Question

The degree of dissociation of HI at a particular temperature is 0.8. Find the volume of 1.5 M sodium thiosulphate solution required to react completely with the iodine present at equilibrium in acidic conditions, when 0.135 mol each of $H_2$ and $I_2$ are heated at 440 K in a closed vessel of capacity 2.0 L :

• A. $V=72mL$
• B. $V=144mL$
• C. $V=288mL$
• D. None of these

Answer 1

The correct option is B $V=144mL$

$H_2+I_2\to 2HI$

$0.135$ mole of iodine will react with $0.135$ mole of hydrogen to give
$2\times 0.135=0.27$ moles of HI

The degree of dissociation of $HI$ is $0.8$
$0.8\times 0.27=0.216$ moles of HI will dissociate to give $\frac{0.216}{2}=0.108$ moles of iodine.

Let V ml of 1.5 M sodium thiosulphate solution is required.
$I_2+2N{a}_2{S}_2{O}_3\to 2NaI+N{a}_2{S}_4{O}_6$

$\frac{1.5\times V}{2}=\frac{0.108}{1}$

$V=0.144L=144mL$