Question

`NH_(4)COONH_(4)(s) hArr 2NH_(3)(g)+CO_(2)(g)`. If equilibrium pressure is `3` atm for the above reaction, `K_(p)` will be
A. 4
B. 27
C. `4//27`
D. `1//27`

Answer 1

Correct Answer - A
Suppose at equilibrium , `p_(CO_(2))=patm`.
Then `p_(NH_(3))=2p ` atm
We are given that `p+2p=3 ` atm
` :. P=1` atm `i.e., p_(CO_(2))=1` atm
`p_(NH_(3))=2` atm ltbr Hence `K_(p)=(p_(NH_(3)))^(2)xxp_(CO_(2))`
`=(2)^(2)xx1=4`