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1 mole of CaCO3 is heated in 11.2 lit vessel so that equilibrium is established at 819K. If KP for CaCO3=CaO CO2 at this temperature is 2 ATM, equilibrium concentration of CO2 is?
Most Upvoted Answer
1 mole of CaCO3 is heated in 11.2 lit vessel so that equilibrium is es...
Kp = Kc.(R∆T)^(∆n)
2×101325pa = Kc.(8.314×819)^(2-1)
Kc = 202,650/(8.314×819)^1= 202,650/6809.2
Kc = 29.76M ~ 30M
Kc = [CO2] = 29.76M~ 30M
2×101325pa = Kc.(8.314×819)^(2-1)
Kc = 202,650/(8.314×819)^1= 202,650/6809.2
Kc = 29.76M ~ 30M
Kc = [CO2] = 29.76M~ 30M
Community Answer
1 mole of CaCO3 is heated in 11.2 lit vessel so that equilibrium is es...
Understanding the Reaction
When 1 mole of CaCO3 is heated, it decomposes into CaO and CO2:
CaCO3 (s) ⇌ CaO (s) + CO2 (g)
At equilibrium, the relationship between the partial pressure of CO2 and the equilibrium constant Kp is given by:
Kp = P(CO2)
Given that Kp at 819 K is 2 ATM, we can conclude:
Setting Up the Problem
- Moles of CaCO3: 1 mole
- Volume of the vessel: 11.2 L
- Temperature: 819 K
- Equilibrium constant (Kp): 2 ATM
Since the reaction produces 1 mole of CO2 for every mole of CaCO3, the equilibrium concentration of CO2 can be determined from the Kp value.
Calculating the Equilibrium Concentration of CO2
- Using Ideal Gas Law:
The ideal gas law states that PV = nRT, where:
- P = pressure of the gas (in ATM)
- V = volume of the vessel (in L)
- n = number of moles of gas
- R = gas constant (0.0821 L·ATM/(K·mol))
- T = temperature (in K)
Rearranging the equation to find n (moles of CO2):
n = PV / RT
- Substituting Known Values:
At equilibrium, P = Kp = 2 ATM, and substituting:
n = (2 ATM * 11.2 L) / (0.0821 L·ATM/(K·mol) * 819 K)
- Final Calculation:
Upon calculating, the moles of CO2 yield approximately 0.53 moles.
Conclusion
Thus, the equilibrium concentration of CO2 in the vessel is approximately 0.53 moles, which corresponds to a partial pressure of 2 ATM as specified by Kp at the given temperature.
When 1 mole of CaCO3 is heated, it decomposes into CaO and CO2:
CaCO3 (s) ⇌ CaO (s) + CO2 (g)
At equilibrium, the relationship between the partial pressure of CO2 and the equilibrium constant Kp is given by:
Kp = P(CO2)
Given that Kp at 819 K is 2 ATM, we can conclude:
Setting Up the Problem
- Moles of CaCO3: 1 mole
- Volume of the vessel: 11.2 L
- Temperature: 819 K
- Equilibrium constant (Kp): 2 ATM
Since the reaction produces 1 mole of CO2 for every mole of CaCO3, the equilibrium concentration of CO2 can be determined from the Kp value.
Calculating the Equilibrium Concentration of CO2
- Using Ideal Gas Law:
The ideal gas law states that PV = nRT, where:
- P = pressure of the gas (in ATM)
- V = volume of the vessel (in L)
- n = number of moles of gas
- R = gas constant (0.0821 L·ATM/(K·mol))
- T = temperature (in K)
Rearranging the equation to find n (moles of CO2):
n = PV / RT
- Substituting Known Values:
At equilibrium, P = Kp = 2 ATM, and substituting:
n = (2 ATM * 11.2 L) / (0.0821 L·ATM/(K·mol) * 819 K)
- Final Calculation:
Upon calculating, the moles of CO2 yield approximately 0.53 moles.
Conclusion
Thus, the equilibrium concentration of CO2 in the vessel is approximately 0.53 moles, which corresponds to a partial pressure of 2 ATM as specified by Kp at the given temperature.
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1 mole of CaCO3 is heated in 11.2 lit vessel so that equilibrium is established at 819K. If KP for CaCO3=CaO CO2 at this temperature is 2 ATM, equilibrium concentration of CO2 is?
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1 mole of CaCO3 is heated in 11.2 lit vessel so that equilibrium is established at 819K. If KP for CaCO3=CaO CO2 at this temperature is 2 ATM, equilibrium concentration of CO2 is? for Class 11 2024 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about 1 mole of CaCO3 is heated in 11.2 lit vessel so that equilibrium is established at 819K. If KP for CaCO3=CaO CO2 at this temperature is 2 ATM, equilibrium concentration of CO2 is? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 1 mole of CaCO3 is heated in 11.2 lit vessel so that equilibrium is established at 819K. If KP for CaCO3=CaO CO2 at this temperature is 2 ATM, equilibrium concentration of CO2 is?.
1 mole of CaCO3 is heated in 11.2 lit vessel so that equilibrium is established at 819K. If KP for CaCO3=CaO CO2 at this temperature is 2 ATM, equilibrium concentration of CO2 is? for Class 11 2024 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about 1 mole of CaCO3 is heated in 11.2 lit vessel so that equilibrium is established at 819K. If KP for CaCO3=CaO CO2 at this temperature is 2 ATM, equilibrium concentration of CO2 is? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 1 mole of CaCO3 is heated in 11.2 lit vessel so that equilibrium is established at 819K. If KP for CaCO3=CaO CO2 at this temperature is 2 ATM, equilibrium concentration of CO2 is?.
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