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KCET · Chemistry · Coordination Compounds

For the equilibrium :
\( \mathrm{CaCO}_{3}(s) \rightleftharpoons \mathrm{CaO}(s)+\mathrm{CO}_{2}(g) ; \mathrm{K}_{p}=1.64 \mathrm{~atm} \)
at \( 1000 \mathrm{~K} \)
\( 50 \mathrm{~g} \) of \( \mathrm{CaCO}_{3} \) in a \( 10 \) litre closed vessel is heated to \( 1000 \mathrm{~K} \). Percentage of \( \mathrm{CaCO}_{3} \) that
remains unreacted at equilibrium is
(Given \( \mathrm{R}=0.082 \mathrm{~L} \mathrm{~atm} K^{-1} \mathrm{~mol}^{-1} \) )

  1. A \( 50 \)
  2. B \( 20 \)
  3. C \( 40 \)
  4. D \( 60 \)
Verified Solution

Answer & Solution

Correct Answer

(D) \( 60 \)

Step-by-step Solution

Detailed explanation

\(\mathrm{CaCO}_{3}(\mathrm{~S}) \rightleftharpoons \mathrm{CaO}(\mathrm{S})+\mathrm{CO}_{2}(g) ; \mathrm{K}_{p}=1.64\)
Number of moles of \(\mathrm{CaCO}_{3}=\frac{50}{100}=0.5 \mathrm{~mol}\)
\(\mathrm{~K}_{p}=\mathrm{P}_{\mathrm{CO}_{3}}\)
From the ideal gas equation
\(\mathrm{pV}=\mathrm{nRT}\)
\(1.64 \times 10=n \times 0.085 \times 100\)
\(n=0.2\)
Therefore, no. of moles of \(\mathrm{CO}_{2}\) formed \(=0.2 \mathrm{~mol}\)
No. of moles of unreacted \(\mathrm{CaCO}_{3}\) left \(=0.5-0.2=0.3 \mathrm{~mol}\)
Percentage of unreacted \(\mathrm{CaCO}_{3}=\frac{0.3}{0.5} \times 100=60 \%\)