For NH 4 HS ( s ) leftharpoons NH 3( g )+ H 2 S ( g ), If Kp=64 atm 2 equilibrium pressure of mixture is

Question

For NH 4 HS ( s ) leftharpoons NH 3( g )+ H 2 S ( g ), If Kp=64 atm 2 equilibrium pressure of mixture is (A) 8 atm (B) 16 atm (C) 64 atm (D) none of these

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/chemistry/40b8ef5e63eb81693ea0abae44f68aa5-.png" /> Total moles of gaseous substances =2 x x H 2 S =x NH 3=(1/2) ; p H 2 S =p NH 3=(P/2) Kp=p H 2 S p NH 3=(P2/4) or 64=(P2/4) or, P2=64 × 4=256 ∴ P=16 atm .

Q. For $N H_{4} H S (s) ⇌ N H_{3} (g) + H_{2} S (g),$ If $K_{p} = 64 a t m^{2}$ equilibrium pressure of mixture is

8 atm

16 atm

64 atm

none of these

Total moles of gaseous substances $= 2 x$

$x_{H_{2} S} = x_{N H_{3}} = \frac{1}{2};$

$p_{H_{2} S} = p_{N H_{3}} = \frac{P}{2}$

$K_{p} = p_{H_{2} S} p_{N H_{3}} = \frac{P ^{2}}{4}$ or $64 = \frac{P ^{2}}{4}$

or, $P^{2} = 64 \times 4 = 256$

$∴ P = 16$ atm .

Q. For NH4​HS(s)⇌NH3​(g)+H2​S(g), If Kp​=64atm2 equilibrium pressure of mixture is