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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

111.2

b

13

c

133.6

d

122.4

answer is C.

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Detailed Solution

\large CaC{O_3}\left( s \right) \rightleftharpoons CaO\left( 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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