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Standard XII

Chemistry

Soaps

2 moles of ...

Question

2 moles of $P C l_{5}$ is taken in an into a container and the dissociation is carried out. $K_{c}$ for the equilibrium is equal to 1, the degree of dissociation of $P C l_{5}$ is equal to:

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0.5

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Solution

The correct option is C 0.5
$P C l_{5} (g) ⇌ P C l_{3} (g) + C l_{2} (g)$
$K_{c} = \frac{[P C l_{3}] [C l_{2}]}{[P C l_{5}]} = 1$
If $α =$ degree of dissociation
$K_{c} = \frac{1 \times 1}{1} = 1$
$[P C l_{5}] = 2 - 2 α = 1$
$[P C l_{3}] = 2 α = 1$
$[C l_{2}] = 2 α = 1.$

Hence, $α = 0.5$

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Similar questions

Q. A 1-litre container contains 2 moles of

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P C l_{5} ⇌ P C l_{3} + C l_{2}

Dissociation constant

(K_{c})

P C l_{5}

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Q. 4 moles of

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At $227^{o} C$ one mole of $C l_{2}$ and 3 moles of $P C l_{5}$ are placed in a 100 litre container and allowed to reach equilibrium. It is found that the equilibrium pressure is 2.05 atmospheres. Assuming ideal behaviour, the degree of dissociation 'x' for $P C l_{5}$ and $K_{p}$ for the reaction $P C l_{5}$ (g) $⇌$ $P C l_{3}$ (g) + $C l_{2}$ (g) is:

Q. If

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2 moles of PCl5 is taken in an into a container and the dissociation is carried out. Kc for the equilibrium is equal to 1, the degree of dissociation of PCl5 is equal to:

The correct option is C 0.5PCl5(g)⇌ PCl3(g)+ Cl2(g)Kc=[PCl3][Cl2][PCl5]=1If α= degree of dissociationKc=1×11=1[PCl5]=2−2α=1[PCl3]=2α=1[Cl2]=2α=1.Hence, α=0.5

2 moles of $P C l_{5}$ is taken in an into a container and the dissociation is carried out. $K_{c}$ for the equilibrium is equal to 1, the degree of dissociation of $P C l_{5}$ is equal to:

The correct option is C 0.5
$P C l_{5} (g) ⇌ P C l_{3} (g) + C l_{2} (g)$
$K_{c} = \frac{[P C l_{3}] [C l_{2}]}{[P C l_{5}]} = 1$
If $α =$ degree of dissociation
$K_{c} = \frac{1 \times 1}{1} = 1$
$[P C l_{5}] = 2 - 2 α = 1$
$[P C l_{3}] = 2 α = 1$
$[C l_{2}] = 2 α = 1.$

Hence, $α = 0.5$