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.

The vapour density of fully dissociated NH_(4)Cl would be

Ammonium carbamate when heated to 200^(@)C gives a mixture of NH_(3) and CO_(2) vapours with a density of 16.0. What is the degree of dissociation of ammonium carbamate? (Given vapour density of ammonium carbamate is 48)

Ammonium carbamate when heated tto 200^(@)C gives a mixture of NH_(3) and CO_(2) vapour with a density of 13. what is the degree of dissociation of ammonium carbamate?

Phosphorous pentachloride when heated in a sealed tube at 700 K it undergoes decomposition as PCl_(5)(g) hArr PCl_(3)(g)+Cl_(2)(g), K_(p)=38 atm Vapour density of the mixture is 74.25 . Percentage dissociation of PCl_(5) may be given as

Phosphorus pentachloride dissociates as follows in a closed reaction vessel. PCl_5(g) hArr PCl_3(g) +Cl_2(g) If total pressure at equilibrium of the reactions mixture is P and degree of dissociation of PCl_5 is x, the partial pressure of PCl_3 will be:

Ammonium carbamate dissociates on heating as: NH_(2)COONH_(4)(g)hArr2NH_(3)(g)+CO_(2)(g) The equilibrium constant K_(p) for the reaction, at some temperature is 3.2xx10^(-5)atm^(3) . Calculate the partial pressure of NH_(3) in the equilibrium system at the same temperature.