Chemical Kinetics Question 316
Question: The instantaneous rate of disappearance of $ MnO_4^{-} $ ion in the following reaction is
$ 4.56\times {10^{-3}}M{s^{-1}}2MnO_4^{-}+10{I^{-}}+16{H^{+}}\to $
$ 2M{n^{2+}}+5I_2+8H_2O $ The rate of appearance $ I_2 $ is:
Options:
A) $ 4.56\times {10^{-4}}M{s^{-1}} $
B) $ 1.14\times {10^{-2}}M{s^{-1}} $
C) $ 1.14\times {10^{-3}}M{s^{-1}} $
D) $ 5.7\times {10^{-3}}M{s^{-1}} $
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Answer:
Correct Answer: B
Solution:
[b] Given $ -\frac{dMnO_4^{-}}{dt}=4.56\times {10^{-3}}M{s^{-1}} $ From the reaction given, $ -\frac{1}{2}\frac{dMnO_4^{-}}{dt}=\frac{4.56\times {10^{-3}}}{2}M{s^{-1}} $
$ -\frac{1}{2}\frac{dMnO_4^{-}}{dt}=\frac{1}{5}\frac{dI_2}{dt} $
$ \therefore -\frac{5}{2}\frac{dMnO_4^{-}}{dt}=\frac{dI_2}{dt} $ On substituting the given value
$ \therefore \frac{dI_2}{dt}=\frac{4.56\times {10^{-3}}\times 5}{2}=1.14\times {10^{-2}}M/s $