Question

The oxidation of iodide ion by perdisulphate ion is described as follows :

$I^{⊝}+S_2{O}_8^{2-}\to I_3^{⊝}+S{O}_4^{2-}$

If the rate disappearance of $S_2{O}_8^{2-}$ ions is $1.5\times {10}^{-3}M{s}^{-1}$, the rate of formation of $S{O}_4^{2-}$ ions is:

• A. $3.0\times {10}^{-3}M{s}^{-1}$
• B. $2.5\times {10}^{-3}M{s}^{-1}$
• C. $2.75\times {10}^{-3}M{s}^{-1}$
• D. $Noneofthese$

Answer 1

The correct option is A $3.0\times {10}^{-3}M{s}^{-1}$

First, write the balanced equation.
$3I^{⊝}+S_2{O}_8^{2-}\to I_3^{⊝}+2S{O}^{3-}4$

$\Rightarrow \frac{-1}{3}\frac{d\left[I^{⊝}\right]}{dt}=\frac{-d\left[S_2{O}_8^{2-}\right]}{dt}=\frac{1}{2}\frac{d\left[S{O}_4^{2-}\right]}{dt}$

Given: $\frac{-d\left[S_2{O}_8^{2-}\right]}{dt}=1.5\times {10}^{-3}M{s}^{-1}$

$\therefore \frac{d\left[I^{⊝}\right]}{dt}=3\times 1.5\times {10}^{-3}=4.5\times {10}^{-3}M{s}^{-1}$

Also, $\frac{d\left[S{O}_4^{2-}\right]}{dt}=2\times 1.5\times {10}^{-3}=3\times {10}^{-3}M{s}^{-1}$

Hence, option A is correct.