When heated, CaCO 3 decomposes as follows: CaCO 3 (s)<==========>CO 2 (g) +CaO (

Question

When heated, CaCO3 decomposes as follows:
CaCO3 (s)<==========>CO2 (g) +CaO (s)
In an experiment, 7.00 g of CaCO3 was placedin a .500-L container and when it was heated to a constanttemperature, 50% of the solid decomposed. What is the value ofKc (equilibrium constant) under these conditionsfor this reaction?

CaCO3 (s)<==========>CO2 (g) +CaO (s)
In an experiment, 7.00 g of CaCO3 was placedin a .500-L container and when it was heated to a constanttemperature, 50% of the solid decomposed. What is the value ofKc (equilibrium constant) under these conditionsfor this reaction?

Explanation / Answer

mass of CaCO3 is , m = 7 g Molar mass of CaCO3 is = 40 + 12 + 3*16 = 100 g /mol Volume V = 0.5 L Molarity M = ( mass / Molar mass ) / Volume in L                   = ( 7 / 100 ) / 0.5                    = 0.14 M Since 50% of the solid decomposes so the concentration is 0.14/ 2 = 0.07 M                                CaCO3 (s) <=====>CO2 (g) +CaO (s) initialconc                  0.14                             0            0 change                          0.07                           0.07      0.07 Equbconc.              0.14-0.07                       0.07        0.07                                  = 0.07 Equilibrium constant Kc = ([CO2 ] [ CaO] ) / [CaCO3]                                      = ( 0.07 * 0.07 ) / 0.07                                      = 0.07

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