Question

Consider the reaction, $3I_{\left(aq\right)}^{-}+S_2{O}_8^{2-}\to I_{3\left(aq\right)}^{-}+2S{O}_{4\left(aq\right)}^{2-}$. At particular time t, $\frac{d\left[S{O}_4^{2-}\right]}{dt}=2.2\times {10}^{-2}M/s$. What are the values of the following at the same time?

$-\frac{d\left[I^{-}\right]}{dt}=?$

Answer 1

Rate of consumption of $I^{-}=-\frac{d\left[I^{-}\right]}{dt}$

When, 3 moles of $I^{-}$ are consumed, 2 moles of $S{O}_4^{2-}$ are formed

$\therefore -\frac{d\left[I^{-}\right]}{dt}=\frac{3}{2}\times \frac{d\left[S{O}_4^{2-}\right]}{dt}$

$=\frac{3}{2}\times 2.2\times {10}^{-2}$

$=3.3\times {10}^{-2}M/s$