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.

Explanation

the ratio of 2 :1 Thus, ` p_( NH_(3)) = 2/3 P, p_(CO_(2))= 1/3 P :. K_(P) = (p_(NH_(3)))^(2) (p_(CO_(2)))^(2) = (2/3P)^(2) (1/3 P) = 4/27 P^(3)`
After adding ` NH_(3) , p_(NH_(3))= P ("Given ") :. P^(2) xx p_(CO_(2)) = K_(p) = 4/27 P^(3) or p_(CO_(2)) = 4/ 27 P`
` :. " Total pressure now " = p_(NH_(3)) + p_(CO_(2)) = P + 4/27 P = 31/27 P`
` :. " Ratio of total pressure now to the original pressure "= 31 /27 .`