Answer: For the given reaction, the value of [tex]K_c[/tex] is greater than 1
Explanation:
For the given chemical equation:
[tex]CaCO3(s)\rightleftharpoons Ca^{2+}(aq.)+CO_3^{2-}(aq.)[/tex]
The expression of [tex]K_c[/tex] for above equation follows:
[tex]K_c=\frac{[Ca^{2+}]\times [CO_3^{2-}]}{[CaCO_3}]\\\\K_c=[Ca^{2+}]\times [CO_3^{2-}][/tex]
The concentration of pure solids and pure liquids are taken as 1 in equilibrium constant expression
As, the denominator is missing and the numerator is the only part left in the expression. So, the value of [tex]K_c[/tex] will be greater than 1.
Hence, for the given reaction, the value of [tex]K_c[/tex] is greater than 1
