Home
Class 11
CHEMISTRY
5.1g of solid NH(4)HS is introduced in a...

`5.1g` of solid `NH_(4)HS` is introduced in a `16.4` lit. vessel & heated upto `500 K` `K_(B)` for equilibrium `NH_(4)HS(s) hArr NH_(3)(g)+H_(2)S(g)` is `0.16`. The maximum pressure developed in the vessel will be `:`

A

`0.8 atm`

B

`0.40 atm`

C

`0.5 atm`

D

None of these

Text Solution

Verified by Experts

The correct Answer is:
3

`{:(,NH_(4)HS(s),hArr,NH_(3)(g),+,H_(2)S(g),),(t=0,0.1"mole",,,,,),("for max. pressure",-,,0.1,0.1,,0.20"gaseous mole"),(Pxx16.4=0.2xx0.82xx500,,rArr,,P=0.5,,):}`
Promotional Banner

Topper's Solved these Questions

  • CHEMICAL EQUILIBRIUM

    RESONANCE|Exercise Solved examples|9 Videos
  • CHEMICAL EQUILIBRIUM

    RESONANCE|Exercise (MSPs)|8 Videos
  • CHEMICAL BONDING

    RESONANCE|Exercise Inorganic chemistry (Chemistry Bonding)|49 Videos
  • D & F-BLOCK ELEMENTS & THEIR IMPORTANT COMPOUNDS

    RESONANCE|Exercise Match the column|1 Videos

Similar Questions

Explore conceptually related problems

For NH_(4)HS(s)hArr NH_(3)(g)+H_(2)S(g) If K_(p)=64atm^(2) , equilibrium pressure of mixture is

Calculate K_(p) for the equilibrium, NH_(4)HS_((s))hArrNH_(3(g))+H_(2)S_((g)) if the total pressure inside reaction vessel s 1.12 atm at 105.^(@)C .

For the reaction NH_(4)HS(s)rArrNH_(3)(g)+H_(2)S(g) ,K_(p)=0.09 . The total pressure at equilibrum is :-

On adding NH_4HS in following equilibrium NH_4HS(s) leftrightarrow NH_3(g)+H_2S(g)

For NH_4HS(s)hArrNH_3(g)+H_2S(g) , if K_p = 64 atm^2 , equilibrium pressure of mixture is

Which of the following statement(s) is/are correct for the following equilibrium? NH_(4)HS(s)hArrNH_(3)(g)+H_(2)(g)