Question

Calcium carbonate decomposes to form calcium oxide and carbon dioxide, like this: (s)(s)(g) At a certain temperature, a chemist finds that a reaction vessel containing a mixture of calcium carbonate, calcium oxide, and carbon dioxide at equilibrium has the following composition: compound amount Calculate the value of the equilibrium constant for this reaction. Round your answer to significant digits.

Answer 1

`Kc=0.0851`

Step-by-step explanation:

Hello there!

In this case, according to the given chemical reaction:

CaCO3(s)→CaO(s)+CO2(g)

In which the amounts are unfortunately given, we can however, assume the information of similar problems so you can further modify the numbers but follow the same work:

CaCO3 = 25.3 g

CaO = 14.9 g

CO2 = 33.7 g

Thus, since just CO2 is involved on the equilibrium expression because CaCO3 and CaO are solid, we can compute the moles of the CO2 at equilibrium and further compute the concentration in the 9.0-L vessel:

`n_(CO_2)=33.7g*(1molCO_2)/(44.01gCO_2)=0.766molCO_2\\\\`

`[CO_2]=(0.766molCO_2)/(9.0L)=0.0851M`

Thus, we proceed with the equilibrium expression to obtain:

`Kc=[CO_2]\\\\Kc=0.0851`

Best regards!