For the reaction
`NH_(2)COONH_(4)(g)hArr 2NH_(3)(g)+CO_(2)(g)` the equilibrium constant `K_(p)=2.92xx10^(-5)atm^(3)`. The total pressure of the gaseous products when 1 mole of reactant is heated, will be

Options

A. 0.0194 atm
B. 0.0388 atm
C. 0.0582 atm
D. 0.0667 atm

Answer

C

Explanation

`{:(NH_(2)COONH_(4)(s),hArr,2NH_(3)(g),+,CO_(2)(g)),("Initial 0",,-,,-),("At eqm. 0,,2x,,x):}`
`K_(p)=(p_(NH_(3)))^(2)(p_(CO_(2)))=(2x)^(2)(x)=4x^(3)`
`4x^(3)=2.92xx10^(-5)`
`x^(3)=(2.92)/(4)xx10^(-5)=(29.2)/(4)xx10^(-6)`
`x=3sqrt(7.3)xx10^(-2)=1.94xx10^(-2)`
Total pressure of gaseous products `=3x`
`=3xx1.94xx10^(-2)=5.82xx10^(-2)atm`
`=0.0582atm`.