If more than one phase is present in the reversible reaction then it is said to be heterogenous system.
Example: `CaO(s)+CO_(2)(g)` Expression of equilibrium constant for the above reaction can be taken as :
`K=([CaO(s)][CO_(2)(g)])/([CaO(s)])`." ".....(i)
Now concentration of `CaO(s)=[CaO(s)]`
`=("moles of CaO")/("volume of CaO")`
as density of `CaO[rho_(CaO(s))]` and molar mass of `CaO[M_(CaO(s))]`are a fixed quantity therefore concentration of pure solid and liquid term is uncharge with respect to time. Hence, equilibrium constant for the equation (i) can be written as :
`K_(C)=[CO_(2)(g))]`
`K_(P)=P_(CO_2)`
As `K_(p) and K_(c)` is not containing solid terms therefore, addition or removel of pure solid and pure liquid has no effect on the equilibrium process.
`CaCO_(3)(s)hArr+CaO(s)+CO_(2)(s)`
At equilibrium in the above case, 'a' moles of `CaCO_(3)`, 'b' moles of CaO and 'c' moles of `CO_(2)` are found then identify the wrong statement:

Options

A. a' will decrease with the additon of inert gas at constant pressure.
B. a' will remain constant with the increase in volume.
C. If volume of the vessel is halved then 'a' increases.
D. ['b' decreases with the increase in pressure.

Answer

B