Question

The instantaneous rate of disappearance of \(\mathrm{MnO_4^-}\) ion in the following reaction is \(4.56\times10^{-3}\ \mathrm{M\,s^{-1}}\)
\(2\mathrm{MnO_4^-} + 10\mathrm{I^-} + 16\mathrm{H^+} \rightarrow 2\mathrm{Mn^{2+}} + 5\mathrm{I_2} + 8\mathrm{H_2O}\)
The rate of appearance \(\mathrm{I_2}\) is :

• A. \(4.56\times10^{-4}\ \mathrm{M\,s^{-1}}\)
• B. \(1.14\times10^{-2}\ \mathrm{M\,s^{-1}}\)
• C. \(1.14\times10^{-3}\ \mathrm{M\,s^{-1}}\)
• D. \(5.7\times10^{-3}\ \mathrm{M\,s^{-1}}\)

Hint:

Divide each species' rate by its stoichiometric coefficient: all give the same reaction rate.

Answer 1

The Correct Option is B

Rate of reaction \(= -\dfrac{1}{2}\dfrac{d[\mathrm{MnO_4^-}]}{dt} = +\dfrac{1}{5}\dfrac{d[\mathrm{I_2}]}{dt}\).
So \(\dfrac{d[\mathrm{I_2}]}{dt} = \dfrac{5}{2}\times\left(-\dfrac{d[\mathrm{MnO_4^-}]}{dt}\right) = 2.5\times4.56\times10^{-3}\).
\(= 11.4\times10^{-3} = 1.14\times10^{-2}\ \mathrm{M\,s^{-1}}\).
Answer: (2) \(1.14\times10^{-2}\ \mathrm{M\,s^{-1}}\)