Answer:
Kp = 0.0146
Explanation:
We can solve this equilibrium problem using an ICE Table. There are 3 stages: I(Initial), C(Change) and E(Equilibrium). Only gases and aqueous species participate in the equilibrium.
NH₄CO₂NH₂ ⇄ 2 NH₃(g) + CO₂(g)
I 0 0
C +2x +x
E 2x x
Total gas pressure is equal to the sum of partial pressures at equilibrium.
pNH₃ + pCO₂ = 0.463 atm
2x + x = 0.463 atm
3 x = 0.463 atm
x = 0.154 atm
Then,
pNH₃ = 0.308 atm
pCO₂ = 0.154 atm
We can replace this data in the Kp expression.
[tex]Kp=pNH_{3}^{2} \times pCO_{2} =0.308^{2} \times 0.154 = 0.0146[/tex]
The equilibrium constant Kp = 0.0147
The balanced equation for the reversible decomposition of Ammonium carbonate:
NH4CO2NH2<------->2NH3(g) + CO2(g)
Total gas pressure of only the solid = 0.463 atm.
At a temperature of = 40°C
Kp = (pNH3)² (CO2)
Using the initial, change and equilibrium table:
Since it is initially only solid, partial pressure= 0
The expected change for NH3(g) and CO2(g) = +2x and +x respectively.
The expected equilibrium for NH3(g) and CO2(g) = 2x and x respectively.
The total gas pressure = pNH₃ + pCO₂
0.463 atm = 2x + x
0.463 atm = 3x
X = 0.463 / 3
X = 0.154atm
If x = 0.154atm, then partial pressure for Co2 = 0.154atm
Therefore, partial pressure for NH3 = 0.463 - 0.154
= 0.309atm
Substitute into the formula for Kp
Kp = (pNH3)² (pCO2)
Kp = 0.309² × 0.154
Kp = 0.095481 × 0.154
Kp = 0.0147
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