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Q.

1 mole CaCO3(s) is heated in 11.2 lit vessel so that equilibrium is established at 819 K. If KP for large CaC{O_3}, rightleftharpoons ,CaO, + ,C{O_2} at this temperature is 2 atm, equilibrium concentration of CO2 (in mol-lit-1)

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a

13

b

111.2

c

133.6

d

122.4

answer is C.

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Detailed Solution

large CaC{O_3}left( s right) rightleftharpoons CaOleft( s right) + C{O_2}left( g right)

Initial moles of CaCO3 = 1

Volume of the vessel = 11.2 Lit

Temperature = 819 K

KP = 2 atm

pCaCO3=pCaO=1as they are solids
large {K_P} = {P_{C{O_2}}}
large therefore boxed{{P_{C{O_2}}} = 2atm}

for ideal gases

PV = nRT

large frac{n}{v} = frac{P}{{RT}}
large therefore left[ {C{O_2}} right] = frac{{{P_{C{O_2}}}}}{{RT}}
large = frac{2}{{left( {0.0821} right)left( {819} right)}}
large = frac{2}{{67.2}}M
large boxed{left[ {C{O_2}} right] = frac{1}{{33.6}}M}

Alternate method

large {left[ {CaCO} right]_s} = {left[ {C{O_2}} right]_s} = 1
large therefore {K_C} = left[ {C{O_2}} right]
large {K_P} = {K_C}{left( {RT} right)^1}left( {because Delta {n_g} = 1} right)
large 2 = left[ {C{O_2}} right]left[ {0.0821 times 819} right]
large boxed{left[ {C{O_2}} right] = frac{1}{{33.6}}M}

 

 

 

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