Consider the equation below. cac03(s) ca0(s) + c02(g) what is the equilibrium constant expression for the given reaction?

First at time t=0, 'a' moles of CaCO3(s) will convert 0 moles of CaO(s) and 0 moles of CO2(g) means reaction is not started yet.

'a' moles 0 moles  +   0 moles       t= 0 sec

At equilibrium stage meas at t= t(equ) x moles of CaO(s) and x moles of CO2(g) will have formed and their is also reduction of x moles from 'a' moles of CaCO3(s)

(a-x) moles 'x' moles + 'x' moles                t= t(equ)

divided by volume we also get concentration form,

(a-x)/V moles dm⁻³  x/V moles dm⁻³  + x/V moles dm⁻³

Now,
Kc = concentration of products/concentration of reactants
Kc= [Ca0][C02] / [CaCO3]
Kc= (x/V)(x/V) / (a-x)/V
Kc= x² / V(a-x)
is the equilibrium constant expression as this expression is not independent of the factor of volume so the change in volume at equilibrium stage disturbs the equilibrium position. 

It's either 3 or that ,,but if not than choose 4