20 gm of CaCO₃ is allowed to dissociate in a 5.6 litre container at 819⁰ C. If 50% of CaCO₃ is dissocitated at equilibrium, the 'K_p' value is

Initial number of  moles of CaCO₃  $=\frac{20}{100}=0.2$

Volume of the container = 5 - 6

Degree of decomposition of CaCO₃$=\frac{50}{100}=0.5$

moles of CaCO₃ decomposed = 0.2 x 0.5 = 0.1

moles of CaCO₃ remain at equilibrium = (0.2 - 0.1) = 0.1

1 mole of CaCO₃ → 1 mole of CO₂

0.1 mole of CaCO₃ → -

Stoichiometry
Initial moles	0.2		0	0
Moles at equlibrium	(0.2 - 0.1)		0.1	0.1

n = 0.1

T = (819 + 273) K

V = 5.6 Lit