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
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
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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