In the equilibrium $N H_{4} H S (s) \Leftrightarrow N H_{3} (g) + H_{2} S (g)$ . If the equilibrium pressure is $2$ atm at $80^{o}$ C $K_{p}$ for the reaction is:

Open in App

Solution

$N H_{4} H S (s) ⇌ N H_{3} (g) + H_{2} S (g)$

$P_{e} q . = 2 a t m$

$K_{p} = P_{N H_{3}} . P_{H_{2} S}$

(1) One mole of $N H_{4} H S$ decompose in 1 mole of $N H_{3}$ and 1 mole of $H_{2} S$
So both the gases have equal partial pressure
$= \frac{2}{2} = 1 a t m$
$K_{p} = 1.1$
$K_{p} = 1 a t m^{2}$

Suggest Corrections

Similar questions

N H_{4} H S (s) \Leftrightarrow N H_{3} (g) + H_{2} S (g)

80^{\circ} C

K_{p}

N H_{4} H S (s) ⇌ N H_{3} (G) + H_{2} S (g)

25^{o} C

0.660

K_{p}

{N H}_{4} H S (s) ⇌ {N H}_{3} (g) + H_{2} S (g)

25^{\circ} C

K_{p}

N H_{4} H S_{(} s) ⇌ N H_{3 (g)} + H_{2} S_{(} g)

1.12

105^{0}

N H_{4} H S (s) ⇌ N H_{3} (g) + H_{2} S (g)

K_{p}

Join BYJU'S Learning Program