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NH(4)COONH(4)(s) hArr 2NH(3)(g)+CO(2)(g)...

`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`

Text Solution

Verified by Experts

The correct Answer is:
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`
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