For the reaction

CaCO₃ (s) <=> CaO (s) + CO₂ (g), Kp = 0.5 atm at 25⁰C . If 20 gm of CaCO₃ is placed in a closed container at 25⁰C, calculate the amount of CO₂ produced at equilibrium.

(a) 20 gm (b) 44gm (c) Data insufficient (d) None of these

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Sir,

I proceeded as shown below , am I right or wrong Please give detailed solution

Kp = Kc (RT)ⁿ Here n = 1

So,

Kc = Kp/RT = 0.5/(0.0821* 298) = 0.0204 -------------------------------- (1)

Now,

CaCO₃ (s) <=> CaO (s) + CO₂ (g)

20/100 0 0

At equilibrium

0.2 - x x x

So, Kc = x -------------(2) as CaO and CaCO₃ are solid

From (1) and (2)

x = 0.0204 moles = 0.0204 * 44 gm = 0.9 gm

Please reply soon

Anurag Kishore , 15 Years ago

Grade 12