Question

For the decomposition of the compound, represented as
$NH _{2} COONH _{4} (s) \rightleftharpoons 2 NH _{3} (g) + CO _{2} (g)$
the $K _{ p }=2.9 \times 10^{-5} atm ^{3}$. If the reaction is started with $1$ mole of the compound, the total pressure at equilibrium would be:

• A. $7.66\times 10^{- 2}atm$
• B. $38.8\times 10^{- 2}atm$
• C. $5.82\times 10^{- 2}atm$
• D. $1.94\times 10^{- 2}atm$

Answer 1

Answer: (C)

In case of heterogeneous equilibrium. The achive mass of solid is unit so not consider in the expression of $K_{p}$ and $K_{c}$ $( NH )_{2}( COONH )_{4}( s ) \rightleftharpoons 2( NH )_{3}( g )+( CO )_{2}( g )$ At equilibrium $\Rightarrow 2 p$ and $p$ will be partial pressures of ammonia and carbon dioxide gas respectively.
$ K _{ P }=\left[ P _{ NH _{3}}\right]^{2}\left[ P _{ CO _{2}}\right]=[2 P ]^{2}[ P ]=4 P ^{3}$
$4 P ^{3}=2 \cdot 9 \times 10^{-5} atm ^{3} $
$ P ^{3}=0 \cdot 725 \times 10^{-5} atm ^{3}=7 \cdot 25 \times 10^{-6} atm ^{3} $
$ P =\sqrt[3]{7 \cdot 25 \times 10^{-6}}=1 \cdot 94 \times 10^{-2} atm$
$ \Sigma P =\left(( P )_{\text {Total }}\right)_{\text {equilibrium }}=2 P + P =3 P $
$=3 \times 1 \cdot 94 \times 10^{-2} $ atm
$=5 \cdot 82 \times 10^{-2} atm$