Question

$ PC{{l}_{5}}PC{{l}_{3}}+C{{l}_{2}} $ in this reversible reaction the moles of $ PC{{l}_{5}}, $ $ PC{{l}_{3}}, $ and $ C{{l}_{2}} $ are a, b and c respectively and total pressure is P then value of $ {{K}_{p}} $ is:

• A. $ \frac{bc}{a}.RT $
• B. $ \frac{b}{(a+b+c)}.P $
• C. $ \frac{bc.P}{a(a+b+c)} $
• D. $ \frac{c}{(a+b+c)}.P $

Answer 1

Answer: (C)

$ {{K}_{P}}=\frac{P'PC{{l}_{3}}\times P'C{{l}_{2}}}{P'PC{{l}_{5}}} $ $ =\frac{\frac{b}{(a+b+c)}.P\times \frac{c}{(a+b+c)}\times P}{\frac{a}{(a+b+c)}P} $ $ {{K}_{p}}=\frac{bc.\,P}{a(a+b+c)} $ $ \underset{Acetylene}{\mathop{CH\equiv CH}}\,\xrightarrow[Ni]{{{H}_{2}}}C{{H}_{2}}=C{{H}_{2}} $