For the decomposition reaction:
`NH_(2)COONH_(4(s))hArr2NH_(3(g))+CO_(2(g))`.
`(K_(p)=2.9xx10^(-5)atm^(2))`
The total pressure of gases at equilibrium when `1 "mole"` of `NH_(2)COONH_(4(s))` was taken to start with would be:

Options

A. `0.0194` atm
B. `0.0388` atm
C. `0.0582` atm
D. `0.0766` atm

Answer

C

Explanation

`{:(NH_(2)COONH_(4)(s),hArr,2NH_(3)(g),+,CO_(2)(g)),(1,,2,,1):}`
`K_(p)=2.9xx10^(-5)"atm"^(3)`
`K_(p) = 2.9xx10^(-5) atm^(3)`
If the P is the total pressure at equilibrium
`K_(p) = ((2p)/(3))^(2) ((p)/(3))`
`:. p^(3) = (27xx2.9xx10^(-5))/(4)= 1.9575`
`P = 3sqrt(1.9575) = 0.058`