Question

Write the expression for the equilibrium constant for the reaction represented by the equation CaCO3(s)⇌Ca2+(aq)+CO32−(aq). Is Kc > 1, < 1, or ≈ 1? Explain your answer.

Answer 1

Answer: For the given reaction, the value of $K_c$ is greater than 1

Explanation:

For the given chemical equation:

$CaCO3(s)\rightleftharpoons Ca^{2+}(aq.)+CO_3^{2-}(aq.)$

The expression of $K_c$ for above equation follows:

$K_c=\frac{[Ca^{2+}]\times [CO_3^{2-}]}{[CaCO_3}]\\\\K_c=[Ca^{2+}]\times [CO_3^{2-}]$

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 $K_c$ will be greater than 1.

Hence, for the given reaction, the value of $K_c$ is greater than 1