Ammonium carbamate decomposes as `NH_(2)COONH_(4)(s) hArr 2NH_(3)(g) +CO_(2)(g)` . The value of `K_(p)` for the reaction is `2.9 xx 10^(-5) atm^(3)`.

Question

Ammonium carbamate decomposes as `NH_(2)COONH_(4)(s) hArr 2NH_(3)(g) +CO_(2)(g)` . The value of `K_(p)` for the reaction is `2.9 xx 10^(-5) atm^(3)`.

Ammonium carbamate decomposes as `NH_(2)COONH_(4)(s) hArr 2NH_(3)(g) +CO_(2)(g)` . The value of `K_(p)` for the reaction is `2.9 xx 10^(-5) atm^(3)`. If we start the reaction with 1 mole of the compound, the total pressure at equilibrium would be
A. 0.0766 atm
B. 0.0194 atm
C. 0.194 atm
D. 0.0582 atm

Monjur Alahi

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2025-01-04 04:42

1 Answers

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Salim Ahmed Kawsar · Accepted Answer · 2023-02-05T15:29:48+06:00

Salim Ahmed Kawsar

Correct Answer - D
`NH_(2)COONH_(4)(s) hArR 2NH_(3)(g)+CO_(2)(g)`

If partial pressure of `CO_(2)` at equilibrium is p
Then partial pressure of `NH_(3)` is 2p
`K_(p)=(p_(NH_(3)))^(2)xx(p_(CO_(2)))=(2p)^(2)(p)=4p^(3)`
Now `4p^(3)=2.9 xx 10^(-5) `or `p=1.935xx10^(-2)`
`=5.81 xx 10^(-2) atm `

4 views Answered 2023-02-05 15:29