2N2O5(g) → 4NO2(g) + O2(g) (this is our reaction and it is already balanced)
2 moles N2O5 disappears (reactant)
4 moles NO2 appears (product)
We can work with equations here that represent the disappearance and appearance of compounds:
The disappearance of N2O5: rate of disappearance
[tex]-\frac{1}{2}\frac{d\lbrack N_2O_5\rbrack}{dt}[/tex]
The appearance of NO2: rate of appearance
[tex]+\frac{1}{4}\frac{d\lbrack N_{}O_2\rbrack}{dt}[/tex]
We can do this:
[tex]-\frac{1}{2}\frac{d\lbrack N_2O_5\rbrack}{dt}=\text{ }+\frac{1}{4}\frac{d\lbrack N_{}O_2\rbrack}{dt}[/tex]
We can do this because the disappearance of reactants involves the appearance of the product.
Then, we clear the rate of N2O5:
[tex]-\frac{d\lbrack N_2O_5\rbrack}{dt}=2x\frac{1}{4}\frac{d\lbrack N_{}O_2\rbrack}{dt}=\frac{1}{2}x2.8x10^{-3}\frac{M}{s}=1.4x10^{-3}\frac{M}{s}[/tex]
Now the rate of disappearance of N2O5 = 1.4x10^-3 M/s
