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.
A
`4`
B
`9`
C
`(4)/(9)`
D
`(2)/(9)`
Text Solution
Verified by Experts
The correct Answer is:
C
Topper's Solved these Questions
CHEMICAL EQUILIBRIUM
NARENDRA AWASTHI|Exercise Level 1 (Q.93 To Q.122)|1 Videos
CHEMICAL EQUILIBRIUM
NARENDRA AWASTHI|Exercise Level 2|1 Videos
ATOMIC STUCTURE
NARENDRA AWASTHI|Exercise Exercise|273 Videos
DILUTE SOLUTION
NARENDRA AWASTHI|Exercise Level 3 - Match The Column|1 Videos
Similar Questions
Explore conceptually related problems
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 :
Solid ammonium carbamate dissociated according to the given reaction NH_2COONH_4(s) hArr 2NH_3(g) +CO(g) Total pressure of the gases in equilibrium is 5 atm. Hence K_p .
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
