(NH4)2CO3 <-> 2NH3 + CO2 + H2O

solid (NH4)2CO3 is placed in an evacuated container at 115C. When equilibrium is reached Kp = 1.93x10^-3. Extra NH3 gas is added, when equilibrium is reestablished, the partial pressure of NH3 is four times the partial pressure of H20. What is the partial pressure of NH3?

I believe this problem is solved this way.

........(NH4)2CO3 ==> 2NH3 + CO2 + H2O
If H2O is p, then CO2 is p and NH3 is 4p.

Substitute into Kp expression and solve for p and 4p.

To find the partial pressure of NH3, we need to calculate the equilibrium partial pressures of NH3, CO2, and H2O. We are given that the equilibrium constant (Kp) for the reaction is 1.93x10^-3.

Step 1: Write the balanced equation:
(NH4)2CO3 ↔ 2NH3 + CO2 + H2O

Step 2: Define the initial and equilibrium conditions:
Since the container is initially evacuated, the initial pressures of NH3, CO2, and H2O are all zero. At equilibrium, the partial pressure of NH3 is four times the partial pressure of H2O.

Let x be the partial pressure of H2O.
The partial pressure of NH3 is given as 4x.

Step 3: Write the equilibrium expression using partial pressures:
Kp = [(P(NH3))^2 × P(CO2) × P(H2O)] / [(P(NH4)2CO3)]

Step 4: Use the given equilibrium constant (Kp) to set up the equation:
1.93x10^-3 = [(4x)^2 × P(CO2) × x] / [(P(NH4)2CO3)²]

Step 5: Solve for the partial pressure of NH3:
Simplify the equation by substituting (NH4)2CO3 with a constant value, let's say 1:
1.93x10^-3 = [(4x)^2 × P(CO2) × x] / 1

1.93x10^-3 = (16x^3 × P(CO2)) / 1

Cross-multiply:
1.93x10^-3 = 16x^3 × P(CO2)

Divide both sides by 16x^3 × P(CO2):
1.93x10^-3 / (16x^3 × P(CO2)) = 1

Simplify:
1.93x10^-3 / 16P(CO2) = x^3

Since we are only interested in the partial pressure of NH3, substitute its value:
4x = (1.93x10^-3 / 16P(CO2))^(1/3)

Now, plug in the value of P(CO2) and solve for x to find the partial pressure of H2O:

However, the value of P(CO2) was not provided in the question. Please provide the required value of P(CO2) to continue with the calculations.