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Standard XII

Chemistry

Le Chatelier's Principle for Del N Lesser Than Zero

Equilibrium c...

Question

Equilibrium constant $K_{p}$ for the reaction
$C a C O_{3 (s)} ⇌ C a O_{(s)} + C O_{2 (g)}$ is $8.21$ atm at $727^{0}$ C,
if $10$ mole of $C a C O_{3 (s)}$ is placed in a $10$ L container, what is the weight (in gm) of CaO formed at equilibrium.

56

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28

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112

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Solution

The correct option is A $56$
${C a C O_{3}}_{(s)} ⇌ C a O_{(s)} + {C O_{2}}_{(g)}$
$Δ n = 1$
$100 g$ $56 g$

$K_{P} = K_{C} (R T)^{Δ n}$
$\Rightarrow 8.21 = K_{C} R T$
$∴ K_{C} = \frac{8.21}{0.0821 \times 1000}$
$\Rightarrow K_{C} = 0.1$
Let the degree of dissociation be $x$
Moles of $C a C O_{3} = 10 m o l e s$ (Given) in $10 L$
$∴ [C a C O_{3}] = \frac{10}{10} = 1$ mole
$∴ C a O$ formed= $1$ mole
$= 56 g$

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Equilibrium constant Kp for the reactionCaCO3(s)⇌CaO(s)+CO2(g) is 8.21 atm at 7270C,if 10 mole of CaCO3(s) is placed in a 10 L container, what is the weight (in gm) of CaO formed at equilibrium.

The correct option is A 56CaCO3(s)⇌CaO(s)+CO2(g)Δn=1 100g 56gKP=KC(RT)Δn⇒8.21=KCRT∴KC=8.210.0821×1000⇒KC=0.1Let the degree of dissociation be xMoles of CaCO3=10moles (Given) in 10L∴[CaCO3]=1010=1mole∴CaO formed= 1 mole =56g

Equilibrium constant $K_{p}$ for the reaction
$C a C O_{3 (s)} ⇌ C a O_{(s)} + C O_{2 (g)}$ is $8.21$ atm at $727^{0}$ C,
if $10$ mole of $C a C O_{3 (s)}$ is placed in a $10$ L container, what is the weight (in gm) of CaO formed at equilibrium.