In the equilibrium NH4HSs - NH3g - H2Sg If the equilibrium pressure is 2 atm at 80-C Kp for the reaction is-

The correct option is C 1The equilibrium reaction is NH_4HS(s) ⇔ NH_3(g) + H_2S(g) .The expression for the equilibrium constant is K_p= P_NH_3× P_H_2 ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Chemistry

Derivation of Kp and Kc

In the equili...

Question

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 $80^{\circ} C$ $K_{p}$ for the reaction is;

0.5

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

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

1.5

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

Open in App

Solution

The correct option is C 1
The equilibrium reaction is $N H_{4} H S (s) \Leftrightarrow N H_{3} (g) + H_{2} S (g)$ .

The expression for the equilibrium constant is $K_{p} = \frac{P_{N H_{3}} \times P_{H_{2} S}}{P_{N H_{4} H S}} = P_{N H_{3}} \times P_{H_{2} S}$ as $P_{N H_{4} H S} = 1$ .

The mole fractions of $N H_{3} (g) a n d H_{2} S (g)$ are 0.5 each.

The partial pressures of $N H_{3} (g) a n d H_{2} S (g)$ are $0.5 \times 2 = 1 a t m$ each.

Substitute the values in the expression for the equilibrium constant.

$K_{P} = 1 \times 1 = 1$ .

So, the correct option is $C$

Suggest Corrections

Similar questions

Q. 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}

K_{p}

for the reaction is:

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

The equilibrium pressure at

25^{o} C

0.660

atm. What is

K_{p}

for the reaction?

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

. The equilibrium pressure at

25^{\circ} C

is 0.66 atm. What is the Kp of the reaction?

Q. Calculate the

K_{p}

for the equilibrium,

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

if the total pressure inside the reaction vessel is

1.12

atm at

105^{0}

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

What is the

K_{p}

of the reaction if the total pressure at equilibrium is 30 atm?

Join BYJU'S Learning Program

In the equilibrium NH4HS(s)⇔NH3(g)+H2S(g) If the equilibrium pressure is 2 atm at 80∘C Kp for the reaction is;

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 $80^{\circ} C$ $K_{p}$ for the reaction is;