| \[4.56\times {{10}^{-3}}M{{s}^{-1}}2MnO_{4}^{-}+10{{I}^{-}}+16{{H}^{+}}\to \] \[2M{{n}^{2+}}+5{{I}_{2}}+8{{H}_{2}}O\] |
| The rate of appearance \[{{I}_{2}}\] is: |
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) \[5.7\times {{10}^{-3}}M{{s}^{-1}}\]
Correct Answer: B
Solution :
[b] Given \[-\frac{dMnO_{4}^{-}}{dt}=4.56\times {{10}^{-3}}M{{s}^{-1}}\] From the reaction given, \[-\frac{1}{2}\frac{dMnO_{4}^{-}}{dt}=\frac{4.56\times {{10}^{-3}}}{2}M{{s}^{-1}}\] \[-\frac{1}{2}\frac{dMnO_{4}^{-}}{dt}=\frac{1}{5}\frac{d{{I}_{2}}}{dt}\] \[\therefore -\frac{5}{2}\frac{dMnO_{4}^{-}}{dt}=\frac{d{{I}_{2}}}{dt}\] On substituting the given value \[\therefore \frac{d{{I}_{2}}}{dt}=\frac{4.56\times {{10}^{-3}}\times 5}{2}=1.14\times {{10}^{-2}}M/s\]
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