Question

Ammonium carbamate (NH2COONH4) is a salt of carbamic acid that is found in the blood and urine of mammals. At 240°C, Kc = 0.006 for the following equilibrium: If 6.01 g of NH2COONH4 is put into a 0.859-L evacuated container, what is the total pressure (in atm) at equilibrium? Enter a number to 2 decimal places.

Answer 1

Missing information:

Equilibrium reaction: NH2COONH4(s)⇌2NH3(g)+CO2(g)

Answer:

14.43 atm

Explanation:

The equilibrium occurs when in a reversible reaction, the velocity of the formation of the products is equal to the velocity of the formation of the reactants. When this occurs, the concentration and partial pressures remain constant. To characterize the equilibrium there's the equilibrium constant, which can be based on the concentration (Kc) or based on the gases partial pressure (Kp).

The conversion between them is:

Kp = Kc*(RT)⁻ⁿ

Where n is the variation of the coefficients of the gas substances (reactants - products) so, n = 0 - (2+1) = -3

Kp = Kc*(RT)³

R is the gas constant (0.082 atm.L/mol.K), and T is the temperature ( 240°C = 513 K), so:

Kp = 0.006*(0.082*513)³

Kp = 446.6

The value of Kp is also the ratio between the multiplication of the partial pressures of the products divided by the partial pressure of the reactants, each one elevated by the coefficient of the substance, and only for gas substances, so:

Kp = (pNH3)²*(pCO2)

Doing an equilibrium chart

NH3 CO2

0 0 Intial

+2x x Reacts (stoichiometry is 2:1)

2x x Equilibrium

446.6 = (2x)²*x

4x³ = 446.6

x³ = 111.65

x = ∛111.65

x = 4.81 atm

pNH3 = 2*4.81 = 9.62 atm

pCO2 = 4.81 atm

By Dalton's law, the total pressure of a gas mixture is the sum of the partial pressure of the substances:

P = 9.62 + 4.81 = 14.43 atm