On decomposition of NH 4 HS, the following equilibrium is established. NH 4 HS ( s ) leftharpoons NH 3( g )+ H 2 S ( g ) . If the total pressure is P atm, then the equilibrium constant K p will be equal to ( P 2/ x ) atm 2 where x is

Question

On decomposition of NH 4 HS, the following equilibrium is established. NH 4 HS ( s ) leftharpoons NH 3( g )+ H 2 S ( g ) . If the total pressure is P atm, then the equilibrium constant K p will be equal to ( P 2/ x ) atm 2 where x is  (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

NH 4 HS ( s ) leftharpoons NH 3(1 mol ) + H 2 S (1 mol ) K P =( P NH 3) ⋅( P H 2 S ) =( P /2) × ( P /2)=( P 2/4) atm 2

Q. On decomposition of $N H_{4} H S$ , the following equilibrium is established. $N H_{4} H S (s) ⇌ N H_{3} (g) + H_{2} S (g) .$ If the total pressure is $P$ atm, then the equilibrium constant $K_{p}$ will be equal to $\frac{P ^{2}}{x}$ atm $^{2}$ where $x$ is __

$N H_{4} H S (s) ⇌ N H_{3} (1 m o l) + H_{2} S (1 m o l)$
$K_{P} = (P_{N H_{3}}) \cdot (P_{H_{2} S})$
$= \frac{P}{2} \times \frac{P}{2} = \frac{P ^{2}}{4} a t m^{2}$

Q. On decomposition of NH4​HS, the following equilibrium is established. NH4​HS(s)⇌NH3​(g)+H2​S(g). If the total pressure is P atm, then the equilibrium constant Kp​ will be equal to xP2​ atm 2 where x is __

NH4​HS(s)⇌NH3​(1mol)+H2​S(1mol) KP​=(PNH3​​)⋅(PH2​S​) =2P​×2P​=4P2​atm2

Q. On decomposition of $N H_{4} H S$ , the following equilibrium is established. $N H_{4} H S (s) ⇌ N H_{3} (g) + H_{2} S (g) .$ If the total pressure is $P$ atm, then the equilibrium constant $K_{p}$ will be equal to $\frac{P ^{2}}{x}$ atm $^{2}$ where $x$ is __

$N H_{4} H S (s) ⇌ N H_{3} (1 m o l) + H_{2} S (1 m o l)$
$K_{P} = (P_{N H_{3}}) \cdot (P_{H_{2} S})$
$= \frac{P}{2} \times \frac{P}{2} = \frac{P ^{2}}{4} a t m^{2}$