Question

CaO can be used as a drying agent. One such application occurs when water is added to dry concrete or cement. The reaction that occurs is

CaO(s)+H2O(l)⇌Ca(OH)2(s)

The product is commonly called slaked lime.

Assuming the commonly used standard-state temperature of 25∘C, calculate ΔSuniv for this reaction using table from the table below.

Substance	S∘ [J/(K⋅mol)]	ΔH∘f (kJ/mol)
CaO(s)	39.9	−635.1
H2O(l)	69.9	−285.8
Ca(OH)2(s)	83.4	−986.1

Answer 1

Concepts and reason

The reaction is as follows:

Calculate the entropy change of the universe by using the following formula.

$\Delta {S_{{\rm{Universe}}}} = {\rm{ }}\Delta {S_{{\rm{System}}}} + \Delta {S_{{\rm{Surroundings}}}}$

Calculate the $\Delta {S_{{\rm{Surroundings}}}}$ as shown below.

$\Delta {S_{{\rm{Surroundings}}}} = - \frac{{\Delta {H^{\rm{o}}}_{{\rm{rxn}}}}}{T}$

Here, $\Delta {H^{\rm{o}}}_{{\rm{rxn}}}$ is enthalpy change of the reaction, T is temperature.

Calculate the entropy change for the reaction (system) by using the following formula.

$\Delta {S^{\rm{o}}}_{{\rm{reaction}}} = {\rm{ }}\sum {n_p}{S^{\rm{o}}}_{\rm{f}}{\rm{(products)}} - \sum {n_r}{S^{\rm{o}}}_{\rm{f}}{\rm{(reactants)}}$

Change of enthalpy during the formation of substance from its constituent elements is called as enthalpy of formation.

Calculate the enthalpy change for the reaction by using the following formula.

$\Delta {H^{\rm{o}}}_{{\rm{reaction}}} = {\rm{ }}\sum {n_p}\Delta {H^{\rm{o}}}_{\rm{f}}{\rm{(products)}} - \sum {n_r}\Delta {H^{\rm{o}}}_{\rm{f}}{\rm{(reactants)}}$

Fundamentals

Entropy is a thermodynamic quantity, it is a measure of randomness of a system.

Change in the enthalpy when one mole of a substance is formed from its pure elements under standard conditions is called as standard enthalpy of formation. The standard conditions are 1atm pressure and 298 K temperature.

Write the expansion of enthalpy formation for the chemical equation as shown below.

$\begin{array}{c}\\\Delta {S^{\rm{o}}}_{{\rm{rxn}}} = \left[ {{n_{{\rm{Ca}}{{\left( {{\rm{OH}}} \right)}_{\rm{2}}}\left( {\rm{s}} \right)}} \times {S^{\rm{o}}}_{\rm{f}}{\rm{Ca}}{{\left( {{\rm{OH}}} \right)}_{\rm{2}}}\left( {\rm{s}} \right)} \right] - \left[ {{n_{{\rm{CaO}}\left( {\rm{s}} \right)}} \times {S^{\rm{o}}}_{\rm{f}}{\rm{CaO}}\left( {\rm{s}} \right) + {n_{{{\rm{H}}_{\rm{2}}}{\rm{O}}\left( {\rm{l}} \right)}} \times {S^{\rm{o}}}_{\rm{f}}{{\rm{H}}_{\rm{2}}}{\rm{O}}\left( {\rm{l}} \right)} \right]\\\\\Delta {S^{\rm{o}}}_{{\rm{rxn}}} = \left[ {1{\rm{ mol}} \times 83.4{\rm{ }}\frac{{\rm{J}}}{{{\rm{K}} \cdot {\rm{mol}}}}} \right] - \left[ {\left( {1{\rm{ mol}} \times 39.9{\rm{ }}\frac{{\rm{J}}}{{{\rm{K}} \cdot {\rm{mol}}}}} \right) + \left( {{\rm{1 mol}} \times {\rm{69}}{\rm{.9 }}\frac{{\rm{J}}}{{{\rm{K}} \cdot {\rm{mol}}}}} \right)} \right]\\\end{array}$

$\begin{array}{c}\\\Delta {S^{\rm{o}}}_{{\rm{rxn}}} = \left[ {1{\rm{ }}\cancel{{{\rm{mol}}}} \times 83.4{\rm{ }}\frac{{\rm{J}}}{{{\rm{K}} \cdot \cancel{{{\rm{mol}}}}}}} \right] - \left[ {\left( {1{\rm{ }}\cancel{{{\rm{mol}}}} \times 39.9{\rm{ }}\frac{{\rm{J}}}{{{\rm{K}} \cdot \cancel{{{\rm{mol}}}}}}} \right) + \left( {{\rm{1 }}\cancel{{{\rm{mol}}}} \times {\rm{69}}{\rm{.9 }}\frac{{\rm{J}}}{{{\rm{K}} \cdot \cancel{{{\rm{mol}}}}}}} \right)} \right]\\\\\Delta {S^{\rm{o}}}_{{\rm{rxn}}} = 83.4{\rm{ }}\frac{{\rm{J}}}{{\rm{K}}} - \left[ {39.9{\rm{ }}\frac{{\rm{J}}}{{\rm{K}}} + 69.9{\rm{ }}\frac{{\rm{J}}}{{\rm{K}}}} \right]\\\\\Delta {S^{\rm{o}}}_{{\rm{rxn}}} = - {\rm{26}}{\rm{.4 }}\frac{{\rm{J}}}{{\rm{K}}}\\\end{array}$

Thus, $\Delta {S_{{\rm{System}}}}$ of the reaction is $- {\rm{26}}{\rm{.4 }}\frac{{\rm{J}}}{{\rm{K}}}$ .

