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Byju's Answer

Standard XII

Chemistry

Atomic Size and Electronegativity

In the reacti...

Question

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:

1.0

atm

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2.0

atm

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3.0

atm

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6.0

atm

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Solution

The correct option is A $1.0$ atm
$P_{p c l 5} = 2 \times 3 = 6 a t m$
$P_{p c l 3} = 6 a t m$
$P_{p c l 2} = 6 a t m$
$2 p c l_{5} ⇌ 2 p c l_{3} + c l_{2}$
$K p = (\frac{Partial pressure of products}{partial pressure of reactants})$
$K p = \frac{(6) (6)^{2}}{(6)^{2}} = 6 a t m$

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Similar questions

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}

Q. For the given reaction,

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

Calculate

K_{p}

, if at equilibrium

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

Q. A reaction at equilibrium involving

2

mol each of

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

is maintained at

250^{o} C

and total pressure of

3

atm. The value of

K_{p}

is:

Q. '

a

' mol of

P {C l}_{5}

undergoes thermal dissociation as

P {C l}_{5} ⇌ P {C l}_{3} + C l_{2}

; the mole fraction

P {C l}_{3}

at equilibrium is 0.25 and the total pressure is 2 atm. The partial pressure is

{C l}_{2}

at equilibrium is:

Q. For the reaction

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

The mole of each component

P C l_{5}, P C l_{3} a n d C l_{2}

at equilibrium were found to be 2. If the total pressure is 3 atm. The value of

K_{P}

will be-

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In the reaction, PCl5⇌PCl3+Cl2, the amount of PCl5,PCl3 and PCl2 at equilibrium are 2 moles each and the total pressure is 3 atm. The equilibrium constant Kp is:

The correct option is A 1.0 atmPpcl5=2×3=6atmPpcl3=6atmPpcl2=6atm2pcl5⇌2pcl3+cl2Kp=(Partial pressure of productspartial pressure of reactants)Kp=(6)(6)2(6)2=6atm

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:

The correct option is A $1.0$ atm
$P_{p c l 5} = 2 \times 3 = 6 a t m$
$P_{p c l 3} = 6 a t m$
$P_{p c l 2} = 6 a t m$
$2 p c l_{5} ⇌ 2 p c l_{3} + c l_{2}$
$K p = (\frac{Partial pressure of products}{partial pressure of reactants})$
$K p = \frac{(6) (6)^{2}}{(6)^{2}} = 6 a t m$