Question

Consider the equation below.

CaC03(S) -------- Ca0(s) + C02(G)

What is the equilibrium constant expression for the given reaction?

Answer 1

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.

Answer 2

Answer is: Keq = [CO₂].

Balanced chemical reaction: CaCO3(s) ⇄ CaO(s) + CO₂(g).

The equilibrium constant (Keq) is a ratio of the concentration of the products to the concentration of the reactants.

Pure liquids (shown in chemical reactions by appending (l) to the chemical formula) and solids (shown in chemical equations by appending (s) to the chemical formula) not go in to he equilibrium constant expression, only gas state (shown in chemical reactions by appending (g) to the chemical formula) reactants and products go in to the equilibrium constant expression.