Q. Ammonium carbonate decomposes as $NH_{2}COONH_{4}(s) \rightleftharpoons 2{NH}_{3}(g) + {CO}_{2}(g)$ 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) 0.0766 atm
B) 0.0582 atm
C) 0.388 atm
D) 0.0194 atm
Explanation
(b)
$NH_{2}COONH_{4}(g) \rightleftharpoons 2{NH}_{3}(g) + {CO}_{2}(g)$
At eq. if partial pressure of $CO_{2} = p$
Then that of $NH_{3} = 2p$
$K_{p} = p_{NH_{3}}^{2} \times p_{CO_{2}} = (2p)^{2} \times p = 4p^{3}$
$= 2.9 \times 10^{- 5}$ or $p^{3} = 0.725 \times 10^{- 5}$
or $p = 1.935 \times 10^{- 2}$
Hence, total pressure $= p = 5.81 \times 10^{- 2} = 0.0581\ atm$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.