Q.A reaction occurs in the following steps i. $\mathrm{NO_2(g) + F_2(g) \longrightarrow NO_2F(g) + F(g)}$ (slow) ii. $\mathrm{F(g) + NO_2(g) \longrightarrow NO_2F(g)}$ (fast) a. Write the equation of overall reaction.
b. Write down rate law.
c. Identify the reaction intermediate.

Adding the two steps cancels the F atom; the slow step (i) controls the rate, and its reactants give the rate law. …

Step 1 (overall reaction). Adding step i. $\mathrm{NO_2 + F_2 \longrightarrow NO_2F + F}$ and step ii. $\mathrm{F + NO_2 \longrightarrow NO_2F}$: the F produced in i. is consumed in ii., leaving

Step 2 (rate law). The slow step is the rate-determining step; its reactants give the rate law directly: rate = $k$[NO$_2$] [F$_2$]. …

Add the elementary steps cancelling species appearing on both sides; take the rate law from the slow step's reac …

• Writing the rate law from the OVERALL equation (k[NO2]^2[F2]) -- only the slow step's reactants appear: k[NO2][F2].
• Naming NO2F the intermediate -- NO2F is a PRODUCT; the intermediate F is made then consumed. …