Question

For the decomposition reaction \[\ce{NH2COONH4 (s) <=> 2NH3 (g) + CO2 (g)}\] the K_p = 2.9 × 10^-5 atm³. The total pressure of gases at equilibrium when 1 mol of \[\ce{NH2COONH4 (s)}\] was taken initially could be ______.

पर्याय

• 0.0194 atm 

• 0.0388 atm 

• 0.0582 atm

• 0.0766 atm

Answer 1

For the decomposition reaction \[\ce{NH2COONH4 (s) <=> 2NH3 (g) + CO2 (g)}\] the K_p = 2.9 × 10^-5 atm³. The total pressure of gases at equilibrium when 1 mol of \[\ce{NH2COONH4 (s)}\] was taken initially could be 0.0582 atm.

Explanation:

\[\ce{\underset{t = 0}{NH2}\underset{1}{COON}H4 (s) <=> \underset{0}{2NH3} (g) + \underset{0}{CO2} (g)}\]

at equation 1 - x      2x      x

`therefore "K"_"P" = ((2x)/(3x) xx "P")^2 xx (x/(3x) xx "P")`

(Active mass of solid is unity)

`"K"_"P" = ((4x^2)/(9x^2) xx "P"^2) xx (x/(3x) xx "P")`

`therefore "K"_"P" = [(4x^3)/(27x^3) xx "P"^3]`

`2.9 xx 10^-5 = (4x^3)/(27x^3) xx "P"^3`

78.3 × 10^-5 = 4 × P³

P³ = 1.97 × 10^-4

P = (197 × 10^-6)^1/3 

P = 0.0582 atm.