For the reaction: CaCO3(s) ⇌ CaO(s) + CO2(g). Kp = 1 atm at 9270C. If 20g of CaCO3 were kept in a 10 liter vessel at 9270C, then calculate percentage of CaCO3 remaining at equilibrium:
Open in App
Solution
Given, KP=1 atm=PCO2
V=10 liters
Temperature=927+273=1200 K
m=20 g
⟹PV=nRT⟹(1)(10)=(n)(0.0821)(1200)⟹n=0.1moles
0.1 moles of CO2 is produced.
So, 0.1 moles of CaCO3 is consumed, i.e.,10 g of CaCO3 is consumed and 50% of CaCO3 is remaining.