Question

Calculate the standard entropy change for the following reaction:

\({\bf{Ca(OH}}{{\bf{)}}_{\bf{2}}}{\bf{(\;s)}} \to {\bf{CaO(s) + }}{{\bf{H}}_{\bf{2}}}{\bf{O(l)}}\)

Short answer

The standard entropy change for above process is \(120.644\;{\rm{J}}/{\rm{Kmol}}\)

Step-by-step solution

Define enthalpy of the reaction

• Enthalpy change is defined as the difference in enthalpies of all the product and the reactants each multiplied with their respective number of moles. It is represented as

• The equation used to calculate enthalpy change is of a reaction is:
• \(\Delta {H_{rxn}} = \sum {\left[ {n \times \Delta {H_{f({\rm{ product }})}}} \right]} - \sum {\left[ {n \times \Delta {H_{f({\rm{ reactant }})}}} \right]} \)

Determine the enthalpy of the reaction.

Reaction is:

\(\begin{array}{l}{\rm{Ca}}{({\rm{OH}})_2}(\;{\rm{S}}) \to {\rm{CaO}}({\rm{S}}) + {{\rm{H}}_2}{\rm{O - - - - - - - }}({\rm{l}})\\\Delta {S^\theta }\left( {{\rm{Ca}}{{({\rm{OH}})}_2}} \right) = 83.39\;{\rm{J}}/{\rm{molK}}\\\Delta {S^\theta }({\rm{CaO}}) = 38.2\;{\rm{J}}/{\rm{molK}}\\\Delta {S^\theta }\left( {{{\rm{H}}_2}{\rm{O}}} \right) = 69.91\;{\rm{J}}/{\rm{molK}}\end{array}\)

Calculate \(\Delta {s^\theta }\) of the reaction:

\(\Delta {S^\theta } = \Delta {S^\theta }(\)Product \() - \Delta {S^\theta }({\mathop{\rm Reactant}\nolimits} )\)

\(\Delta {S^\theta } = \left[ {\Delta {S^\theta }({\rm{CaO}}) + \Delta {S^\theta }\left( {{{\rm{H}}_2}{\rm{O}}} \right)} \right] - \left[ {\Delta {S^\theta }\left( {{\rm{Ca}}{{({\rm{OH}})}_2}} \right)} \right]\)

\(\Delta {S^\theta } = [38.2 + 69.91] - [83.39]\)

\(\Delta {S^\theta } = (108.11 - 83.39){\rm{J}}\mid {\rm{kmol}}\)

\(\Delta {s^\theta } = + 24.72\;{\rm{J}}/{\rm{kmo}}\quad \)(Answer)

Hence; the standard entropy changed is \( + 24.72\;{\rm{J}}/{\rm{kmol}}\)