5-1 g NH4SH is introduced in 3-0 L evacuated flask at 327oC- 30R - 0-082 L atm mol-1-molar mass of S-32 g mol-1-molar mass N- 14 g mol-1-

Correction option is C) Moles of nbsp;NH_4SH=5.1/51=0.1 mol NH_4SH(s)⇌ NH_3(g)+H_2S(g) Initially 0.1 0 0 At eq ...

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Byju's Answer

Standard XII

Chemistry

Avogadro's Law & Gay Lussac's Law

5.1 g NH4SH i...

Question

$5.1 g N H_{4} S H$ 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 $N = 14 g m o l^{- 1} .$

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

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$4.9 \times 10^{- 4} a t m^{2}$

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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
$0.242 a t m^{2}$

Correction option is C)
Moles of $N H_{4} S H = \frac{5.1}{51} = 0.1 m o l$
$N H_{4} S H (s) ⇌ N H_{3} (g) + H_{2} S (g)$

$I n i t i a l l y 0.1 00$

$A t e q u i l 0.1 (1 - α) 0.1 α 0.1 α$

$α = 30 % = 0.3$

So, number of moles equilibrium due to gaseous product

$= 0.1 \times 0.3 + 0.1 \times 0.3$
$= 0.03 + 0.03$

Now use PV = nRT
At equlibrium,

$P_{t o t a l} \times 3 l i t = (0.03 + 0.03) \times .082 \times 600$

$P_{t o t a l} = 0.984 a t m$

At equilibrium,

$P_{N H_{3}} = P_{H_{2} S} = \frac{P_{t o t a l}}{2} = 0.492$

$K_{p} = P_{N H_{3}} \times P_{H_{2} S} = 0.492 \times 0.492$

$K_{p} = 0.242 a t m^{2}$

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5.1

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

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(R = 0.082 L a t m m o l^{- 1}

,molar mass of

S = 32 g m o l /^{01}

, molar mass of

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5.1 g NH4SH is introduced in 3.0 L evacuated flask at 327oC. 30% of the solidNH4SH decomposed to NH3 and H2S as gases. The KP of the reaction at 327oC is (R=0.082 L atm mol−1, molar mass of S=32 g mol−1, molar mass N=14 g mol−1.