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The equilibrium pressure of NH4CN((s)...

The equilibrium pressure of `NH_4CN_((s)) harr NH_(3(g))+HCN_((g))` is 4 atm. In an experiment, if `NH_4CN_((s))` is allowed to decompose in presence of `NH_3` at 3 atm, then

A

pressure of `HCN` = 2atm (at old equilibrium state)

B

pressure of `HCN` = 1 atm (at new equilibrium state)

C

`K_p = 4`

D

pressure of `NH_3` = 4 atm (at new equilibrium state)

Text Solution

Verified by Experts

The correct Answer is:
A, B, C, D
Old equilibrium, `NH_(4)CN_((s)) harr NH_(3) + HCN` `K_(P) = P^(2) = 4 implies P = 2`atm
New equilibrium `NH_(4)CN_((s)) harr underset(3+P.)(NH_(3))+underset(P.)(HCN)`
`(3+P.)P. = K_(P) = 4 implies` P. =1atm
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