Verified Solution • 2025–26 Session moderate

QJEEM19Ch70f2: JEE Main 2019 Chemistry - Class 11 JEEM19Ch70f2 Solution: 5.1 g NH 4 SH is introduced in 3.0 L evacuated flask at 327ºC.…

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

Multiple Choice Options

  • A 0.242 ×\times 10 −- 4 atm2
  • B 1 ×\times 10–4 atm2
  • C 4.9 ×\times 10 −- 3 atm2
  • D 0.242 atm2

Step-by-Step Verified Solution

   NH4SH(s)   ⇌\rightleftharpoons    NH3(g) + H2S(g)

n=5.151=.1 mole        0                        0  n = {{5.1} \over {51}} = .1\,mole\,\,\,\,\,\,\,\,0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0\,\,

.1(−1−α)                    .1α                    .1α.1\left( { - 1 - \alpha } \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,.1\alpha \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,.1\alpha

α  =  30%=.3\alpha \,\, = \,\,30\% = .3

so number of moles at equilibrium

       .1 (1−.3)     .1  ×  .3         .1  × .3\,\,\,\,\,\,\,.1\,(1 - .3)\,\,\,\,\,.1\,\, \times \,\,.3\,\,\,\,\,\,\,\,\,.1\,\, \times \,.3

=    .07                =.03           =.03= \,\,\,\,.07\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = .03\;\,\,\,\,\,\,\,\,\, = .03

Now use PV = nRT at equilibrium

Ptotal ×\times 3 lit = (.03 + .03) ×\times .082 ×\times 600

Ptotal = .984 atm

At equilibrium

PNH3 = PH2S = Ptotal2{{{P_{total}}} \over 2} = .492

So kp = PNH3 . PH2S = (.492) (.492)

kp = .242 atm2

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