$3.2$ moles of hydrogen iodide were heated in a sealed bulb at $444^{\circ} \mathrm{C}$ till the equilibrium…
$3.2$ moles of hydrogen iodide were heated in a sealed bulb at $444^{\circ} \mathrm{C}$ till the equilibrium state was reached. Its degree of dissociation at this temperature was found to be $22 \%$ The number of moles of hydrogen iodide present at equilibrium
are
$2.496$
$1.87$
2
4
Solution
$2 \mathrm{HI} ightleftharpoons \mathrm{H}_{2}+\mathrm{I}_{2}$.It is $22 \%$ decomposed
$\therefore \frac{3.20 \times 22}{100}=0.704$
(3.2-0.704) is equal to HI present at equilibrium which is $=2.496$
^