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: