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 ...

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

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 $N = 14 g m o l^{- 1} .$

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

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

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

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

$0.242 a t m^{2}$

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

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

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

Open in App

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,

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

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

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

Suggest Corrections

152

Similar questions

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

. 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}

Q. The osmotic pressure of solution containing 34.2 g of cane sugar (molar mass = 342 g

m o l^{- 1}

) in 1L of solution at

20^{o} C

is :
[Given. R = 0.082 L atm

K^{- 1} m o l^{- 1}

]

Q. In a closed cylinder of capacity

24.6 L

, the following reaction occurs at

27^{\circ} C

A_{2} (s) ⇌ B_{2} (s) + 2 C (g)

At equilibrium,

1 g

B_{2} (s)

(molar mass

= 50 g m o l^{- 1})

is present. The equilibrium constant

K_{p}

for the equilibrium in

a t m^{2}

unit is:

[R = 0.082 L a t m K^{- 1} m o l^{- 1}]

Q. The equilibrium constant

K_{p}

( in atm) for the reaction is

2.46

7

atm and

300 K

A_{2} (g) ⇌ B_{2} (g) + C_{2} (g)

Calculate the average molar mass (in g/mol) of an equilibrium mixture.
Given: Molar mass of

A_{2}, B_{2}

and

C_{2}

are

70, 49

and

21 g/mol

respectively.
(Given

R = 0.082 {L atm K}^{- 1} {mol}^{- 1}, \sqrt{0.1} \approx 0.3

Q. A solution is prepared by dissolving

0.6 g

of urea (molar mass =

60 g m o l^{- 1}

) and

1.8 g

of glucose (molar mass =

180 g m o l^{- 1}

) in

100 m L

of water at

27^{0} C

. The osmotic pressure of the solution is :

(R = 0.08206 L a t m K^{- 1} m o l^{- 1})

Join BYJU'S Learning Program

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.