# Give the form of the rate law for this reaction?

## The reaction $2 N {O}_{2} \left(g\right) + {F}_{2} \left(g\right) \to 2 N {O}_{2} F \left(g\right)$ Proceeds by the following mechanism: $N {O}_{2} \left(g\right) + {F}_{2} \left(g\right) \implies N {O}_{2} F \left(g\right) + F \left(g\right)$ (slow) $N {O}_{2} \left(g\right) + F \left(g\right) \implies N {O}_{2} F \left(g\right)$ (fast) Give the form of the rate law for this reaction

Sep 22, 2016

Well, you are given the mechanism for the overall reaction, so you cannot say that the rate law is $r \left(t\right) = k {\left[N {O}_{2}\right]}^{2} \left[{F}_{2}\right]$. That would have implied that this is an elementary reaction, when it is a complex reaction.

Instead, you are given which step is slow. The slow step takes the most time, so it has the greatest contribution to the time it takes for the reaction to occur. That means it also has the greatest contribution to the rate, because it has more weight as a slower rate.

We say then that the time contributed by the fast reaction is negligible, and that the first mechanistic step is rate-determining.

Therefore, the rate law for step 1 is approximated as the rate law of the overall reaction. For an elementary step (which is how it is given in a mechanism), we say it, as-written, tells you the order of each reactant.

That means with stoichiometric coefficients of $1$ each, the orders are both $1$, and the overall rate law is approximated as

$\textcolor{b l u e}{r \left(t\right) = {k}_{1} \left[N {O}_{2}\right] \left[{F}_{2}\right]}$

when ${t}_{\text{step2}}$ $\text{<<}$ ${t}_{\text{step1}}$.