Question

The initial concentration of the reaction CaCO₃(s) −−−CaO(s) + CO₂(g) is 2mole/lt. what is the value of Kc if 0.1 mole of reactant is consumed to get the equilibrium.​

Answer 1

CaCO

3

(s)

⇌CaO

(s)

+CO

2

(g)

Δn=1

100g 56g

K

P

=K

C

(RT)

Δn

⇒8.21=K

C

RT

∴K

C

=

0.0821×1000

8.21

⇒K

C

=0.1

Let the degree of dissociation be x

Moles of CaCO

3

=10moles (Given) in 10L

∴[CaCO

3

]=

10

10

=1mole

∴CaO formed= 1 mole

=56g