Question

For the reaction 5H₂O₂ + 2MnO₄^- + 6H⁺ → 2Mn²⁺ +5O₂ + 8H2O

•The experimentally determined rate of disappearance of MnO₄^- is 2.2x10^-3 M/s 

•Write out the relative rates for all chemicals involved in the reaction.

Answer 1

Answer 2

Given −dMnO−4dt=4.56×10−3Ms−1$-\frac{dMn{O}_4^{-}}{dt}=4.56\times {10}^{-3}M{s}^{-1}$
From the reaction given,
−12dMnO−4dt=4.56×10−32Ms−1$-\frac{1}{2}\frac{dMn{O}_4^{-}}{dt}=\frac{4.56\times {10}^{-3}}{2}M{s}^{-1}$
−12dMnO−4dt=15dI2dt$-\frac{1}{2}\frac{dMn{O}_4^{-}}{dt}=\frac{1}{5}\frac{d{I}_2}{dt}$
∴−52dMnO−4dt=dI2dt$\therefore -\frac{5}{2}\frac{dMn{O}_4^{-}}{dt}=\frac{d{I}_2}{dt}$
On substituting the given value
∴dI2dt=4.56×10−3×52=1.14×10−2M/s