Home
Class 12
CHEMISTRY
Ammonium sulphide and ammonium selenide ...

Ammonium sulphide and ammonium selenide on heating dissociates as
`(NH_4)S(s) hArr 2NH_(3)(g) + H_(2)S(g) , k_(p1) = 9 xx 10^(-3) atm^(3)`
`(NH_4)_(2) Se(s) hArr 2NH_(3)(g) + H_(2)Se(g), K_(p2) = 4.5 xx 10^(-3) atm^(3)`.
The total pressure over the solid mixture at equilibrium is

A

0.15 atm

B

0.3 atm

C

0.45 atm

D

0.6 atm

Text Solution

Verified by Experts

The correct Answer is:
C
Doubtnut Promotions Banner Mobile Dark
|

Topper's Solved these Questions

  • JEE MOCK TEST 7

    NTA MOCK TESTS|Exercise CHEMISTRY - Subjective Numerical|5 Videos
  • JEE MOCK TEST 6

    NTA MOCK TESTS|Exercise CHEMISTRY|25 Videos
  • JEE MOCK TEST 8

    NTA MOCK TESTS|Exercise CHEMISTRY|25 Videos

Similar Questions

Explore conceptually related problems

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

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

Knowledge Check

  • Ammonium sulphide and ammonium selenide on heating dissociates as (NH_4)_2S_((s)) hArr 2NH_(3(g)) + H_2S_((g)) , K_(p_1) = 9xx10^(-3) "atm"^3 (NH_4)_2Se_((s)) hArr 2NH_(3(g)) + H_2Se_((g)) , K_(p_2) = 4.5xx10^(-3) "atm"^3 The total pressure over the solid mixture at equilibrium is

    A
    0.15 atm
    B
    0.3 atm
    C
    0.45 atm
    D
    0.6 atm
  • Ammonium sulphide and ammonium selenide on healing dissociates as (NH_(4))_(2) S_(s) Leftrightarrow 2NH_(3(g)) +H_(2)S_(g)), K_(p_(1))=9 xx 10^(-3)" atm"^3 (NH_(4))_(2) Se_(s) Leftrightarrow 2NH_(3(g)) +H_(2)Se_(g), K_(p_(2))=4.5 xx 10^(-3)" atm"^3 The total pressure over the solid mixture al equilibrium is

    A
    0.15 atm
    B
    0.3 atm
    C
    0.45 atm
    D
    0.6 atm
  • Two solid compounds X and Y dissociates at a certain temperature as follows X(s) hArr A(g)+2B(g),K_(p1)=9xx10^(-3)atm^(3) Y(s) hArr 2B(g)+C(g),K_(p2)=4.5xx10^(-3)atm^(3) The total pressure of gases over a mixture of X and Y is :

    A
    `4.5 atm`
    B
    `0.45 atm`
    C
    `0.6 atm`
    D
    None of these
  • Similar Questions

    Explore conceptually related problems

    One "mole" of NH_(4)HS(s) was allowed to decompose in a 1-L container at 200^(@)C . It decomposes reversibly to NH_(3)(g) and H_(2)S(g). NH_(3)(g) further undergoes decomposition to form N_(2)(g) and H_(2)(g) . Finally, when equilibrium was set up, the ratio between the number of moles of NH_(3)(g) and H_(2)(g) was found to be 3 . NH_(4)HS(s) hArr NH_(3)(g)+H_(2)S(g), K_(c)=8.91xx10^(-2) M^(2) 2NH_(3)(g) hArr N_(2)(g)+3H_(2)(g), K_(c)=3xx10^(-4) M^(2) Answer the following: To attain equilibrium, how much % by weight of folid NH_(4)HS got dissociated?

    One "mole" of NH_(4)HS(s) was allowed to decompose in a 1-L container at 200^(@)C . It decomposes reversibly to NH_(3)(g) and H_(2)S(g). NH_(3)(g) further undergoes decomposition to form N_(2)(g) and H_(2)(g) . Finally, when equilibrium was set up, the ratio between the number of moles of NH_(3)(g) and H_(2)(g) was found to be 3 . NH_(4)HS(s) hArr NH_(3)(g)+H_(2)S(g), K_(c)=8.91xx10^(-2) M^(2) 2NH_(3)(g) hArr N_(2)(g)+3H_(2)(g), K_(c)=3xx10^(-4) M^(2) Answer the following: Assuming the volume due to solid NH_(4)HS is negligible what will be the density of the gaseous mixture in the above equilibrium system?

    For the reaction NH_(2)COONH_(4)(g)hArr 2NH_(3)(g)+CO_(2)(g) the equilibrium constant K_(p)=2.92xx10^(-5)atm^(3) . The total pressure of the gaseous products when 1 mole of reactant is heated, will be

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

    NH_(4)COONH_(2) (s) hArr 2 NH_(3) (g) + CO_(2)(g) . if equilibrium pressure is 3 atm for the above reaction , K_(p) for the reaction is