Question

Solid $N{H}_{4}I$ on rapid heating in a closed vessel at $347K$ develops a constant pressure of $275mmHg$ owing to the partial decomposition of $N{H}_{4}I$ into $N{H}_3$ and $HI$, but the pressure gradually increases further (when the excess solid residue remains in the vessel) owing to the dissociation of $HI$. Calculate the final pressure in $atm$, developed at equilibrium.

$K_p$ for the dissociation of $HI$ is $0.015$ at ${357}^{\circ }C$.

Answer 1

$N{H}_{4}I\rightleftharpoons N{H}_3+HI$ ...(1)
$PP$
For this equilibrium $2P=275$
$\therefore P=137.5$
Thus, $K_p$ for equation (1) $=P_{N{H}_3}\times P_{HI}=137.5\times 137.5$ ...(2)
For $2HI\rightleftharpoons H_2+I_2$
$100$
$(1-x)\frac{x}{2}\frac{x}{2}$
$\therefore K_p=K_c=\frac{x^2}{4{(1-x)}^2}={\left(0.015\right)}^2$
$\therefore x=0.2$
i.e., degree of dissociation of $HI$ is $0.2$. Again since, $HI$ decomposes, reaction (1) proceeds in forward direction, i.e.,
$N{H}_{4}I\rightleftharpoons N{H}_3+HI$ ...(3)
$(P+A)(P-a)$
$2HI\rightleftharpoons H_2+I_2$ ....(4)
$(P-a)\frac{a}{2}\frac{a}{2}[\because a=0.2\times P]$
Also, $K_p$ for final equilibrium equation (3) or (1) is
$K_p=(P+A)(P-a)=(P+A)[P-0.2\times P]$
$(P+A)\times 0.8P=137.5\times 137.5$
$\therefore (137.5+A)\times 0.8\times 137.5=137.5\times 137.5$
$\therefore A=34.375$
$a=137.5\times 0.2=27.5$
$\therefore A=34.375$
$N{H}_{3}HI{H}_2{I}_2$
Total pressure at equilibrium $=P+A+P-a+a/2+a/2$
$=2P+A$
$=2P+A=2\times 137.5+34.375$
$=309.375atm$