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What is the equilibrium expression for t...

What is the equilibrium expression for the reaction`P_(4)(s)+50_(2)(g)hArrP_(4)O_(10)(s)`

A

`K_(c) = [O_(2)]^(5)`

B

`K_(c) = [P_(4)O_(10)]//5[P_(4)][O_(2)]`

C

`K_(c) = [P_(4)O_(10)]//[P_(4)][O_(2)]^(5)`

D

`K_(c) = 1//[O_(2)]^(5)`

Text Solution

Verified by Experts

The correct Answer is:
D
`P_(4)(s) + 5O_(2)(g)hArrP_(4)O_(10)(s)`
`K_(c) = ([P_(4)O_(10)(s)])/([P_(4)(s)][O_(2)(g)]^(5))`
we know that concentration of a solid compoundent is always taken as unity `K_(c) = (1)/([O_(2)]^(5))`
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The equilibrium constant for the reaction? P_(4(s)) +5 O_(2(g)) rarr P_(4)O_(10(s)) (A) k_(p) =(1)/ (p_(o_(2))^(5)) (B) k_(p) = p_(o_(2))^(5) (C) k_(p) = (P_(P_(4)o_(10))) / (p_(p_(4)) p_(o_(2))^(5)) (D) k_(p) = (p_(p_(4)) * p_(o_(2))^(5)) / (P_(p_(4)o_(10))) ]

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A2Z-CHEMICAL EQUILIBRIUM-Section D - Chapter End Test
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