Question

Write the expression for the equilibrium constant for the reaction represented by the equation

CaCO3(s) ⇌ Ca2+(aq) + CO3 −(aq). Is Kc > 1, < 1, or ≈ 1? Explain your answer.

Answer 1

Kc = [Ca2+][CO3-]Kc >1 since the concentration of the CaCO3 (denominator) is not included because it is not in same aqueous state as the ions

Answer 2

The expression for the equilibrium is: Kc=[Ca 2+][CO3 -2] and the value of Kc depends of the concentrations.

Step-by-step explanation:

When writting a Kc expression it's important to know the state (gas, solid, liquid or aqueous) of every component of the ecuation.

This will help you to know how to express the activities (usualy equal to the concentrations) of each component. Being them:

• Gases: concentration in fuction of presure
• Liquid: concentration equal to 1
• Solid: concentration equal to 1
• Aqueous: concentration in molarity (for ex: 0.1 M)

Liquids and solids are separated of the aqueous medium and that is why they don't go in the ecuation, even if they form part of the reaction.

Going to the equilibrium given, both Ca2+ and CO3 2- are aqueous so they are included in the Kc, but as said before CaCO3 is solid and it's not included.

Cause all of this we get the following expression: Kc=[Ca 2+][CO3 -2]

About the value of Kc, it depends of the concentration of the ions.

If nothing is said in the problem, we can asume that [Ca 2+]=[CO3 -2] beacuse 1 molecule of CaCO3 gives you 1 Ca2+ and 1 CO3 -2. So:

• If [Ca 2+] < 1 M --> Kc< 1

• If [Ca 2+] = 1 M --> Kc=1

• If [Ca 2+] > 1 M --> Kc>1