Question: In 1790, Nicolas Leblanc found a way to form from . His process, now obsolete, consisted of three steps obsolete, consisted of three steps:

$$\begin{gathered} \mathbf{2}\mathbf{NaCl}\mathbf{\left(}\mathbf{s}\mathbf{\right)}\mathbf{+}{\mathbf{H}}_{\mathbf{2}}{\mathbf{SO}}_{\mathbf{4}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{\to }{\mathbf{Na}}_{\mathbf{2}}{\mathbf{SO}}_{\mathbf{4}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{+}\mathbf{2}\mathbf{HCl}\mathbf{\left(}\mathbf{g}\mathbf{\right)} \\ {\mathbf{Na}}_{\mathbf{2}}{\mathbf{SO}}_{\mathbf{4}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{+}\mathbf{2}\mathbf{C}\mathbf{\left(}\mathbf{s}\mathbf{\right)}\mathbf{\to }{\mathbf{Na}}_{\mathbf{2}}\mathbf{S}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{+}\mathbf{2}{\mathbf{CO}}_{\mathbf{2}}\mathbf{\left(}\mathbf{g}\mathbf{\right)} \\ {\mathbf{Na}}_{\mathbf{2}}\mathbf{S}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{+}{\mathbf{CaCO}}_{\mathbf{3}}\mathbf{\left(}\mathbf{s}\mathbf{\right)}\mathbf{\to }{\mathbf{Na}}_{\mathbf{2}}{\mathbf{CO}}_{\mathbf{3}}\mathbf{\left(}\mathbf{s}\mathbf{\right)}\mathbf{+}\mathbf{CaS}\mathbf{\left(}\mathbf{s}\mathbf{\right)} \end{gathered}$$

1. Write a balanced overall equation for the process.
2. Calculate the ${\mathbf{H}}^{\mathbf{0}}$of $\mathbf{CaS}$if ${\mathbf{H}}_{\mathbf{rxn}}^{\mathbf{\circ }}$is$\mathbf{351}\mathbf{.}\mathbf{8}\mathbf{}\mathbf{kJ}\mathbf{/}\mathbf{mol}$
3. Is the overall process spontaneous at standard-state conditions and$\mathbf{298}\mathbf{}\mathbf{K}$?
4. How many grams of${\mathbf{Na}}_{\mathbf{2}}{\mathbf{CO}}_{\mathbf{3}}$form $\mathbf{250}\mathbf{}\mathbf{g}$of$\mathbf{NaCl}$if the process is 73% efficient?

Nicolas Leblanc, is a French surgeon and chemist who invented the process for producing soda ash (sodium carbonate) from common salt in 1790. (Sodium chloride). This process, which bears his name, went on to become one of the most important industrial-chemical processes of the nineteenth century.

Given the balanced three-step reactions, the following process reaction can be obtained:

$$\begin{gathered} 2\mathrm{NaCl}\left(\mathrm{s}\right)+{\mathrm{H}}_2{\mathrm{SO}}_4\left(\mathrm{aq}\right)\to {\mathrm{Na}}_2{\mathrm{SO}}_4\left(\mathrm{aq}\right)+2\mathrm{HCl}\left(\mathrm{g}\right) \\ {\mathrm{Na}}_2{\mathrm{SO}}_4\left(\mathrm{aq}\right)+2\mathrm{C}\left(\mathrm{s}\right)\to {\mathrm{Na}}_2\mathrm{S}\left(\mathrm{aq}\right)+2{\mathrm{CO}}_2\left(\mathrm{g}\right) \\ {\mathrm{Na}}_2\mathrm{S}\left(\mathrm{aq}\right)+{\mathrm{CaCO}}_3\left(\mathrm{s}\right)\to {\mathrm{Na}}_2{\mathrm{CO}}_3\left(\mathrm{s}\right)+\mathrm{CaS}\left(\mathrm{s}\right) \end{gathered}$$

As a result, the balanced equation would be:

