HW 7C.7

[Angela Wu-2H]

The question states:

Step 1: NO + Br2 ---> NOBr2 (slow)
Step 2: NOBr2 + NO ---> NOBr + NOBr (fast)

Write the rate law for the formation of NOBr implied by this mechanism.

I know how to write the rate law (rate=k[NO][Br2]), but I was just curious as to how to write the overall reaction, just to prepare for the final and for future classes.

Can someone tell me how I'd write the overall reaction?

[Uisa_Manumaleuna_3E]

When you right the overall reaction, all you do is look at the two steps and add up all the reactants on one side and all the products on the other. Any intermediates (anything that was made as a product of one step that was later used up as a reactant in the following step) should be cancelled out from the overall reaction. I hope that made some sense!

[Kylie Lim 4G]

To write the overall reaction, cancel out any species that appears in both the reactants and products side, and then add the steps together. Cancelling out NOBr2 and then adding both reactions together gives you 2NO + Br2 ---> 2NOBr.