Question

The following equation represents a reversible decomposition:

\({\mathbf{CaC}}{{\mathbf{O}}_3}{\text{ }}\left( {\mathbf{s}} \right) \rightleftharpoons {\mathbf{CaO}}\left( {\mathbf{s}} \right){\text{ }} + {\text{ }}{\mathbf{C}}{{\mathbf{O}}_2}{\text{ }}\left( {\mathbf{g}} \right):\)

Under what conditions will decomposition in a closed container proceed to completion so that no \({\bf{CaC}}{{\bf{O}}_3}\) remains?

Short answer

The amount of \({\rm{CaC}}{{\rm{O}}_3}\) Will not be able to generate the required \({{\rm{P}}_{\rm{C}}}{{\rm{O}}_2}\) to establish an equilibrium.

Step-by-step solution

Using the Le Châtelier's Principle:

When\({\rm{CaC}}{{\rm{O}}_3}\)decomposes,\({{\rm{P}}_{\rm{C}}}{{\rm{O}}_2}\) becomes less than\({{\rm{K}}_p}\) making the amount of\({\rm{CaC}}{{\rm{O}}_3}\)lesser.

Meaning, the amount of\({\rm{CaC}}{{\rm{O}}_3}\)Will not be able to generate the required\({{\rm{P}}_{\rm{C}}}{{\rm{O}}_2}\) to establish an equilibrium.