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Q. For a reaction $NH_4COONH_{2(s)} \rightleftharpoons 2NH_{3(g)} +CO_{2(g)}$, the equilibrium pressure is $3\, atm$. $K_p$ for the reaction will be

Equilibrium

Solution:

$NH_4COONH_{2(s)} \rightleftharpoons \underset{\text{2p}}{2NH_{3(g)}}+ \underset{\text{p}}{CO_{2(g)}}$
When volume and temperature are constant, the number of moles of a gas is proportional to its partial pressure.
So, $2p + p = 3$
$3p=3$,
$\therefore p=1\,atm$
$K_p=(2p)^2 \times p=4p^3$
$=4 \times (1)^3=4\,atm^3$