The rate equation for the hydrogen-iodine reaction indicates that if the hydrogen concentration is doubled, and the iodine vapor concentration is trebled, the reaction will proceed 2 X 3 = 6 times as fast. If the hydrogen concentration is halved, but the iodine concentration is doubled, the rate will be unchanged. Although only half as many molecules are available for reaction, collisions of any one of them with molecules will be twice as frequent, so the effects of the two concentration changes will cancel. Since the overall rate depends on the product of two concentrations, this is a second-order reaction.

It is a little trickier to see why the reverse reaction, decomposition of HI into and , also should be a second-order process, proportional to the square of the HI concentration:

2HI +
= [HI] [HI]
= [HI]