Equilibrium constant K p for the reaction CaCO 3 ⇌ CaO + CO 2 is 0.82 atm at 7 2 7 ∘ C. If 1 mole of CaCO 3 is placed in a closed container of 20 L and heated to this temperature, what amount of CaCO 3 would dissociate at equilibrium?

CaCO 3 ⇌ CaO + CO 2 K p = p CO 2 = 0.82 atm 0.82 atm × 20 L = n CO 2 × 0.082 × 1000 2 0 0 1 0 0 0 = 1 5 = n CO 2 No. of moles CO 2 = no. of moles of CaCO 3 decomposed = 1 5 mole Amount of CaCO 3 decomposed = 1 5 × 100 = 20 g

Chemistry - Chemical Equilibrium Question with Solution | TestHub

ChemistryChemical EquilibriumMiscellaneous/MixedEasy2 minPYQ_2020

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Equilibrium constant K_pfor the reaction ${CaCO}_{3} ⇌ CaO + {CO}_{2}$ is 0.82 atm at $72 7^{\circ} C.$
If 1 mole of CaCO₃is placed in a closed container of 20 L and heated to this temperature, what amount of CaCO₃would dissociate at equilibrium?

Options:

Answer:

Solution:

${CaCO}_{3} ⇌ CaO + {CO}_{2}$
$K_{p} = p_{{CO}_{2}} = 0.82 atm$
$0.82 atm \times 20 L = n_{{CO}_{2}} \times 0.082 \times 1000$
$\frac{200}{1000} = \frac{1}{5} = n_{{CO}_{2}}$
No. of moles CO₂= no. of moles of CaCO₃decomposed = $\frac{1}{5}$ mole
Amount of CaCO₃decomposed $= \frac{1}{5} × 100 = 20 g$

Stream:NTA_ABHYASSubject:ChemistryTopic:Chemical EquilibriumSubtopic:Miscellaneous/Mixed

⏱ 2mℹ️ Source: PYQ_2020

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