Question

The equilibrium constant for the reaction \[CaC{{O}_{3}}(s)\rightleftharpoons CaO(s)+C{{O}_{2}}(g)\] is ………..

Answer 1

Hint: Equilibrium is defined as that state at which the rate of forward reaction is equal to the rate of backward reaction. Equilibrium constant describes the reaction quotient at equilibrium state and is represented by \[K\].

Complete answer:
Mathematically, equilibrium constant \[(K)\]is described as the ratio of the molar concentration of product species over the reactant species at equilibrium state, considering the stoichiometric coefficients of the reaction into account as exponents of the molar concentrations. For the general reaction \[aA+bB\rightleftharpoons cC+dD\], \[K\] is represented as
\[K=\dfrac{{{[C]}^{c}}{{[D]}^{d}}}{{{[A]}^{a}}{{[B]}^{b}}}\]
Considering the above equation into account, the equilibrium constant for the given reaction is given as
\[K=\dfrac{[CaO(s)][C{{O}_{2}}(g)]}{[CaC{{O}_{3}}(s)]}----\left( 1 \right)\]
It is important to mention that molar concentration of solid and liquids are constant and are equal to 1. Therefore,
\[[CaO(s)]=1\], \[[CaC{{O}_{3}}(s)]=1\]
Put these values in equation 1, then we get
\[K=[C{{O}_{2}}(g)]\]
Therefore, the equilibrium constant for the above reaction is equal to \[[C{{O}_{2}}(g)]\]i.e. molar concentration of carbon dioxide.

Additional information:
The value of equilibrium constant provides an idea at which the equilibrium state lies. If the molar concentration of reactants and products are the same at equilibrium, then the value of equilibrium constant is 1. Greater the value of \[K\], more will be the formation of the product as equilibrium predominantly lies towards the product.

Note:
It is important to note that the equilibrium constant for the above reaction is equal to \[[C{{O}_{2}}(g)]\]. The molar concentration of solids and liquids are constant. Greater the value of \[K\], more will be the formation of the product as equilibrium predominantly lies towards the product.