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JEE Main 2017 · Chemistry · Chemical Equilibrium · medium

60 g of CaCO3 is heated in a rigid evacuated 41 L vessel at 1000 K until equilibrium CaCO3(s)rightleftharpoons CaO(s)+CO2(g) with Kp=0.50 atm is…

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60 g60\ \mathrm{g} of CaCO3\mathrm{CaCO_3} is heated in a rigid evacuated 41 L41\ \mathrm{L} vessel at 1000 K1000\ \mathrm{K} until equilibrium CaCO3(s)⇌CaO(s)+CO2(g)\mathrm{CaCO_3(s)\rightleftharpoons CaO(s)+CO_2(g)} with Kp=0.50 atmK_p=0.50\ \mathrm{atm} is reached. The vessel volume is then doubled at the same temperature and equilibrium is re-established. The mass of CaCO3\mathrm{CaCO_3} left is (R=0.082 L atm K−1mol−1R=0.082\ \mathrm{L\,atm\,K^{-1}mol^{-1}}, CaCO3=100\mathrm{CaCO_3}=100)
  1. A. 25 g25\ \mathrm{g}
  2. B. 35 g35\ \mathrm{g}
  3. C. 0 g0\ \mathrm{g}
  4. D. 10 g10\ \mathrm{g}
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Answer: (D) 10 g10\ \mathrm{g}

For the heterogeneous equilibrium Kp=pCO2=0.50K_p=p_{\mathrm{CO_2}}=0.50 atm, independent of amounts as long as some solid CaCO3\mathrm{CaCO_3} remains. In 4141 L: n=pV/RT=0.50×41/(0.082×1000)=0.25n=pV/RT=0.50\times41/(0.082\times1000)=0.25 mol, so 2525 g decomposed and 3535 g is left. After doubling the volume to 8282 L the pressure must still be 0.500.50 atm: n=0.50×82/82=0.50n=0.50\times82/82=0.50 mol of CO2\mathrm{CO_2}, i.e. 5050 g of CaCO3\mathrm{CaCO_3} decomposed in total. Since 60>5060>50, solid remains: 60−50=1060-50=10 g. (3535 g is the value before expansion.)
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