For the reaction CaCO3(s)→ CaO(s)+ CO₂(g), the partial pressure of C*O_{2} at 1100K is found to be 2.5 * 10 ^ 4 Pa at equilibrium, Calculate Kp for the reaction
Answers
Answer:
Given, K
P
=1 atm=P
CO
2
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 CO
2
is produced.
So, 0.1 moles of CaCO
3
is consumed, i.e.,10 g of CaCO
3
is consumed and 50% of CaCO
3
is remaining.
Explanation:
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Given, KP=1 atm=PCO2
Given, KP=1 atm=PCO2V=10 liters
Given, KP=1 atm=PCO2V=10 litersTemperature=927+273=1200 K
Given, KP=1 atm=PCO2V=10 litersTemperature=927+273=1200 Km=20 g
Given, KP=1 atm=PCO2V=10 litersTemperature=927+273=1200 Km=20 g⟹PV=nRT⟹(1)(10)=(n)(0.0821)(1200)⟹n=0.1moles
Given, KP=1 atm=PCO2V=10 litersTemperature=927+273=1200 Km=20 g⟹PV=nRT⟹(1)(10)=(n)(0.0821)(1200)⟹n=0.1moles0.1 moles of CO2 is produced.
Given, KP=1 atm=PCO2V=10 litersTemperature=927+273=1200 Km=20 g⟹PV=nRT⟹(1)(10)=(n)(0.0821)(1200)⟹n=0.1moles0.1 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.