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 10 litre 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)`).

Options

A. 40
B. 50
C. 60
D. 20

Answer

C

Explanation

`K_(p)=1.64` atm
`K_(p)=K_(c )(RT)^(Delta n)`
`1.64=K_(c )=(0.082xx1000)^(1)`
or `K_(c )=(1.64)/(0.082xx1000)=0.02 M`
As `K_(c )=[CO_(2)]`
`therefore [CO_(2)]` at equilibrium `=0.02 M`.
Now, moles of `CaCO_(3)` given `=(50g)/(100 g mol^(-1))`
= 0.5 mol
Volume of vessel = 10 L
`therefore [CaCO_(3)]` at the beginning of reaction
`=(0.5)/(10)=0.05 M`
According to equation,
`CaCO_(3)(s)hArr CaO(s)+CO_(2)(g)`
1 mol of `CaCO_(3)` gives 1 mol of `CO_(2)`.
`therefore 0.02` mol of `CO_(2)` is produced from 0.02 mol of `CaCO_(3)`. This means, moles of `CaCO_(3)` that react = 0.02 mol.
Moles of `CaCO_(3)` unreacted at equilibrium
`=0.05-0.02=0.03` mol
% of `CaCO_(3)` that remains unreacted
`=(0.03)/(0.05)xx100=60%`.