Ammonium carbonate (solid) decomposes when heated to produce three gaseous products - ammonia, water, and carbon dioxide.

1. Write the balanced equation for the reaction.
2. Suppose that ammonium carbonate is heated to 500 K in a sealed vessel. At equilibrium there is still some solid in the vessel, and the partial pressure of the ammonia gas is 4.00 atm. Find both the Kp and the Kc of the reaction at 500 K.

3. It is found that the mass loss of the ammonium carbonate on being heated is 1.000 gram. What is the volume of the sealed vessel?

(NH4)2CO3(s) ==> 2NH3(g) + H2O(g) + CO2(g)

equilibrium:
NH3 = 4.00 atm
Therefore, CO2 = 2.00 atm
and H2O = 2.00 atm
Ptotal = 4.00 + 2.00 + 2.00 = 8.00 atm.

Kp =PNH3^2*PCO2*PH2O = (4)^2*2*2 = 64
Kc = Kp(RT)-delta n
= 64(0.08206*773)-4 = 3.95E-6

3.
moles (NH4)2CO3 = 1.000 g/molar mass = 0.01041. Therefore, moles NH3 = 0.02082, moles H2O = 0.01041, moles CO2 = 0.01041 and total moles = 0.04163.
Use PV = nRT to calculate volume.
8*V = 0.04163*0.08206*773, solve for V.

1. The balanced equation for the decomposition reaction of ammonium carbonate is:

(NH4)2CO3(s) → 2NH3(g) + H2O(g) + CO2(g)

2. To find the Kp and Kc values at 500 K, we need to determine the equilibrium concentrations of the gases. The given information states that there is still some solid ammonium carbonate present at equilibrium, indicating that the reaction has not gone to completion.

Let's assume that x moles of (NH4)2CO3 decomposes. According to stoichiometry, 2x moles of NH3 and CO2 will be formed, and x moles of H2O will be formed.

The initial concentration of NH3, H2O, and CO2 is 0, as they are all products of the reaction.

The total pressure at equilibrium is the sum of the individual partial pressures of the gases. Given that the partial pressure of NH3 is 4.00 atm, we can express the equilibrium molar concentration as [NH3]=4/(RT/V), where R is the ideal gas constant, T is the temperature in Kelvin (500K), and V is the volume of the vessel in liters.

Next, we need to determine the concentration of H2O and CO2 in equilibrium. Since their coefficients are equal to x and 2x, respectively, we can write their concentration as [H2O] = x/(RT/V) and [CO2] = 2x/(RT/V).

Now, we can substitute these concentrations into the equilibrium expression:

Kp = P(NH3)^2 * P(H2O) * P(CO2)
= [NH3]^2 * [H2O] * [CO2]
= (4/(RT/V))^2 * (x/(RT/V)) * (2x/(RT/V))
= 16 * x^3 / (R^3 * T^3 / V^3)

Kc is the equilibrium constant in terms of concentration. Since pressure is directly proportional to concentration for ideal gases, we can say that Kc and Kp have the same value.

Therefore, Kp = Kc = 16 * x^3 / (R^3 * T^3 / V^3)

3. To find the volume of the sealed vessel, we need to use the mass loss of the ammonium carbonate. The molar mass of (NH4)2CO3 is approximately 96 g/mol, so the number of moles of ammonium carbonate decomposed can be calculated as:

moles of (NH4)2CO3 = mass loss / molar mass
= 1.000 g / 96 g/mol

Since 2 moles of NH3 are produced for every mole of (NH4)2CO3 decomposed, the number of moles of NH3 can be calculated as:

moles of NH3 = 2 * moles of (NH4)2CO3
= 2 * (1.000 g / 96 g/mol)

Finally, we can use the ideal gas law to find the volume of the sealed vessel. Assuming ideal gas behavior, we can use the equation:

PV = nRT

Since we already have the number of moles (moles of NH3), we can rearrange the equation to solve for V:

V = (nRT) / P
= [(2 * moles of NH3) * (R) * (T)] / P

Substituting the appropriate values and converting to liters gives the volume of the sealed vessel.

1. To write the balanced equation for the reaction, we need to know the chemical formula for ammonium carbonate and the products of its decomposition.

The chemical formula for ammonium carbonate is (NH4)2CO3.

The products of its decomposition are ammonia (NH3), water (H2O), and carbon dioxide (CO2).

The balanced equation for the reaction can be written as:

(NH4)2CO3(s) → 2NH3(g) + H2O(g) + CO2(g)

2. To find the Kp and Kc of the reaction at 500 K, we need to use the given information of the partial pressure of ammonia gas (4.00 atm) and the fact that there is still some solid in the vessel at equilibrium.

Kp is the equilibrium constant expressed in terms of partial pressures, while Kc is the equilibrium constant expressed in terms of molar concentrations.

First, we need to determine the initial moles of ammonia gas (NH3) produced from the given mass loss of ammonium carbonate (1.000 gram).

To do that, we need to calculate the number of moles of ammonium carbonate that decomposes. The molar mass of ammonium carbonate is 96.09 g/mol.

Moles of (NH4)2CO3 = mass / molar mass = 1.000 g / 96.09 g/mol

Next, we use the balanced equation to determine the moles of ammonia gas produced. According to the equation, for every 1 mole of ammonium carbonate that decomposes, 2 moles of ammonia gas are produced.

Moles of NH3 = (2 * moles of (NH4)2CO3)

Now we can calculate the concentration of ammonia gas (NH3) in moles per liter:

Concentration of NH3 = moles of NH3 / volume of vessel

Since the vessel is sealed, the volume of the vessel remains constant. Therefore, the molar concentration of NH3 is constant as well.

Now we can calculate the equilibrium concentration of ammonia gas (NH3), which is also equal to the value of Kc:

Kc = [NH3]^2 / [NH4)2CO3]

Substitute the calculated concentration of NH3 and the value for partial pressure of ammonia (4.00 atm) to find Kc.

To calculate Kp, we use the ideal gas law:

PV = nRT

Rearrange the formula to solve for the number of moles:

n = PV / RT

Since we have the value for pressure (4.00 atm), the volume of the vessel (which we need to calculate), the gas constant R (0.0821 L atm/(mol K)), and the temperature (500 K), we can calculate the number of moles of ammonia gas (NH3).

Finally, we substitute the values for the moles of ammonia gas and the moles of ammonium carbonate into the equation for Kp:

Kp = (P(NH3))^2 / ((P(NH4)2CO3)^1)

3. To calculate the volume of the sealed vessel, we need to use the ideal gas law equation:

PV = nRT

Since the vessel is sealed, the pressure inside the vessel remains constant at the given value (4.00 atm). The number of moles of ammonia gas (NH3) calculated in the previous step is also known. The value of the gas constant R is 0.0821 L atm/(mol K), and the temperature is given as 500 K.

Rearranging the formula to solve for volume, we have:

V = (nRT) / P

Substitute the values for the moles of ammonia gas, the gas constant R, the temperature, and the given pressure to find the volume of the sealed vessel.