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 2.0L.

• A. 144 mL
• B. 167 mL
• C. 186 mL
• D. 200 mL

Answer 1

The correct option is A 144 mL

$H_2+I_2\to 2HI$

0.135 0.135 0.270
And HI dissociates 80%, so moles of $I_2$ formed iare $0.270\times 0.8/2=0.108$ mole.
$2N{a}_2{S}_2{O}_3+I_2\to N{a}_2{S}_4{O}_6+2NaI$
So moles of thiosulphate requires is $1.5\times V/1000=0.108\times 2;V=144ml$.