We first verify that the reaction is balanced. We have 2 nitrogens and 4 oxygens from each of the reaction, so if it is balanced.
The average rate of the reaction (vm) will be:
[tex]v_m=-\frac{1}{StoichiometriCoefficient}\times\frac{\Delta\lbrack NO\rbrack}{\Delta t}[/tex]Where,
Delta[NO] corresponds to the concentration difference
Delta[t] is the time difference
The Stoichiometric coefficient of NO is 2
We put the minus sign because it corresponds to a reactant.
We replace known data:
[tex]\begin{gathered} v_m=-\frac{1}{2}\times\frac{0.0225M-0.0500M}{650.0s-5.0s} \\ v_m=-\frac{1}{2}\times\frac{0.1750M}{645.0s}=-1.356\times10^{-4}\frac{M}{s} \\ \end{gathered}[/tex]So, the average rate of reaction is -1.356x10^-4M/s
