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Acidity of Alcohols and Phenols

Kp for a reac...

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Kp for a reaction PCl5 gives PCl3+Cl2 at 250^°C is 0.82.calculate the degree of dissociation at given temperature under a total pressure of 5atm

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a = \sqrt{\frac{K_{p}}{P + K_{p}}}

P C l_{5} ⇌ P C l_{3}

C l_{2}

a

is degree of dissociation at temperature when equilibrium constant is

K_{p}

Q. Under what pressure must an equimolar mixture of

P C l_{3}

C l_{2}

250^{\circ} C

in order to obtain

P C l_{5}

1 a t m

K_{p}

for dissociation of

P C l_{5} = 1.78)

.

4.5 g

P C l_{5}

were completely vaporized at

250

and the vapour occupied

1.7 l i t r e

at 1 atmospheric pressure. Calculate the degree of dissociation(%), for the reaction :

P C l_{5} ⇌ P C l_{3} + C l_{2}

. Write answer as nearest integer after diving degree of dissociation calculated by 10.

Q. A reaction at equilibrium involving 2 moles each of

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

C l_{2}

is maintained at

250^{o} C

and a total pressure of 3 atm. The value of

K_{p}

P C l_{5} ⇌ P C l_{3} + C l_{2}

α = {[K_{p} / (P + K_{p})]}^{1 / 2}

P C l_{5} ⇌ P C l_{3} + C l_{2}

α

is degree of dissociation at temperature when equilibrium constant is .

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