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Q.

When 20 g of CaCO3 were put in 10-liter flask and heated to 800°C,35% of CaCO3 remained unreacted at equilibrium. KP for decomposition of CaCO3 is

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a

0.145atm

b

2.146atm

c

1.145 atm

d

3.145 atm

answer is A.

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Detailed Solution

CaCO3(s)CaO(s)+CO2(g)

20 g (0.2 moles)

0.2 × 6510065% reacted

KCCO2=0.1310=0.013

Kp=KC(RT)1=0.013×0.0821×1073

KP=1.145 atm

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