31.2g of ammonium carbamate `(NH_(2)COONH_(4))` was heated in a flask of 5.0L to a temperature of `77^(@)C` .The flask initial contained 1.70g of `NH_(3)` . `NH_(2)COONH_(4)` dissociated to the extent of 50% when equilibrium reaches. Calculate `K_(p)` (in terms of `10^(-2)atm^(3)` for dissociation of `NH_(2)COONH_(4)`

Solid ammonium carbamate (NH_(4)COONH__(4)) was taken in excess in closed container according to the following reaction NH_(2)COONH_(4)(s)hArr2NH_(3)(g)+CO_(2)(g) . If equilibrium pressure is 4 atm, it's equilibrium constnat K_(P) is ?

What weight of solid ammonium carbamate (NH_(2)COONH_(4)) , when vaporised at 200^(@)C will have a volume of 8.96 litre at 1.0 atm pressure. Assume that the solid completely decomposes into CO_(2) and NH_(3) at 200^(@)C and 1.0 atm :

The value of K_(P) for NH_(2)COONH_(4(s))hArr 2NH_(2(g))+CO_(2(g)) if the total prassure at equilibrium si P :-

Some solid NH_(4)HS is placed in flask containing 0.5 atm of NH_(3) . What would be the pressure of NH_(3) and H_(2)S when equilibrium is reached. NH_(4)HS(g) hArr NH_(3)(g)+H_(2)S(g), K_(p)=0.11

Ammonium carbamate decomposes as NH_(2)COONH_(4)(s) hArr 2NH_(3)(g) +CO_(2)(g) . The value of K_(p) for the reaction is 2.9 xx 10^(-5) atm^(3) . If we start the reaction with 1 mole of the compound, the total pressure at equilibrium would be

NH_(2)COONH_(2)(s) harr 2NH_(3)(g) + CO_(2)(g) If equilibrium pressure is 3 atm for the above reaction, K_(p) will be

A definite amount of solid NH_(4)HS is placed in a flask already containing ammonia gas at a certain temperature and 0.1 atm pressure. NH_(4)HS decompses to give NH_(3) and H_(2)S and at equilibrium total pressure in flask is 1.1 atm. If the equilibrium constant K_(P) for the reaction NH_(4)HS(s) iff NH_(3)(g)+H_(2)S(g) is represented as zxx10^(-1) then find the value of z.