For (NH)4HS(.s.)leftharpoons(NH)3(.g.)+H2S(.g.) , the observed pressure for reaction mixture in equilibrium is 1.12atm at 106° C . The value of Kp for the reaction is:

Question

For (NH)4HS(.s.)leftharpoons(NH)3(.g.)+H2S(.g.) , the observed pressure for reaction mixture in equilibrium is 1.12atm at 106° C . The value of Kp for the reaction is: (A) 3.136atm2 (B) 0.3136atm2 (C) 31.36atm2 (D) 6.98atm2

Tardigrade Admin · Accepted Answer

undersetPressure at equilibrium(NH)4 HS (. s .)leftharpoons undersetP(NH)3 (. g .)+ undersetPH2 S (. g .) ∴ Total pressure at equilibrium ⇒ 2P=1.12atm P=1.12/2atm ∵ Kp=P'NH3× P'H2 S ∴ Kp=(1 . 12/2)× (1 . 12/2)=0.3136atm2

Q. For $(N H)_{4} H S (s) ⇌ (N H)_{3} (g) + H_{2} S (g)$ , the observed pressure for reaction mixture in equilibrium is $1.12 a t m$ at $10 6^{\circ} C$ . The value of $K_{p}$ for the reaction is:

$3.136 a t m^{2}$

$0.3136 a t m^{2}$

$31.36 a t m^{2}$

$6.98 a t m^{2}$

$P ress u re a t e q u i l ib r i u m (N H)_{4} H S (s) ⇌ P (N H)_{3} (g) + P H_{2} S (g)$
$∴$ Total pressure at equilibrium $\Rightarrow 2 P = 1.12 a t m$
$P = 1.12/2 a t m$
$∵ K_{p} = P_{N H_{3}}^{'} \times P_{H_{2} S}^{'}$
$∴ K p = \frac{1.12}{2} \times \frac{1.12}{2} = 0.3136 a t m^{2}$

Q. For (NH)4​HS(s)⇌(NH)3​(g)+H2​S(g) , the observed pressure for reaction mixture in equilibrium is 1.12atm at 106∘C . The value of Kp​ for the reaction is:

3.136atm2

0.3136atm2

31.36atm2

6.98atm2