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 ...

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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

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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 $C$

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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

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. The equilibrium pressure at

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

Q. Calculate the

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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;