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Q.

1 mole CaCO_3(s) is heated in 11.2 lit vessel so that equilibrium is established at 819 K. If K_P 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

$\frac{1}{3}$

b

$\frac{1}{11.2}$

c

$\frac{1}{33.6}$

d

$\frac{1}{22.4}$

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

K_P = 2 atm

p_{{CaCO}_{3} =} p_{CaO} = 1 (as 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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1 mole CaCO3(s) is heated in 11.2 lit vessel so that equilibrium is established at 819 K. If KP for at this temperature is 2 atm, equilibrium concentration of CO2 (in mol-lit-1)