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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 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
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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?.
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