Question

When heated, ammonium carbamate decomposes as follows: NH4CO2NH2(s) ⇆ 2NH3(g) + CO2(g) .At a certain temperature the equilibrium pressure of the system is 0.318 atm. Calculate the KP for the reaction.

Answer 1

NH4CO2NH2(s) ⇆ 2NH3(g) + CO2(g)

For given reaction at equilibrium,

N$H_{4}$C$O_{2}$N$H_{2}$(s) ⇄ 2N$H_{3}$(g) + C$O_{2}$

As there only N$H_{3}$(g) and C$O_{2}$ are in gaseous form so only these gases would generated pressure at equilibrium.

Total moles of gases = 2+1 = 3

Mole fraction of N$H_{3}$(g) = $\frac{2}{3}$

Mole fraction of C$O_{2}$ = $\frac{1}{3}$

For given reaction,

$K_{p}$ = $P_{NH_{3} }$ × $P_{CO_{2} }$

                        = [ 0.318 × $\frac{2}{3}$ ] × [ 0.318 × $\frac{1}{3}$ ]

                        = 0.212 × 0.106

$K_{p}$ = 0.0225

Answer 2

p total = 3p. p= o. 318/3 =0.106 therefore kp=4p cube = 4.76*10power-3. MARK ME BRAINLIEST IF YOU LIKED AND UNDERSTOOD