Q 91CP - Chemistry: Molecular Nature Of Matter And Change

Question

Question: Limestone ${\mathbf{CaCO}}_{\mathbf{3}}$ is the second most abundant mineral on Earth after SiO2. For many uses, it is first decomposed thermally to quicklime$\mathbf{CaO}$.$\mathbf{MgO}$ is prepared similarly from${\mathbf{MgCO}}_{\mathbf{3}}$.

(a) At what T is each decomposition spontaneous? 

(b) Quicklime reacts with ${\mathbf{SiO}}_{\mathbf{2}}$ to form a slag ${\mathbf{CaSiO}}_{\mathbf{3}}$, a byproduct of steelmaking. In 2003, the total steelmaking capacity of the U.S. steel industry was 2,370,000 tons per week, but only 84% of this capacity was utilized. If $\mathbf{50}\mathbf{.}\mathbf{0}\mathbf{ }\mathbf{kg}$of slag is produced per ton of steel, what mass (in kg) of limestone was used to make slag in 2003?

Step-by-Step Solution

Answer

1. The decomposition spontaneous is $\mathrm{T}=671 \mathrm{K}$ for ${\mathrm{MgCO}}_3$.
2. the mass of limestone is used is $4.2\times {10}^9 \mathrm{kg} {\mathrm{CaCO}}_3$.

Define decomposition spontaneous

The truth is that decomposition reactions rarely happen on their own. They usually require some sort of impetus to get started. The driving force in the transformation of water into hydrogen and oxygen is energy in the form of electricity.

Step 2: For the decomposition spontaneous

The change in enthalpy and entropy of calcium carbonate breakdown are computed as,

$CaC{O}_3\left(s\right) \stackrel{∆}{\to }CaO\left(s\right)+C{O}_2\left(g\right)$

$\begin{array}{rcl}∆{\mathrm{H}}_{\mathrm{rxn}}^{\circ }& =& \left[\left(1 \mathrm{mol}\right)\times ∆{\mathrm{H}}_{\mathrm{i}}^{\circ }\left(\mathrm{CaO}\right)+\left(1 \mathrm{mol}\right)\times ∆{\mathrm{H}}_{\mathrm{i}}^{\circ }\left({\mathrm{CO}}_2\right)\right]-\left[\left(1 \mathrm{mol}\right)\times ∆{\mathrm{H}}_{\mathrm{i}}^{\circ }\left({\mathrm{CaCO}}_3\right)\right]\\ & =& \left[\left(1 \mathrm{mol}\right)\times \left(-631.5 \mathrm{kJ}/\mathrm{mol}\right)+\left(1 \mathrm{mol}\right)\times \left(-393.5 \mathrm{kJ}/\mathrm{mol}\right)\right]-\left[\left(1 \mathrm{mo}\right)\mathrm{l}\times \left(-1206.9 \mathrm{kJ}/\mathrm{mol}\right)\right]\\ & =& 178.3 \mathrm{kJ}\end{array}$

$\begin{array}{rcl}∆{\mathrm{S}}_{\mathrm{rxn}}^{\circ }& =& \underset{}{\overset{}{\sum \mathrm{m}∆{\mathrm{S}}_{\mathrm{products}}^{\circ }-}}\sum \mathrm{m}∆{\mathrm{S}}_{\mathrm{reactants}}^{\circ }\\ & =& \left[\left(1 \mathrm{mol}\right)\times \left(38.2 \mathrm{J}/\mathrm{Kmol}\right)+\left(1 \mathrm{mol}\right)\times \left(213.7 \mathrm{J}/\mathrm{Kmol}\right)\right]-\left[\left(1 \mathrm{mo}\right)\mathrm{l}\times \left(92.9 \mathrm{J}/\mathrm{Kmol}\right)\right]\\ & =& 150.9\mathrm{J}/\mathrm{K}\end{array}$

The rate of breakdown is computed as,

$$\begin{gathered} ∆{\mathrm{G}}_{\mathrm{rxn}}^{\circ }=0=∆{\mathrm{H}}_{\mathrm{rxn}}^{\circ }-\mathrm{T}∆{\mathrm{S}}_{\mathrm{rxn}}^{\circ } \\ \mathrm{T}=\frac{∆{\mathrm{H}}_{\mathrm{ma}}^{\circ }}{∆{\mathrm{S}}_{\mathrm{ma}}^{\circ }}=\frac{178.3 \mathrm{kJ}}{159.0 \mathrm{J}/\mathrm{Kmol}}\times \frac{1000 \mathrm{J}}{1 \mathrm{kJ}}=1121 \mathrm{K} \mathrm{for} {\mathrm{CaCO}}_3 \end{gathered}$$

