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For the decomposition reaction NH(2)COON...

For the decomposition reaction `NH_(2)COONH_(4)(s) hArr2NH_(3)(g)+CO_(2)(g)`, the `K_(p)=3.2xx10^(-5)atm^(3)`. The total pressure of gases at equilibrium when `1.0` mol of `NH_(2)COONH_(4)(s)` was taken to start with will be:

A

`0.25atm`

B

`0.12atm`

C

`0.04atm`

D

`0.06atm`

Text Solution

Verified by Experts

The correct Answer is:
D
We have
`NH_(2)COONH_(4)(S)hArr2NH_(3)(g)+CO_(2)(g)hArr2NH_(3)(g)+CO_(2)(g)`
`K_(p)+(2p)^(2)p=4p^(3)=3.2xx10^(-5)atm^(3)`
This gives
`p=2xx10^(-2)atm`
`p_(total)=2p+p=3p=6xx10^(-2)atm`
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