Chemistry, asked by kajalsureshks, 1 year ago

0.84 moles of PCl5 and 0.18 moles of PCl3 are mixed in a one litre flask at 27°C. It is later found
that 0.72 moles of PCl5 are present at equilibrium. The K. for the reaction.
PCl5(g) =PCl3(g) + Cl2(g) is​

Answers

Answered by gopalberma
1

Answer:

Hello dear,

● Answer - 0.133

◆ Explanation-

The reaction is given by,

Reaction PCl5 ---> PCl3 + Cl2

Initially 0.8 0.2 0

Equilibrium 0.8-x 0.2+x x

We have, at equilibrium

[PCl3] = 0.2+x = 0.4

x = 0.2

Hence,

[PCl5] = 0.8-x = 0.6

[PCl3] = x = 0.2+x = 0.4

[Cl2] = x = 0.2

For this reaction, equilibrium constant is given by,

Kc = [PCl3][Cl2] / [PCl5]

Kc = 0.4×0.2 / 0.6

Kc = 2/15

Kc = 0.133

Hope this is useful...

Answered by devkum0654
0

Explanation:

PCl5(g) =PCl3(g) + Cl2(g)

initial mole 0.84 0.18 0

mole at eqbm 0.84-a 0.18+a a

=0.12. =0.30 =0.12

Given 0.84 -a =0.72

a =0.12

now Kc=0.30×0.12

________ =0.05

0.72

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