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Standard XII

Chemistry

Dalton's Law of Partial Pressures

Ammonium hydr...

Question

Ammonium hydrogen sulphide dissociates according to the equation $N H_{4} H S (s) ⇌ N H_{3} (g) + H_{2} S (s)$ .
The total pressure at equilibrium at $400 K$ is found to be $1 a t m$
The equilibrium constant $K_{p}$ of the reaction is:

1 a t m^{2}

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0.25 a t m^{2}

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0.5 a t m

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0.25 a t m

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Solution

The correct option is B $0.25 a t m^{2}$
Answer is option (B)
GIVEN:
Total pressure of ammonia + hydrogen sulfide = $1 a t m$
Therefore, pressure of ammonia=pressure of hydrogen sulphide= $= \frac{1}{2} = 0.5 a t m$
The values of $K_{P}$ can be calculated as,
$K_{P} = ({N H}_{3} a t m) (H_{2} S a t m)$
$K_{P} = 0.25 {a t m}^{2}$

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Ammonium Hydrogen sulphide dissociates according to the equation,

$N H_{4} H S$ (s) $⇋$ $N H_{3}$ (g) + $H_{2} S$ (g) if the observed pressure of the mixture is 1.12 atm at $106^{\circ}$ C. What is the KP of the reaction -

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Ammonium hydrogen sulphide dissociates according to the equation NH4HS(s)⇌NH3(g)+H2S(s).The total pressure at equilibrium at 400 K is found to be 1 atmThe equilibrium constant Kp of the reaction is:

The correct option is B 0.25 atm2Answer is option (B)GIVEN:Total pressure of ammonia + hydrogen sulfide = 1atmTherefore, pressure of ammonia=pressure of hydrogen sulphide= =12=0.5atmThe values of KP can be calculated as,KP=(NH3atm)(H2Satm) KP=0.25atm2

Ammonium hydrogen sulphide dissociates according to the equation $N H_{4} H S (s) ⇌ N H_{3} (g) + H_{2} S (s)$ .
The total pressure at equilibrium at $400 K$ is found to be $1 a t m$
The equilibrium constant $K_{p}$ of the reaction is: