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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.

User Plusor
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1 Answer

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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

User Crazymatt
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