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The dissociation of ammonium hydrogen su...

The dissociation of ammonium hydrogen sulphide in a closed container produces a pressure of `10 atm` at `200^@C`. The value of `K_p` is

A

`25`

B

`50`

C

`100`

D

`75`

Text Solution

Verified by Experts

The correct Answer is:
A
`{:(,NH_(4)^(+)HS^(-)(s)hArrNH_(3)(g)+H_(2)S(g)),("Initial moles (assume)"," a 0 0"),("Change"," -x +x +x"),("Equilibrium moles",bar(" a-x x x")):}`
Total moles of gases `=x+x=2x`
(`NH_4HS` is a solid. Thus, its moles are not counted. )
`P_(NH_3)=`(Mole fraction `_NH_3`) (Total pressure)
`=(("Moles"_(NH_3))/("Total moles"))("Total pressure")`
`=((x)/(2x))*P`
Similarly,
`P_(H_2S)=((x)/(2x))*P`
`K_P=P_(NH_3)P_(H_2S)=((x)/(2x)*P)((x)/(2x)*P)`
`=(P/2)^2`
Since `P=10atm`, we have
`K_P=((10)/(2))^2=(5)^2=25`
Note that `NH_4HS` being a solid does not appear in `K_p` expression.
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