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What is the equilibrium expression for t...

What is the equilibrium expression for the reaction `P_(4(s)) + 5O_(2(g))hArrP_(4)O_(10(s))` ?

A

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

B

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

C

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

D

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

Text Solution

Verified by Experts

The correct Answer is:
D
`K_(p) = K_(c)xx(RT)^(Delta n)`
`.: delta n = 1 - 2 = - 1`
`:. K_(p) = K_(c)xx(RT)^(-1)` Or `(K_(p))/(K_(c)) = (1)/(RT)`
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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))) ]

For the reaction 2NO_(2(g))+(1)/(2)O_(2)hArrN_(2)O_(5(g)) if the equilibrium constant is K_(P) ,then the equilibrium constant for the reaction 2N_(2)O_(5(g))hArr4NO_(2(g))+O_(2(g)) would be :-

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