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25/10/2021 1:23 pm
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In the dissociation of PCl5 as
PCl5(g) ⇌ PCl3(g) + Cl2(g)
if the degree of dissociation is α at equilibrium pressure P, then the equilibrium constant for the reaction is
(a) Kp = \(\frac{α^2}{1 + α^2P}\)
(b) Kp = \(\frac{α^2P^2}{1 - α^2}\)
(c) Kp = \(\frac{P^2}{1 - α^2}\)
(d) Kp = \(\frac{α^2P}{1 - α^2}\)
1 Answer
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25/10/2021 1:29 pm
Correct answer: (d) Kp = \(\frac{α^2P}{1 - α^2}\)
Explanation:
PCl5(g) ⇌ PCl3(g) + Cl2(g)
Initial 1 0 0
At. eqn. 1-α α α
Partial pressure \(\frac{1-α}{1+α}.P\) \(\frac{α}{1+α}.P\) \(\frac{α}{1+α}.P\)
Total mole = 1 - α + α + α = 1 + α
Kp = \(\frac{\frac{α}{1+α}.P. \frac{α}{1+α}.P}{\frac{1-α}{1+α}.P}\)
= \(\frac{α^2P}{1 - α^2}\)
This post was modified 3 years ago by Mayank1444