JEE Main 27-Aug-2021 Shift 2 Solved Paper

51g of NH4‌HS=1‌mol
5.1g of NH4‌HS=

1

51

×5.1=0.1‌mol
NH4‌HS(s)⇌NH3(g)+H2S(g)

At‌t=0,	‌	‌	0.1
At‌t=t,	‌	‌	0.1(1−α)	‌	‌	0.1α	‌	‌	0.1α

It is given that dissociation is 20% from 100 moles.
∴ 20 moles get dissociated.
20% dissociation from 1‌mol‌

20

100

=0.2 moles get dissociated.
α = 0.2
∵ KC=

[NH3][H2S]

[NH4‌HS]

=

[NH3][H2S]

1

(Concentration of solid is assumed as 1)
[NH3]=

Number‌of‌moles‌of‌NH3(g)

Volume(in‌L)

=

0.1α

2

[H2S]=

Number‌of‌moles‌of‌H2‌S

Volume(in‌L)

=

0.1α

2

KC=0.1α2×

0.1α

2

=

0.1×0.2

2

×

0.1×0.2

2

=10−4
∴Kp=KC(RT)Δn
[Δn =change in the number of gaseous moles = 2]
Kp=10−4×(0.08×300)2
Kp=10−4×(24)2=0.06
x×10−2=0.06
⇒x=

0.06

10−2

=0.06×102=6
x=6 is the answer.