Question

Calcium carbonate decomposes to form calcium oxide and carbon dioxide, like this:

CaCO3(s)→CaO(s)+CO2(g)
At a certain temperature, a chemist finds that a 9.0L reaction vessel containing a mixture of calcium carbonate, calcium oxide, and carbon dioxide at equilibrium has the following composition:
Compound Amount
CaCO3 25.3 g
CaO 14.9 g
CO2 33.7 g
Calculate the value of the equilibrium constant Kc for this reaction. Round your answer to 2 significant digits.

Answer 1

Answer: The value of the equilibrium constant Kc for this reaction is 0.088

Explanation:

`Molarity=(x)/(M* V_s)`

where,

x = given mass

M = molar mass

`V_s` = volume of solution in L

Equilibrium concentration of
`CaCO_3` =
`(25.3)/(100* 9.0)=0.028M`

Equilibrium concentration of
`CaO` =
`(14.9)/(56* 9.0)=0.029M`

Equilibrium concentration of
`CO_2` =
`(33.7)/(44* 9.0)=0.085M`

The given balanced equilibrium reaction is,

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

The expression for equilibrium constant for this reaction will be,

`K_c=([CaO]* [CO_2])/([CaCO_3])`

Now put all the given values in this expression, we get :

`K_c=(0.029* 0.085)/(0.028)=0.088`