relating rates

[Farcus]

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ok

2NO2 ->2NO + O2

relate the decomposition o fNO2 to rate of formation of O2

why is the answer 1/2 delta[NO2]/delta[T] = delta[O2]/Delta[T]

I thought you are suppose to divide 2 by each side so wouldn't the right side get the 1/2 rather than the left side?

 

[TieuBachHo]

ok

2NO2 ->2NO + O2

relate the decomposition o fNO2 to rate of formation of O2

why is the answer 1/2 delta[NO2]/delta[T] = delta[O2]/Delta[T]

I thought you are suppose to divide 2 by each side so wouldn't the right side get the 1/2 rather than the left side?

OK, the dissociation of NO2 is twice as fast as the formation of O2 because the rxn is 2 to 1. In other words, the rate for NO2 dissociation is double the rate of O2 formation. So, delta[NO2]/delta[T] = 2 (delta[O2]/Delta[T]) right? Or 1/2 delta[NO2]/delta[T] = delta[O2]/Delta[T].

Think of it this way, the reaction rate is an inverse of its stoichiometry.