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On decomposition of NH(4)HS , the follow...

On decomposition of `NH_(4)HS` , the following equilibrium is estabilished: `NH_(4) HS(s)hArrNH_(3)(g) + H_(2)S(g)` If the total pressure is P atm, then the equilibrium constant `K_(p)` is equal to

A

`P` atm

B

`P^(2)`atm

C

`P^(2)//4 atm^(2)`

D

`2P` atm

Text Solution

Verified by Experts

The correct Answer is:
C
`NH_(4)HS(s)hArrNH_(3)(g)+H_(2)S(g)`
Total pressure is `P`.
So, `P_(NH_(3)=P_(H_(2)S=(P)/(2)`
`K_(p)=P_(NH_(3))xxP_(H_(2)S=(P^(2))/(4)`.
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