Q. Ammonium carbonate decomposes as $NH_{2}COONH_{4}(s) \rightleftharpoons 2NH_{3}(g) + CO_{2}(g)$ For the reaction, $K_{p} = 2.9 \times 10^{- 5}\ atm^{3}$. If we start with 1 mole of the compound, the total pressure at equilibrium would be
A) 0.766 atm
B) 0.0582 atm
C) 0.0388 atm
D) 0.0194 atm
Explanation
(b)
$NH_{2}COONH_{4}(s) \rightleftharpoons 2NH_{3}(g) + CO_{2}(g)$
At equilibrium if partial pressure of $CO_{2}\ \ \ \ p$ then that of $NH_{3}$ $2p$
$K_{p}\ \ \ \ p_{NH_{3}}^{2}\ \ \ \ p_{CO_{2}}\ \ \ \ \ \ (2p)^{2}\ \ \ p\ \ \ \ \ \ \ 4p^{3}\$
$2.9\ \ \ \ \ 10^{- 5}\ \ \ 4p^{3}$
$p^{3}\ \ \ \ 0.725\ \ \ \ 10^{- 5}$
$or\ \ p^{3}\ \ \ \ 7.25\ \ \ \ 10^{- 6}\ \ \ or\ \ \ p\ \ \ 1.935\ \ \ \ 10^{- 2}$
hence, total pressure $p\ \ \ \ \ \ \ 3p$
$5.81\ \ \ \ \ \ \ 10^{- 2}\ \ \ \ 0.0581$ atm.
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.