1. Tardigrade
  2. Question
  3. Chemistry
  4. For the equilibrium: CaCO3(s) rightharpoons CaO(s)+CO2(g); Kp = 1.64 atm at 1000 K 50 g of CaCO3 in a 10 litre closed vessel is heated to 1000 K. Percentage of CaCO3 that remains unreacted at equilibrium is (Given R = 0.082 L atm K-1 mol-1)

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Q. 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 $R \,= \,0.082\, L \,atm \,K^{-1} mol^{-1}$)

KCETKCET 2014Equilibrium

Solution:

$CaCO _{3}(s) \rightleftharpoons CaO (s)+ CO _{2}(g)$

$K_{p}=p_{ CO _{2}}=\frac{n}{V} R T$

$\therefore 1.64=\frac{n}{10} \times 0.082 \times 1000$

$\therefore n=\frac{1.64 \times 10}{0.082 \times 1000}=0.2$

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 $CaCO _{3}=\frac{0.3}{0.5} \times 100$

$=60 \%$