At 1173 K, Keq = 0.0108 for the following reaction: CaCO3(s) ⇄ CaO(s) + CO2(g) The reaction takes place in a 10.0 L vessel at 1173 K. If a mixture of 15.0 g CaCO3, 15.0 g CaO, and 4.25 g CO2 is allowed to approach equilibrium, what will happen to the amount of CaCO3? Group of answer choices
- It will remain the same
- It will increase
- Not enough information is provided to answer this question
- It will decrease

K is the constant of a certain reaction when it is in equilibrium, while Q is the quotient of activities of products and reactants at any stage other than equilibrium of a reaction.

K>Q , reaction will move forward by making more product.
K<Q , reaction will move backward by making more reactant.

[tex]CaCO_3(s)\rightleftharpoons CaO(s) + CO_2(g)[/tex]

Equilibrium constant of the reaction = [tex]K_{eq}=0.0108 [/tex]

Concentration of [tex]CaO=\frac{15.0 g}{56 g/mol\times 10.0L}=0.027 M[/tex]

Concentration of [tex]CO_2=\frac{4.25 g}{44 g/mol\times 10.0L}=0.0096 M[/tex]

Concentration of [tex]CaCO_3=\frac{15.0 g}{100 g/mol\times 10.0L}=0.015 M[/tex]

[tex]Q=\frac{[CaO][CO_2]}{[CaCO_3]}[/tex]

[tex]=\frac{0.027 mol/L\times 0.0096 mol/L}{0.015 mol/L}=0.0174[/tex]

[tex]K_{eq}<Q[/tex]

This means that equilibrium will move in backward direction by which amount of calcium carbonate will increase.