$2\mathrm{NaCl}\left(\mathrm{s}\right)+{\mathrm{H}}_2{\mathrm{SO}}_4\left(\mathrm{aq}\right)+2\mathrm{C}\left(\mathrm{s}\right)+{\mathrm{CaCO}}_3\left(\mathrm{s}\right)\to 2\mathrm{HCl}\left(\mathrm{g}\right)+2{\mathrm{CO}}_2\left(\mathrm{g}\right)+{\mathrm{Na}}_2{\mathrm{CO}}_3\left(\mathrm{s}\right)+\mathrm{CaS}\left(\mathrm{s}\right)$

Because enthalpy is a state function,

$∆{\mathrm{H}}_{\mathrm{rxn}}=\underset{}{\overset{}{\sum {\mathrm{H}}_{\mathrm{products}}-\sum {\mathrm{H}}_{\mathrm{reactants}}}}$

The overall process equation is then,

$$\begin{gathered} ∆H_{rxn}=\left[2mol\times ∆H\left(HCl\right)+2mol\times ∆H\left(C{O}_2\right)+1mol\times ∆H\left(N{a}_{2}C{O}_3\right)+1mol\times ∆H\left(CaS\right)\right] \\ -\left[2mol\times ∆H\left(NaCl\right)+2mol\times ∆H\left(H_{2}S{O}_4\right)+1mol\times ∆H\left(C\right)+1mol\times ∆H\left(CaC{O}_3\right)\right] \end{gathered}$$Assign $∆{\mathrm{H}}_{\mathrm{f}}^{\circ }\left(\mathrm{CaS}\right)=\mathrm{X}$

role="math" localid="1663327971935" $$\begin{gathered} ∆{\mathrm{H}}_{\mathrm{rxn}}=\left[2\mathrm{mol}\times \left(-92.31\mathrm{kJ}/\mathrm{mol}\right)+2\mathrm{mol}\times \left(-393.50\mathrm{kJ}/\mathrm{mol}\right)+1\mathrm{mol}\times \left(-1130.80\mathrm{kJ}/\mathrm{mol}\right)+1\mathrm{mol}\times \mathrm{XkJ}/\mathrm{mol}\right] \\ -\left[2\mathrm{mol}\times \left(-411.10\mathrm{kJ}/\mathrm{mol}\right)+2\mathrm{mol}\times \left(-907.51\mathrm{kJ}/\mathrm{mol}\right)+1\mathrm{mol}\times \left(0\right)+1\mathrm{mol}\times \left(-1206.90\mathrm{kJ}/\mathrm{mol}\right)\right] \\ ∆{\mathrm{H}}_{\mathrm{rxn}}=\left[\left(-2102.42+X\right)kJ\right]-\left[2936.61kJ\right]=351.8\mathrm{kJ} \end{gathered}$$Then X can be easily calculated:

$$\begin{gathered} -2102.42+\mathrm{X}=351.8+2936.61 \\ \mathrm{X}=2584.81-\left(-2102.42\right) \\ \mathrm{X}=-482.39\mathrm{kJ} \end{gathered}$$

Thus,

$∆{\mathrm{H}}_{\mathrm{f}}^{\circ }\left(\mathrm{CaS}\right)=-482.39\mathrm{kJ}/\mathrm{mol}$

To estimate the process's spontaneity at $298\mathrm{K}$, entropy should be calculated in the same way that enthalpy is:$$\begin{gathered} ∆S_{rxn}=\left[2mol\times ∆S\left(HCl\right)+2mol\times ∆S\left(C{O}_2\right)+1mol\times ∆S\left(N{a}_{2}C{O}_3\right)+1mol\times ∆S\left(CaS\right)\right] \\ -\left[2mol\times ∆S\left(NaCl\right)+2mol\times ∆S\left(H_{2}S{O}_4\right)+1mol\times ∆S\left(C\right)+1mol\times ∆S\left(CaC{O}_3\right)\right] \end{gathered}$$

