4. Armonium carbonate decomposes as
NH,COONH,(S) = 2NH,(g) + CO2(g)
For the reacion, K = 2.97.10 atm”. If we start with i
mole of the compound, the total pressure at equilibrium
would be
(1) 0.766 atm
(2) 0.0582 atm
(3) 0.0388 atm
(4) 0.0194 atm
Answers
The total pressure at equilibrium would be P = 0.0582 atm
Explanation:
Eq constant = kp = 2.9 x 10^-5 atm^3
NH2COONH4 (S) ⇆ 2NH3 (g) + CO2(g)
Initial 1 mole 0 0
At eq (1-x) 2x x
Total moles of gaseous substance at eq = 3x
Equilibrium pressure = P
Mole fraction of NH3 - x NH3 = 2/3 P
Mole fraction of CO2 = xCO2 = P/3
We know that
Equilibrium constant Kp = (xNH3)^2 (xCo2)
= 2.9 x 10^-5 = (2/3P)^2(P/3)
= 4/2xp^3
P = 0.0582 atm
Thus the total pressure at equilibrium would be P = 0.0582 atm
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