(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

is there any correction required

Question

Aarti Gupta · Answer

Yes the equality for the above equation will be the same. Thanks & Regards Aarti Gupta askiitians Faculty

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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 is there any correction required

(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

is there any correction required

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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 is there any correction required

(c) 4NH3 + SO2 ? 4NO2 + 6H2OThe equality in this case is-1/4 d[NH3]/dt = -1/5 d[O2]/dt = 1/4 d[NO2]/dt = 1/6 d[H2O]/dtSo -d[NH3 ]/dt = -4/3 d[O2]/dt = d[NO2 ]/dt = d[H2O]/dtis there any correction required

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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

is there any correction required