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Derivation of Kp and Kc

The equilibri...

Question

The equilibrium constant $K_{p}$ for the reaction $P C l_{5 (g)} ⇋ P C l_{3 (g)} + C l_{2 (g)}$ is 16. If the volume of the container is reduced to half of its original volume, the value of $K_{p}$ for the reaction at the same temperature will be :

A

32

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B

64

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C

16

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D

4

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Solution

The correct option is C 16
If the volume of the container is reduced to half of its original volume, the value of $K_{p}$ for the reaction at the same temperature will be same i.e. 16. $K_{p}$ value changes if and only if temperature change. It does change with change in volume, pressure or concentration.

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Similar questions

Q. The equilibrium constant

(K_{p})

for the reaction

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

is 16. If the volume of the container is reduced to one half its original volume, the value of

K_{p}

for the reaction at the same temperature will be:

Q. The equilibrium constant

(K_{p})

for the reaction

P C l_{5} (g) ⇌ P C l_{3} (g)

is 16. If the volume of the container is reduced to one-half its original volume, the value of

K_{p}

for the reaction at the same temperature will be:

Q. For the reaction

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

K_{c}

250^{\circ} C

is 26. The value of

K_{p}

at the same temperature will be:

Q. For the following reaction at

250^{o} C

K_{c}

26

, then the value of

K_{p}

at the same temperature will be:

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

For the reaction $P C l_{3} (g) + C l_{2} (g) ⇌ P C l_{5} (g)$ at $2500^{\circ} C$ , the value of $K_{C}$ is $26$ , then the value of $K_{P}$ at the same temperature will be

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