Question

When limestone (solid CaCO2) is heated, it decomposes into lime (solid CaO) and carbon dioxide gas. This is an extremely useful industrial process of great antiquity, because powdered lime mixed with water is the basis for mortar and concrete – the lime absorbs CO, from the air and turns back into hard, durable limestone. Suppose some calcium carbonate is sealed into a limekiln of volume 550. L and heated to 1190. °C. When the amount of Caco, has stopped changing, it is found that 2.29 kg have disappeared. Calculate the pressure equilibrium constant K this experiment suggests for the equilibrium between Caco, and Cao at 1190. °C. Round your answer to 2 significant digits. Note for advanced students: it's possible there was some error in this experiment, and the value it suggests for K, does not match the accepted value. K = 1 x 6 ?

Answer 1

Solution :

The mass of CaCO3 which is disappeared = 2.29 kg = 2290 g

Molar mass of CaCO3 = 100 g mol-1

Thus, number of moles = 2290 g / 100 g mol-1 = 22.90 mol

Since, volume is 550 L

Hence, concentration = 22.90 mol / 550 L = 0.0416 M

The equilibrium reaction is given as,

CaCO3 (s)  ==== CaO (s) + CO2 (g)

A ---------------------0 ------------0 (initial)

-X ------------------- +X ----------+X (change)

Since, dissaperred concentration = X

Thus, X = 0.0416

Since,

Kc = [CO2]

(Because both CaCO3 and CaO are solids)

Thus,

Kc = X = 0.0416

Kp = Kc (RT) ^Δn

(Δn = 1)

Kp = Kc (RT)^1

Kp = Kc x 0.0821 x (1190+273)K

Kp = 0.0416 x 0.0821 x 1463

Kp = 4.996 = 5.0

Kp = 5.0