1.

For the equilibrium: CaCO_(3(s)) iff CaO_((s)) + CO_(2(g)) , K_p = 1.64 atm at 1000 K, 50 g of CaCO_3 in a 10litre closed vessel is heated to 1000 K. Percentage of ĆaCO_3 that remains unreacted at equilibrium is (Given R=0.082 L atm K^(-1) "mol"^(-1)).

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Solution :`CaCO_3 iff CaO + CO_2`
`Kp = pCO_2`
No. of MOLES =n
`1.64 xx 10 = 0.082 xx 1000 xx n`
`n=(1.64 xx 10)/(0.082 xx 1000)=0.2`
`:.` No. of moles `CO_2 = 0.2`
50g of `CaCO_3 = 0.5` moles of `CaCO_3` gives 0.2 moles of `CO_2`
`IMPLIES` percentage of `CaCO_3` unreacted = 0.3 MOLE = 60%


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