$5.1$ g $N H_{4} S H$ is introduced in 3.0 L evacuated flask at $327^{0} C$ . 30% of the solid $N H_{4} S H$ decomposed to $N H_{3} a n d H_{2} S$ as gases. The $K_{p}$ of the reaction at $327^{0} C$ . is $(R = 0.082 L a t m m o l^{- 1}$ ,molar mass of $S = 32 g m o l /^{01}$ , molar mass of $N = 14 g m o l^{- 1}$ .

1 \times 10^{- 4} a t m^{2}

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4.9 \times 10^{- 3} a t m^{2}

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0.242 a t m^{2}

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0.242 \times 10^{- 4} a t m^{2}

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Solution

The correct option is C $0.242 a t m^{2}$
$N H_{4} S H (s) ⇌ N H_{3} (g) + H_{2} S (g)$
$n = \frac{5.1}{51} = .1 m o l e 0$ 0
$.1 (- 1 - α)$ $.1 α$ $.1 α$
$α = 30 % = .3$
So number of moles equilibrium
$.1 (1 - .3) .1 \times .3 .1 \times .3$
= .07 =.03 =03
Now use $P V = n R T$ at equlibrium
$P_{t o t a l} \times 3 l i t = (.03 + .03) \times .082 \times 600$
$P_{t o t a l} = .984 a t m$
At equilibrium
$P_{N H_{3}} = P_{H_{2} S} = \frac{P_{t o t a l}}{2} = .492$
So $k_{p} = P_{N H_{3}}, P H_{2} S = (. .492 (.492)$
$k_{p} = .242 a t m^{2}$

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Similar questions

$5.1 g N H_{4} S H$ is introduced in 3.0 L evacuated flask at $327^{o} C$ . 30% of the solid $N H_{4} S H$ decomposed to $N H_{3} a n d H_{2} S$ as gases. The $K_{P}$ of the reaction at $327^{o} C$ is
$(R = 0.082 L a t m m o l^{- 1}$ ,
molar mass of $S = 32 g m o l^{- 1}$ ,
molar mass $N = 14 g m o l^{- 1} .$

When 3.06 g of solid $N H_{4}$ HS is introduced into a two-litre evacuated flask at 27 $^{o} C$ , 30 $%$ of the solid decomposes into gaseous ammonia and hydrogen sulphide. Choose the most appropriate one regarding $K_{c}$ , $K_{p}$ and degree of dissociation $α$ :

I_{2}

gets

5 %

dissociated into atoms when1.0 mole of

I_{2} (g)

is introduced into an evacuated flask of 1 L capacity at 150K.The $K_{p}$ for the reaction $I_{2} (g) ⇌ 2 I (g)$ at 150K is:

Q. When

3.06 g

{N H}_{4} H S (s)

is introduced into a

t w o l i t r e

evacuated flask at

27^{o} C

30

% of the solid decomposes into gaseous ammonia and hydrogen sulphide.
(i) Calculate

K_{c}

and

K_{p}

for the reaction at

27^{o} C

.
(ii) What would happen to the equilibrium when more

{N H}_{4} H S (s)

is introduced into the flask?

Equal volumes of two gases A and B diffuse through a porous pot in 20s and 10s, respectively .If the molar mass of A be $80 g m o l^{- 1}$ ,Find the molar mass of B.

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5.1g NH4SH is introduced in 3.0 L evacuated flask at 3270C. 30% of the solid NH4SH decomposed to NH3andH2S as gases. The Kp of the reaction at 3270C. is(R=0.082 L atm mol−1,molar mass of S=32gmol/01, molar mass of N=14gmol−1.