Using Appendix B and NIST database

$$\begin{gathered} ∆S_{\mathrm{rxn}}=\left[2\mathrm{mol}\times \left(186.79J/K\mathrm{mol}\right)+2\mathrm{mol}\times \left(213.70J/K\mathrm{mol}\right)+1\mathrm{mol}\times \left(139J/K\mathrm{mol}\right)+1\mathrm{mol}\times \left(56.4J/K\mathrm{mol}\right)\right] \\ -\left[2\mathrm{mol}\times \left(72.12J/K\mathrm{mol}\right)+2\mathrm{mol}\times \left(17.0J/K\mathrm{mol}\right)+1\mathrm{mol}\times \left(\left(5.686J/K\mathrm{mol}\right)\right)+1\mathrm{mol}\times \left(92.90J/K\mathrm{mol}\right)\right] \\ ∆S_{\mathrm{rxn}}=731.008\mathrm{J}/\mathrm{K} \end{gathered}$$Under normal circumstances,

$$\begin{gathered} ∆{\mathrm{G}}_{\mathrm{rxn}}=∆{\mathrm{H}}_{\mathrm{rxn}}-\mathrm{T}∆\mathrm{S} \\ ∆{\mathrm{G}}_{\mathrm{rxn}}=-482.39\mathrm{kJ}/\mathrm{mol}-298\mathrm{K}\times 731.008\times {10}^{-3}\mathrm{kJ}/\mathrm{Kmol} \\ ∆{\mathrm{G}}_{\mathrm{rxn}}=-700.23\mathrm{kJ} \end{gathered}$$

At standard-state conditions, the overall process is spontaneous, because the change in Gibb's free energy is negative.

Given the overall balanced reaction in a), calculate the mass of $\mathrm{NaCl}$is at first is converted into amount of substance:

$\left(\mathrm{NaCl}\right)=\frac{\mathrm{m}\left(\mathrm{NaCl}\right)}{{\mathrm{M}}_{\mathrm{R}}\left(\mathrm{NaCl}\right)}\mathrm{n}\left(\mathrm{NaCl}\right)=\frac{250\mathrm{g}}{58.44\mathrm{g}/\mathrm{mol}}\mathrm{n}\left(\mathrm{NaCl}\right)=4.2779\mathrm{mol}$

According to the equation,$\mathrm{n}\left(\mathrm{NaCl}\right)=2\times \mathrm{n}\left({\mathrm{Na}}_2{\mathrm{CO}}_3\right)$

Thus,

$$\begin{gathered} \mathrm{n}\left({\mathrm{Na}}_2{\mathrm{CO}}_3\right)=\frac{1}{2}\times 4.2779 \\ \mathrm{n}\left({\mathrm{Na}}_2{\mathrm{CO}}_3\right)=2.13895 \end{gathered}$$

Because the process's efficiency is only 73% the real amount of produced ${\mathrm{Na}}_2{\mathrm{CO}}_3$

$$\begin{gathered} \mathrm{n}{\left({\mathrm{Na}}_2{\mathrm{CO}}_3\right)}_{\mathrm{actual}}=\mathrm{n}\left({\mathrm{Na}}_2{\mathrm{CO}}_3\right)\times \frac{73}{100} \\ \mathrm{n}{\left({\mathrm{Na}}_2{\mathrm{CO}}_3\right)}_{\mathrm{actual}}=1.5614 \end{gathered}$$

Then there's the actual mass of ${\mathrm{Na}}_2{\mathrm{CO}}_3$created is

$$\begin{gathered} \mathrm{m}{\left({\mathrm{Na}}_2{\mathrm{CO}}_3\right)}_{\mathrm{actual}}=1.5614\mathrm{mol}\times 105.998\mathrm{g}/\mathrm{mol} \\ \mathrm{m}{\left({\mathrm{Na}}_2{\mathrm{CO}}_3\right)}_{\mathrm{actual}}=165.4909\mathrm{g} \end{gathered}$$

The actual weight of ${\mathrm{Na}}_2{\mathrm{CO}}_3$created is$165.4909\mathrm{g}$