Solid ammonium carbamate dissociated according to the given reaction NH 2 COONH 4( s ) leftharpoons 2 NH 3( g )+ CO 2( g ) Total pressure of the gases in equilibrium is 5 atm. Hence K p

Question

Solid ammonium carbamate dissociated according to the given reaction NH 2 COONH 4( s ) leftharpoons 2 NH 3( g )+ CO 2( g ) Total pressure of the gases in equilibrium is 5 atm. Hence K p  (A) 18.5 (B) 25 (C) 1/ 5 (D) 12.5

Tardigrade Admin · Accepted Answer

NH2COONH4(s) leftharpoons underset2 P2NH3 undersetPCO2 Total pressure =2 P+P=3 P=5 a t m ⇒ P=(5/3) a t m KP=PC O2 × PN H32 =(5/3) atm ×((2 × 5/3) atm )=18.51

Q. Solid ammonium carbamate dissociated according to the given reaction
$N H_{2} COON H_{4} (s) ⇌ 2 N H_{3} (g) + C O_{2} (g)$
Total pressure of the gases in equilibrium is $5 a t m$ . Hence $K_{p}$

18.5

25

1/ 5

12.5

$N H_{2} COON H_{4} (s) 2 P 2 N H_{3} P C O_{2}$
Total pressure $= 2 P + P = 3 P = 5 a t m$
$\Rightarrow P = \frac{5}{3} a t m$
$K_{P} = P_{C O_{2}} \times P_{N H_{3}}^{2}$
$= \frac{5}{3} a t m \times (\frac{2 \times 5}{3} a t m) = 18.51$

Q. Solid ammonium carbamate dissociated according to the given reaction NH2​COONH4​(s)⇌2NH3​(g)+CO2​(g) Total pressure of the gases in equilibrium is 5atm. Hence Kp​

18.5

25

1/ 5

12.5

NH2​COONH4​(s)​2P2NH3​​PCO2​​ Total pressure =2P+P=3P=5atm ⇒P=35​atm KP​=PCO2​​×PNH3​2​ =35​atm×(32×5​atm)=18.51

Q. Solid ammonium carbamate dissociated according to the given reaction
$N H_{2} COON H_{4} (s) ⇌ 2 N H_{3} (g) + C O_{2} (g)$
Total pressure of the gases in equilibrium is $5 a t m$ . Hence $K_{p}$

$N H_{2} COON H_{4} (s) 2 P 2 N H_{3} P C O_{2}$
Total pressure $= 2 P + P = 3 P = 5 a t m$
$\Rightarrow P = \frac{5}{3} a t m$
$K_{P} = P_{C O_{2}} \times P_{N H_{3}}^{2}$
$= \frac{5}{3} a t m \times (\frac{2 \times 5}{3} a t m) = 18.51$