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Q. $20\, g$ of sample of $CaCO _{3}( s )$ of $50 \%$ impurity is placed in a sealed vessel of $1\, L$ capacity at $400\, K$ when $50 \% T$ dissociation takes place. If $CO _{2}$ gas at $0.5\, atm$ is already present, percent dissociation of $CaCO _{3}$ is
$CaCO _{3}( s ) \rightleftharpoons CaO ( s )+ CO _{2}( g )$

Equilibrium

Solution:

image

$x=0.05\, mol\, (50 \%$ of $0.1\, mol )$

$p_{C O_{2}}=\frac{n}{V} RT=\frac{0.05}{1} \times 0.0821 \times 400$

$=1.642\, atm$

Thus, $K _{ p }=p_{C O_{2}}=1.642$

image

$K_{p}=(0.5+x)=1.642$

$\therefore x=1.142$

$p=\frac{n R T}{V}$

$1.142\, atm=\frac{n \times 0.0821 \times 400}{1}$

$n =0.0348$ (dissociated)

Thus, percentage $=\frac{0.0348}{0.1} \times 100=34.8 \%$