Q.

20 gm of CaCO3 is allowed to dissociate in a 5.6 litre container at 8190 C. If 50% of CaCO3 is dissocitated at equilibrium, the 'Kp' value is

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answer is 2.

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Detailed Solution

Initial number of moles of CaCO3 =20100=0.2Volume of the container = 5 - 6Degree of decomposition of CaCO3=50100=0.5moles of CaCO3 decomposed = 0.2 x 0.5 = 0.1moles of CaCO3 remain at equilibrium = (0.2 - 0.1) = 0.11 mole of CaCO3 → 1 mole of CO20.1 mole of CaCO3 → - n = 0.1T = (819 + 273) KV = 5.6 Lit

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