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

`NH_(4)COONH_(2(s)) hArr 2NH_(3(g)) + CO_(2(g))`. If equilibrium pressure is 6 atm for the above reaction : `K_(p)` will be :

A

32

B

27

C

`4//27`

D

`1//27`

Text Solution

Verified by Experts

The correct Answer is:
A
`P_(NH_(3)) : P_(CO_(2)) = 2:1 therefore p_(NH_(3)) = 2` atm, `p_(CO_(2))` = atm `rArr K_(p) = [p_(NH_(3))]^(2)[p_(CO_(2))] = 2^(2) xx 1 = 4`
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