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Standard XII

Chemistry

Reaction Quotient

n moles of ...

Question

$n$ moles of $P {C l}_{3}$ and $n$ moles of ${C l}_{2}$ are allowed to react at constant temperature $T$ under a total pressure $P$ as $P {C l}_{5} ⇌ P {C l}_{3} + {C l}_{2}$ .If $y$ moles of $P {C l}_{5}$ are formed at equilibrium. Find $K_{p}$ .

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Solution

If initial concentration of n for $P {C l}_{3}$ and for ${C l}_{2}$ is n
Then $[P {C l}_{3}] = n α$
$[{C l}_{2}] = n α$ where $α = n - y$
$[P {C l}_{3}] = γ$
$K_{p} = \frac{[1 - α^{2}]}{α^{2} p}$
$= \frac{[1 - {(n - y)}^{2}]}{{(n - y)}^{2} p}$
$= \frac{1 - n^{2} - γ^{2} + 2 n y}{{(n - y)}^{2} p}$
Since $1 - n^{2} << - 0 ∴$ neglect it
Hence $K_{p} = \frac{2 n y - y^{2}}{{(n - y)}^{2} p}$
$K_{p} = \frac{(2 n - y) y}{{(n - y)}^{2} p}$

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Similar questions

Q. n mole of

P C l_{3}

and n mole

C l_{2}

are allowed to react at constant temperature T to have a total equilibrium pressure

P

, as :

P C l_{3} (g) + C l_{2} (g) ⇌ P C l_{5} (g)

. If y mole of

P C l_{3}

are formed at equilibrium, find

k_{p}

for the given reaction.

$P C l_{5} ⇌ P C l_{3} + C l_{2}$ in the reversible reaction, the moles of $P C l_{5}$ , $P C l_{3}$ and $C l_{2}$ are a, b and c respectively at equilibrium and total pressure is P, then value of $K_{p}$ is:

Q. In the reaction,

P C l_{5} ⇌ P C l_{3} + C l_{2}

, the amount of

P C l_{5}, P C l_{3}

and

P C l_{2}

at equilibrium are

2

moles each and the total pressure is

3

atm. The equilibrium constant

K_{p}

is:

P {C l}_{5} ⇌ P {C l}_{3} + {C l}_{2}

in the reversible reaction the moles of

P {C l}_{5}, P {C l}_{3}, {C l}_{2}

are

a, b

and

c

respectively and total pressure is

P

then value of

K_{p}

is:

Q. A reaction at equilibrium involving 2 moles each of

P C l_{5}, P C l_{3},

and

C l_{2}

is maintained at

250^{o} C

and a total pressure of 3 atm. The value of

K_{p}

is :
Given:

P C l_{5} ⇌ P C l_{3} + C l_{2}

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n moles of PCl3 and n moles of Cl2 are allowed to react at constant temperature T under a total pressure P as PCl5⇌PCl3+Cl2.If y moles of PCl5 are formed at equilibrium. Find Kp.

If initial concentration of n for PCl3 and for Cl2 is n Then [PCl3]=nα[Cl2]=nα where α=n−y[PCl3]=γKp=[1−α2]α2p=[1−(n−y)2](n−y)2p=1−n2−γ2+2ny(n−y)2pSince 1−n2<<−0∴neglect itHence Kp=2ny−y2(n−y)2pKp=(2n−y)y(n−y)2p

$n$ moles of $P {C l}_{3}$ and $n$ moles of ${C l}_{2}$ are allowed to react at constant temperature $T$ under a total pressure $P$ as $P {C l}_{5} ⇌ P {C l}_{3} + {C l}_{2}$ .If $y$ moles of $P {C l}_{5}$ are formed at equilibrium. Find $K_{p}$ .