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Equilibrium constant Kp for the reactionCaCO3( s)⇌CaO(s)+CO2( g) is 0.82 atm at 727°CIf 1 mole of CaCO3 is placed in a closed container of 20L and heated to this temperature, what amount of CaCO3 in grams would dissociate at equilibrium ?

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answer is 20.

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Detailed Solution

CaCO3( s)⇌CaO(s)+CO2( g)Kp=PCO2=0.82 atmnCO2=PVRT=0.82×200.082×1000=0.2 mole Mole of CaCO3 dissociated =nCO2=0.2 Amount dissociated =0.2×100=20 g

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Equilibrium constant Kp for the reactionCaCO3( s)⇌CaO(s)+CO2( g) is 0.82 atm at 727°CIf 1 mole of CaCO3 is placed in a closed container of 20L and heated to this temperature, what amount of CaCO3 in grams would dissociate at equilibrium ?