Question

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

Answer 1

The amount of calcium carbonate will increase.

Step-by-step explanation:

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.

`CaCO_3(s)\rightleftharpoons CaO(s) + CO_2(g)`

Equilibrium constant of the reaction =
`K_(eq)=0.0108`

Concentration of
`CaO=(15.0 g)/(56 g/mol* 10.0L)=0.027 M`

Concentration of
`CO_2=(4.25 g)/(44 g/mol* 10.0L)=0.0096 M`

Concentration of
`CaCO_3=(15.0 g)/(100 g/mol* 10.0L)=0.015 M`

`Q=([CaO][CO_2])/([CaCO_3])`

`=(0.027 mol/L* 0.0096 mol/L)/(0.015 mol/L)=0.0174`

`K_(eq)<Q`

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