Solid ammonium carbamate dissociated according to the given reactionNH2COONH4(s) 2NH3(g) + CO2(g)If total pressure at equilibrium is 6 atm then Kp will

The given reaction is:

NH2COONH4(s) ⇌ 2NH3(g) + CO2(g)

According to the problem, the total pressure at equilibrium is 6 atm.

Since the reaction involves the dissociation of a solid into gases, we can assume that all of this pressure is due to the gases NH3 and CO2.

The stoichiometry of the reaction tells us that for every mole of NH2COONH4 that dissociates, 2 moles of NH3 and 1 mole of CO2 are produced. Therefore, the partial pressures of NH3 and CO2 at equilibrium are in the ratio 2:1.

Let's denote the partial pressure of CO2 as P(CO2) and the partial pressure of NH3 as P(NH3).

Since the total pressure is 6 atm, we can write:

P(CO2) + P(NH3) = 6 atm

And since the partial pressures are in the ratio 2:1, we can write:

P(NH3) = 2 * P(CO2)

Substituting the second equation into the first, we get:

P(CO2) + 2 * P(CO2) = 6 atm

Solving for P(CO2), we find:

P(CO2) = 2 atm

Substituting this back into the equation P(NH3) = 2 * P(CO2), we find:

P(NH3) = 4 atm

The expression for the equilibrium constant Kp for the reaction is:

Kp = [P(NH3)]^2 * [P(CO2)]

Substituting the values we found, we get:

Kp = (4 atm)^2 * (2 atm) = 32

So, the Kp for the reaction is 32.