Question

if the rate of appearance of no2 is equal to 8.00mol/min at a particular moment, what is the rate of dissapearce of N2O5 at that moment

Answer 1

`r_(N_2O_5)=-4.00mol/min\`

Step-by-step explanation:

Hello.

In this case, since the chemical reaction is:

`2N_2O_5\rightarrow 4NO_2+O_2`

We can write the rate ratio between N2O5 and NO2 as shown below:

`(1)/(-2) r_(N_2O_5)= (1)/(4) r_(NO_2)`

Since N2O5 is consumed and NO2 produced at a rate of 8.00 mol/s, therefore, the rate of disappearance is N2O5 is:

`r_(N_2O_5)= (-2)/(4) r_(NO_2)\\\\ r_(N_2O_5)=(-1)/(2) *8.00 mol/min\\\\ r_(N_2O_5)=-4.00mol/min`

Which is negative since N2O5 is a reactant.

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