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Equilibrium constant K(p) for the reacti...

Equilibrium constant `K_(p)` for the reaction:
`CaCO_(3)(s) If 1 mole of `CaCO_(3)` is placed in a closed container of 20L and heated to this temperature, what amount of `CaCO_(3)` in grams would dissociate at equilibrium ?

Text Solution

Verified by Experts

The correct Answer is:
20
`CaCO_(3)(s) `K_(p) = P_(CO_(2)) = 0.82` atm,
`n_(CO_(2)) =(PV)/(RT) = (0.82 xx 20)/(0.082 xx 1000) = 0.2` mole
Mole of `CaCO_(3)` dissociated `=n_(CO_(2)) =0.2`
Amount dissociated `=0.2 xx 100 = 20g`
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Knowledge Check

  • Equilibrium constant K_(p) for the reaction CaCO_(3)(s) hArr CaO(s) + CO_2(g) is 0.82 atm at 727^@C . If 1 mole of CaCO_(3) is placed in a closed container of 20 L and heated to this temperature, what amount of CaCO_(3) would dissociate at equilibrium?

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    B
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