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(a) H2O2 + 2H+ + 2I- ? I3- + 2H2O; The equality in this case is -d[H2 O2 ]/dt = -1/2 d[H+ ]/dt = -1/2 d[I3- ]/dt = 1/2 d[H2O]/dt So, -d[I- ]/dt = -3d[H2 O2 ]/dt = -3/2 d[H+]/dt = 3d[I3- ]/dt = 3/2 d[H2O]/dt how is 3 coming as the coefficient in -3d[H2O2]/dt

(a) H2O2 + 2H+ + 2I- ? I3- + 2H2O;

The equality in this case is

-d[H2 O2 ]/dt = -1/2 d[H+ ]/dt = -1/2 d[I3- ]/dt = 1/2 d[H2O]/dt

So,

-d[I- ]/dt = -3d[H2 O2 ]/dt = -3/2 d[H+]/dt = 3d[I3- ]/dt = 3/2 d[H2O]/dt


how is 3 coming as the coefficient in -3d[H2O2]/dt

Grade:12

1 Answers

Pankaj
askIITians Faculty 131 Points
9 years ago

For the equation:

H2O2 + 2H+ + 2I---> I3- + 2H2O
The equality should be:
-d[I- ]/dt = -2d[H2 O2 ]/dt = - d[H+]/dt = 2d[I3- ]/dt = d[H2O]/dt

Thanks and Regards

Pankaj Singh

askIITians Faculty


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