$N H_{4} H S (s) ⇌ N H_{3} (g) + H_{2} S (g)$
The equilibrium pressure at $25^{o} C$ is $0.660 a t m$ . What is $K_{p}$ for the reaction?

0.109

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

0.218

No worries! We‘ve got your back. Try BYJU‘S free classes today!

1.89

No worries! We‘ve got your back. Try BYJU‘S free classes today!

2.18

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A $0.109$
$K_{p}$ = 0.11 $A t m^{2}$
The solid reactant has zero partial pressure
So $K_{p}$ = $P_{N}$ $_{H}$ $_{3}$ . $P_{H}$ $_{2}$ $_{S}$
Here P denotes the partial pressure of each gas.
$P_{N}$ $_{H}$ $_{3}$ + $P_{H}$ $_{2}$ $_{S}$ =0.660 atm
$P_{N}$ $_{H}$ $_{3}$ = 0.33 atm
$P_{H}$ $_{2}$ $_{S}$ = 0.33 atm
So now they are in 1 : 1 ratio
So $K_{p}$ = 0.109 $a t m^{2}$

Suggest Corrections

Similar questions

{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)

4 {b a r}^{2}

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

30 a t m

K_{p}

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

K_{p}

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

80^{\circ} C

K_{p}

Join BYJU'S Learning Program