Question

The reduction of peroxydisulphate ion by $I^{-}$ ion is expressed by $S_{2}O^{-2}_{8}+3I^{-}\to2SO^{2-}_{4}+I^{-}_{3}$ If rate of disappearance of $I^{-}$ is $9/2 \times 10^{-3} mol L^{-1} S^{-1,}$ what is the rate of formation of $SO^{2-}_4$ during same time?

• A. $10^{-3} mol/L^{-1} s^{-1}$
• B. $2 \times 10^{-3} mol/L^{-1} s^{-1}$
• C. $3 \times 10^{-3} mol/L^{-1} s^{-1}$
• D. $4 \times 10^{-3} mol/L^{-1} s^{-1}$

Answer 1

Answer: (C)

$\frac{1}{2}\frac{d\left[SO^{2-}_{4}\right]}{dt}=\frac{1}{3}\left(\frac{d\left[I^{-}\right]}{dt}\right)$
$\frac{1}{2}\frac{d\left[SO^{2-}_{4}\right]}{dt}=\frac{1}{3}\times\frac{9}{2}\times10^{-3}$
$\therefore \frac{d\left[SO^{2-}_{4}\right]}{dt}=3\times10^{-3}mol L^{-1}s^{-1}$