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(c) 4NH3 + SO2 ? 4NO2 + 6H2O The equality in this case is -1/4 d[NH3]/dt = -1/5 d[O2]/dt = 1/4 d[NO2]/dt = 1/6 d[H2O]/dt So -d[NH3 ]/dt = -4/3 d[O2]/dt = d[NO2 ]/dt = d[H2O]/dt could u pls explain how -4/3 n 1 are coming as d coefficients of d[O2]/dt n d[H2O]/dt in d ans

(c) 4NH3 + SO2 ? 4NO2 + 6H2O

The equality in this case is

-1/4 d[NH3]/dt = -1/5 d[O2]/dt = 1/4 d[NO2]/dt = 1/6 d[H2O]/dt

So -d[NH3 ]/dt = -4/3 d[O2]/dt = d[NO2 ]/dt = d[H2O]/dt


could u pls explain how -4/3 n 1 are coming as d coefficients of d[O2]/dt n d[H2O]/dt in d ans

Grade:12

1 Answers

Pankaj
askIITians Faculty 131 Points
9 years ago
Hi Neha.
The equality should be:

-d[NH3 ]/dt = -4/5 d[O2]/dt = d[NO2 ]/dt = 2/3d[H2O]/dt

Please either provide URL of the question if it is present on our site or post the proper question so that i can provide you the meaningful help.


Thanks and Regards

Pankaj Singh

askIITians Faculty

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