Chemistry, asked by priyankabharadwaj11c, 27 days ago

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

Answered by lnandini93
0

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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Answered by aadianshuman1234
0

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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.

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