Chemical EquilibriumHard
Question
Solid ammonium carbamate dissociates as: NH2COONH4(s) $\rightleftharpoons$2NH3(g) + CO3(g). In a closed vessel, solid ammonium carbamate is in equilibrium with its dissociation products. At equilibrium ammonia is added such that the partial pressure of NH3 at new equilibrium now equals the original total pressure. The ratio of total pressure at new equilibrium to that of original total pressure is
Options
A.1: 1
B.27: 31
C.31: 27
D.3: 4
Solution
NH2COONH4(s) $\rightleftharpoons$2NH3(g) + CO3(g)
Equ. partial pressure 2P0 P0
New Equ. partial pressure 3P0 P0
Now, $K_{p} = \left( 2P_{0} \right)^{2}.P_{0} = \left( 3P_{0} \right)^{2}.P \Rightarrow P = \frac{4}{9}P_{0}$
Now, $\frac{3P_{0} + P}{3P_{0}} = \frac{31}{27}$
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