Home
Class 12
CHEMISTRY
A(s)hArr2B(g)+C(g) The above equilibri...

`A(s)hArr2B(g)+C(g)`
The above equilibrium was established by initially taking `A(s)` only. At equilibrium, `B` is removed so that its partial pressure at new equilibrium becomes `1//3^(rd)` of original total preswsure. Ratio of total pressure at new equilibrium and at initial equilibrium will be :

A

`2//3`

B

`14//13`

C

`5//3`

D

`17//19`

Text Solution

Verified by Experts

The correct Answer is:
C
`A(s)hArr2B(g)+C(g)` ………..
`A(s)hArr2B(g)+c(g)`
`{:(eq.1,2p,p),(eq.2,p,p'):}`
`P_(T_(1)) = 3p`
`K_(p)=4p^(2)`
`(p)^(2)p'=4p^(3)`
`p'=4p`
`P_(T_(2)) = p+4p=5p`
`(P_(T_(2)))/(P_(T_(1))) = (5p)/(3p) = 5//3`
Promotional Banner

Topper's Solved these Questions

  • TEST PAPERS

    RESONANCE|Exercise PART - III CHEMISTRY SEC - 1|12 Videos

  • TEST PAPERS

    RESONANCE|Exercise PART - III CHEMISTRY SEC - 2|18 Videos

  • SURFACE CHEMISTRY

    RESONANCE|Exercise Section - 5|1 Videos

  • TEST SERIES

    RESONANCE|Exercise CHEMISTRY|50 Videos

Similar Questions

Explore conceptually related problems

Solid Ammonium carbamate dissociates as: NH_(2)COONH_(4)(s)hArr2NH_(3)(g)+CO_(2)(g). In a closed vessel, solid ammonium carbonate is in equilibrium with its dissociation products. At equilibrium, ammonia is added such that the partial pressure of NH_(3) at new equilibrium now equals the original total pressure. Calculate the ratio of total pressure at new equilibrium to that of original total pressure. Also find the partial pressure of ammonia gas added.

K_(p) for the reaction A hArr B is 4. If initially only A is present then what will be the partial pressure of B after equilibrium ?

A solid A dissociates to give B and C gases as shown, A(s) hArr 2B(g)+3C(g) At equilibrium some B(g) is introduced keeping volume constant so that pressure of B of new equilibrium becomes equal to 8/(5sqrt2) times original total pressure calculate ratio of initial equilibrium pressure of C to that of its final pressure

K_p for the equation A(s)⇋ B(g) + C(g) + D(s) is 9 atm^2 . Then the total pressure at equilibrium will be

For the reaction : A(g)hArr2B(g),K_(p)=4.0 bar. The equilibrium pressure pf A(g) is 1.0 bar. Now the equilibrium mixture is compressed reversibly and isothermally such that the final total pressure of system becomes 8 bar. The partial pressure of A(g) (in bar)at this new equilibrium is :