Equilibrium constant K p for the reaction CaCO 3 ⇌ CaO + CO 2 is 0.82 atm at 7 2 7 ∘ C. If 1 mole of CaCO 3 is placed in a closed container of 20 L and heated to this temperature, what amount of CaCO 3 would dissociate at equilibrium?

Step-by-step solution

CaCO 3 ⇌ CaO + CO 2 K p = p CO 2 = 0.82 atm 0.82 atm × 20 L = n CO 2 × 0.082 × 1000 2 0 0 1 0 0 0 = 1 5 = n CO 2 No. of moles CO 2 = no. of moles of CaCO 3 decomposed = 1 5 mole Amount of CaCO 3 decomposed = 1 5 × 100 = 20 g