Calculate the enthalpy change for reaction as shown below.

$\begin{array}{c}\\\Delta {H^{\rm{o}}}_{{\rm{rxn}}} = {\rm{ }}\left[ {{n_{{\rm{Ca}}{{\left( {{\rm{OH}}} \right)}_{\rm{2}}}\left( {\rm{s}} \right)}} \times \Delta {H^{\rm{o}}}_{\rm{f}}{\rm{Ca}}{{\left( {{\rm{OH}}} \right)}_{\rm{2}}}\left( {\rm{s}} \right)} \right] - \left[ {{n_{{\rm{CaO}}\left( {\rm{s}} \right)}} \times \Delta {H^{\rm{o}}}_{\rm{f}}{\rm{CaO}}\left( {\rm{s}} \right) + {n_{{{\rm{H}}_{\rm{2}}}{\rm{O}}\left( {\rm{l}} \right)}} \times \Delta {H^{\rm{o}}}_{\rm{f}}{{\rm{H}}_{\rm{2}}}{\rm{O}}\left( {\rm{l}} \right)} \right]\\\\\Delta {H^{\rm{o}}}_{{\rm{rxn}}} = 1{\rm{ mol}} \times \left( { - 986.1{\rm{ }}\frac{{{\rm{kJ}}}}{{{\rm{mol}}}}} \right) - \left[ {1{\rm{ mol}} \times \left( { - 635.1{\rm{ }}\frac{{{\rm{kJ}}}}{{{\rm{mol}}}}} \right) + 1{\rm{ mol}} \times \left( { - {\rm{285}}{\rm{.8 }}\frac{{{\rm{kJ}}}}{{{\rm{mol}}}}} \right)} \right]\\\end{array}$ $\begin{array}{c}\\\Delta {H^{\rm{o}}}_{{\rm{rxn}}} = 1{\rm{ }}\cancel{{{\rm{mol}}}} \times \left( { - 986.1{\rm{ }}\frac{{{\rm{kJ}}}}{{\cancel{{{\rm{mol}}}}}}} \right) - \left[ {1{\rm{ }}\cancel{{{\rm{mol}}}} \times \left( { - 635.1{\rm{ }}\frac{{{\rm{kJ}}}}{{\cancel{{{\rm{mol}}}}}}} \right) + 1{\rm{ }}\cancel{{{\rm{mol}}}} \times \left( { - {\rm{285}}{\rm{.8 }}\frac{{{\rm{kJ}}}}{{\cancel{{{\rm{mol}}}}}}} \right)} \right]\\\\\Delta {H^{\rm{o}}}_{{\rm{rxn}}} = - 986.1{\rm{ kJ}} - \left[ { - 635.1{\rm{ kJ}} - {\rm{285}}{\rm{.8 kJ}}} \right]\\\\\Delta {H^{\rm{o}}}_{{\rm{rxn}}} = - {\rm{65}}{\rm{.2 kJ}}\\\end{array}$

And

$\begin{array}{l}\\\Delta {S_{{\rm{Surroundings}}}} = - \frac{{\Delta {H^{\rm{o}}}_{{\rm{rxn}}}}}{T}\\\\\Delta {S_{{\rm{Surroundings}}}} = - \frac{{\left( { - {\rm{65}}{\rm{.2 kJ}}} \right)}}{{\left( {25 + 273.15} \right){\rm{ K}}}}\\\\\Delta {S_{{\rm{Surroundings}}}} = - \frac{{\left( { - {\rm{65}}{\rm{.2 kJ}}} \right)}}{{298.15{\rm{ K}}}}\\\\\Delta {S_{{\rm{Surroundings}}}} = 0.21868\frac{{{\rm{kJ}}}}{{\rm{K}}}\\\end{array}$

The unit conversion is as follows:

$1{\rm{ kJ}} = 1000{\rm{ J}}$

$\begin{array}{l}\\\Delta {S_{{\rm{Surroundings}}}} = 0.21868\frac{{\cancel{{{\rm{kJ}}}}}}{{\rm{K}}} \times \frac{{1000{\rm{ J}}}}{{1{\rm{ }}\cancel{{{\rm{kJ}}}}}}\\\\\Delta {S_{{\rm{Surroundings}}}} = 218.68\frac{{\rm{J}}}{{\rm{K}}}\\\end{array}$

Calculate the value of $\Delta {S_{{\rm{Universe}}}}$ as shown below.

$\begin{array}{c}\\\Delta {S_{{\rm{Universe}}}} = {\rm{ }}\Delta {S_{{\rm{System}}}} + \Delta {S_{{\rm{Surroundings}}}}\\\\ = - 26.4{\rm{ }}\frac{{\rm{J}}}{{\rm{K}}}{\rm{ + 218}}{\rm{.68 }}\frac{{\rm{J}}}{{\rm{K}}}\\\\{\rm{ = 192}}{\rm{.28 }}\frac{{\rm{J}}}{{\rm{K}}}\\\end{array}$

Thus, the value of $\Delta {S_{{\rm{Universe}}}}$ is ${\rm{192}}{\rm{.28 }}\frac{{\rm{J}}}{{\rm{K}}}$ .

Ans:

The value of $\Delta {S_{{\rm{Universe}}}}$ for the reaction is ${\rm{192}}{\rm{.28 }}\frac{{\rm{J}}}{{\rm{K}}}$ .