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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 CaOleft( 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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