The change in enthalpy and entropy of magnesium carbonate breakdown are estimated as,

$MgC{O}_3\left(s\right) \stackrel{∆}{\to }MgO\left(s\right)+C{O}_2\left(g\right)$

$\begin{array}{rcl}∆{\mathrm{H}}_{\mathrm{rxn}}^{\circ }& =& \left[\left(1 \mathrm{mol}\right)\times ∆{\mathrm{H}}_{\mathrm{i}}^{\circ }\left(MgO\right)+\left(1 \mathrm{mol}\right)\times ∆{\mathrm{H}}_{\mathrm{i}}^{\circ }\left({\mathrm{CO}}_2\right)\right]-\left[\left(1 \mathrm{mol}\right)\times ∆{\mathrm{H}}_{\mathrm{i}}^{\circ }\left(Mg{\mathrm{CO}}_3\right)\right]\\ & =& \left[\left(1 \mathrm{mol}\right)\times \left(-601.2 \mathrm{kJ}/\mathrm{mol}\right)+\left(1 \mathrm{mol}\right)\times \left(-393.5 \mathrm{kJ}/\mathrm{mol}\right)\right]-\left[\left(1 \mathrm{mo}\right)\mathrm{l}\times \left(-1112.9 \mathrm{kJ}/\mathrm{mol}\right)\right]\\ & =& 117.3 \mathrm{kJ}\end{array}$

$\begin{array}{rcl}∆{\mathrm{S}}_{\mathrm{rxn}}^{\circ }& =& \underset{}{\overset{}{\sum \mathrm{m}∆{\mathrm{S}}_{\mathrm{products}}^{\circ }-}}\sum \mathrm{m}∆{\mathrm{S}}_{\mathrm{reactants}}^{\circ }\\ & =& \left[\left(1 \mathrm{mol}\right)\times \left(26.9 \mathrm{J}/\mathrm{Kmol}\right)+\left(1 \mathrm{mol}\right)\times \left(213.7 \mathrm{J}/\mathrm{Kmol}\right)\right]-\left[\left(1 \mathrm{mo}\right)\mathrm{l}\times \left(65.86 \mathrm{J}/\mathrm{Kmol}\right)\right]\\ & =& 174.74 \mathrm{J}/\mathrm{K}\end{array}$

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

$$\begin{gathered} ∆{\mathrm{G}}_{\mathrm{rxn}}^{\circ }=0=∆{\mathrm{H}}_{\mathrm{rxn}}^{\circ }-\mathrm{T}∆{\mathrm{S}}_{\mathrm{rxn}}^{\circ } \\ \mathrm{T}=\frac{∆{\mathrm{H}}_{\mathrm{ma}}^{\circ }}{∆{\mathrm{S}}_{\mathrm{ma}}^{\circ }}=\frac{117.3 \mathrm{kJ}}{174.4.0 \mathrm{J}/\mathrm{Kmol}}\times \frac{1000 \mathrm{J}}{1 \mathrm{kJ}}=671 \mathrm{K} \mathrm{for} {\mathrm{MgCO}}_3 \end{gathered}$$

Therefore, the decomposition spontaneous is $671 \mathrm{K}$ for ${\mathrm{MgCO}}_3$

Step 3: For the mass of limestone

Calculate the mass of calcium carbonate as follows:

$CaO\left(s\right)+Si{O}_2\left(s\right)\to CaSi{O}_3\left(s\right)$

$\begin{array}{rcl}\mathrm{Mass} \mathrm{of} {\mathrm{CaCO}}_3& =& \left(\frac{2.236\times {10}^6 \mathrm{ton}}{\mathrm{weeks}}\right)\left(\frac{84\%}{100\%}\right)\left(\frac{52 \mathrm{weeks}}{1 \mathrm{year}}\right)\left(\frac{50 \mathrm{kg} \mathrm{slug}}{1 \mathrm{ton}}\right)\left(\frac{{10}^3 \mathrm{g}}{1 \mathrm{kg}}\right)\\ & =& \left(\frac{1 \mathrm{mol} \mathrm{slag}}{116.1 \mathrm{g} \mathrm{slag}}\right)\left(\frac{1 \mathrm{mol} {\mathrm{CaCo}}_3}{1 \mathrm{mol} \mathrm{slag}}\right)\left(\frac{100.09 \mathrm{g} {\mathrm{CaCO}}_3}{1 \mathrm{mol} {\mathrm{CaCO}}_3}\right)\left(\frac{1 \mathrm{kg}}{{10}^3 \mathrm{g}}\right)\\ & =& 4.2\times {10}^9 \mathrm{kg} {\mathrm{CaCO}}_3\end{array}$

Therefore, the mass of limestone is used is $4.2 \mathrm{kg}$