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Q. On decomposition of $NH _{4} HS$, the following equilibrium is established. $NH _{4} HS ( s ) \rightleftharpoons NH _{3}( g )+ H _{2} S ( g ) .$ If the total pressure is $P$ atm, then the equilibrium constant $K _{ p }$ will be equal to $\frac{ P ^{2}}{ x }$ atm $^{2}$ where $x$ is __

Equilibrium

Solution:

$NH _{4} HS ( s ) \rightleftharpoons NH _{3}(1\, mol ) + H _{2} S (1 \,mol )$
$K _{ P }=\left( P _{ NH _{3}}\right) \cdot\left( P _{ H _{2} S }\right)$
$=\frac{ P }{2} \times \frac{ P }{2}=\frac{ P ^{2}}{4} \,atm ^{2}$