Question

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

• A. 50
• B. 20
• C. 40
• D. 60

Answer 1

Answer: (D)

60

Answer 2

The correct option is(D): 60.

$CaCO _{3}(s) \rightleftharpoons CaO (s)+ CO _{2}(g)$
$K_{p}=p_{ CO _{2}}=\frac{n}{V} R T$
$\therefore 1.64=\frac{n}{10} \times 0.082 \times 1000$
$\therefore n=\frac{1.64 \times 10}{0.082 \times 1000}=0.2$

i.e., Number of moles of $CO _{2}=0.2$
$CaCO _{3} \rightleftharpoons CaO + CO _{2}$

Initially $50\, g =\frac{50}{100}=0.5\, mol\,\,\, 0$

At equilibrium $0.2\, mol\,\,\,0.2$
$\therefore$ Unreacted $CaCO _{3}=0.5-0.2=0.3\, mol$

and $\%$ of unreacted $CaCO _{3}=\frac{0.3}{0.5} \times 100$
$=60 \%$