nΔS}}_{{}^{\text{f}}\text{(reactants)}}^{\text{o}}}\\ \\ {\text{ΔS}}^{\text{ο}}{}_{\text{rxn}}\text{=}\text{[(1 mol)(38}\text{.2J/mol}\text{.}\text{K)}\text{+}\text{(1 mol)(213}\text{.7 J/mol}\text{.K)]}\text{-}\text{[(1 mol)(92}\text{.9 J/mol)]}\\ \\ {\text{ΔS}}^{\text{ο}}{}_{\text{rxn}}\text{=}\text{159}\text{.0 J/K}\text{.}\end{array}$

When the reaction is spontaneous, the standard free energy is zero.  The temperature at which decomposition occurs is calculated as,

    $\begin{array}{l}{\text{ΔG}}^{\text{ο}}{}_{\text{rxn}}\text{=}\text{0}\text{=}{\text{ΔH}}_{\text{rxn}}^{\text{o}}\text{-}{\text{TΔS}}_{\text{rxn}}^{\text{o}}\\ \\ \text{T}\text{=}\frac{{\text{ΔH}}_{\text{rxn}}^{\text{o}}}{{\text{ΔS}}_{\text{rxn}}^{\text{o}}}\text{=}\left(\frac{\text{178}\text{.3 kJ}}{\text{159}\text{.0 J/K}}\right)\left(\frac{{\text{10}}^{\text{3}}\text{J}}{\text{1kJ}}\right)\text{=}{\text{1121 K for CaCO}}_{\text{3}}\end{array}$

The change in enthalpy and entropy of decomposition of magnesium carbonate are calculated as,

$\begin{array}{l}{\text{MgCO}}_{\text{3(s)}}\underset{}{\overset{\text{0}}{\to }}{\text{MgO}}_{\text{(s)}}{\text{ + CO}}_{\text{2(g)}}\\ \\ {\text{ΔH}}^{\text{ο}}{}_{\text{rxn}}\text{=}{\displaystyle \sum _{}^{}{\text{mΔH}}_{{}^{\text{f}}\text{(products)}}^{\text{o}}\text{-}}{\displaystyle \sum _{}^{}{\text{nΔH}}_{{}^{\text{f}}\text{(reactants)}}^{\text{o}}}\\ \\ {\text{ΔH}}^{\text{ο}}{}_{\text{rxn}}{\text{=[(1mol)(ΔH}}_{{}^{\text{f}}}^{\text{o}}\text{MgO)}\text{+}\text{(1mol}{\text{)(ΔH}}_{{}^{\text{f}}}^{\text{o}}{\text{CO}}_{\text{2(g)}}\text{)]}\text{-}\text{[(1mol}{\text{)(ΔH}}_{{}^{\text{f}}}^{\text{o}}{\text{MgCO}}_{\text{3(s)}}\text{)]}\\ \\ {\text{ΔH}}^{\text{ο}}{}_{\text{rxn}}\text{=}\text{[(1 mol)(-601}\text{.2kJ/mol)}\text{+}\text{(1 mol)(-393}\text{.5kJ/mol)]}\text{-}\text{[(1 mol)(-1112kJ/mol)]}\\ \\ {\text{ΔH}}^{\text{ο}}{}_{\text{rxn}}\text{=}\text{117}\text{.3 kJ}\\ \\ {\text{ΔS}}^{\text{ο}}{}_{\text{rxn}}\text{=}{\displaystyle \sum _{}^{}{\text{mΔS}}_{{}^{\text{f}}\text{(products)}}^{\text{o}}\text{-}}{\displaystyle \sum _{}^{}{\text{nΔS}}_{{}^{\text{f}}\text{(reactants)}}^{\text{o}}}\\ \\ {\text{ΔS}}^{\text{ο}}{}_{\text{rxn}}\text{=}\text{[(1 mol)(26}\text{.9J/mol}\text{.}\text{K)}\text{+}\text{(1 mol)(213}\text{.7 J/mol}\text{.K)]}\text{-}\text{[(1 mol)(65}\text{.86 J/mol)]}\\ \\ {\text{ΔS}}^{\text{ο}}{}_{\text{rxn}}\text{=}\text{174}\text{.74 J/K}\text{.}\end{array}$

When the reaction is spontaneous, the standard free energy is zero.  The temperature at which decomposition occurs is calculated as,

    $\begin{array}{l}{\text{ΔG}}^{\text{ο}}{}_{\text{rxn}}\text{=}\text{0}\text{=}{\text{ΔH}}_{\text{rxn}}^{\text{o}}\text{-}{\text{TΔS}}_{\text{rxn}}^{\text{o}}\\ \\ \text{T}\text{=}\frac{{\text{ΔH}}_{\text{rxn}}^{\text{o}}}{{\text{ΔS}}_{\text{rxn}}^{\text{o}}}\text{=}\left(\frac{\text{117}\text{.3 kJ}}{\text{174}\text{.74 J/K}}\right)\left(\frac{{\text{10}}^{\text{3}}\text{J}}{\text{1kJ}}\right)\text{=}{\text{671 K for MgCO}}_{\text{3}}.\end{array}$

(b)

Interpretation Introduction

Interpretation:

The mass of limestone used to make slag has to be calculated.

Explanation of Solution

The mass of calcium carbonate is calculated as,

$\begin{array}{l}{\text{CaO}}_{\text{(s)}}{\text{ + SiO}}_{\text{2(s)}}\underset{}{\overset{}{\to }}{\text{CaSiO}}_{\text{3(s)}}\\ \\ {\text{Mass (kg) of CaCO}}_{\text{3}}\text{ =}\left(\frac{\text{2}{\text{.236x10}}^{\text{6}}\text{ tons}}{\text{week}}\right)\left(\frac{\text{84\%}}{\text{100\%}}\right)\left(\frac{\text{52 weeks}}{\text{yr}}\right)\left(\frac{\text{50 kg slag}}{\text{ton}}\right)\left(\frac{{\text{10}}^{\text{3}}\text{g}}{\text{1kg}}\right)\\ \\ =\left(\frac{\text{1 mol slag}}{\text{116}\text{.17 g slag}}\right)\left(\frac{\text{1mol}{\text{CaCO}}_{\text{3}}}{\text{1}\text{mol slag}}\right)\left(\frac{\text{100}{\text{.09 g CaCO}}_{\text{3}}}{\text{1mol}{\text{CaCO}}_{\text{3}}}\right)\left(\frac{\text{1kg}}{{\text{10}}^{\text{-3}}\text{g}}\right)\\ \\ \text{= 4}{\text{.2x10}}^{\text{9}}{\text{ kg CaCO}}_{\text{3}}.\end{array}$