About rate of reaction (chem)

Step 2 is the rate determining step, because step 2 is the slowest step in the mechanism.
Thus, Rate = k'[H2(g)][I(g)]2 ...... (1)

I(g) is the intermediate of the reaction. Therefore, it should be represented by the concentration(s) of the reactant(s).

Step 1 shows an equilibrium, and the equilibrium equation is:
Kc = [I(g)]2/[I2(g)]
and thus, [I(g)]2 = Kc[I2(g)] ...... (2)

Substitute (2) into (1):
Rate = k'[H2(g)](Kc[I2(g)])
Rate = k'Kc[H2(g)][I2(g)]

The rate equation is thus:
Rate = k[H2(g)][I2(g)]
(where k = k'KC)
The reaction is first order with respect to H2(g) and to I2(g).
=

For a multi-step reaction, the slowest step is usually the rate determining step. You can see it in this way, the rate of the reaction is the rate of formation of the product. When the reaction proceeds, I is formed very fast but the formation of HI is slow. a lot of I present but the formation of HI is slow. It just like a bottleneck. If we could increase the rate of the 2nd reaction, i.e. open up the bottleneck then make the formation of HI faster. The overall reaction will increase significantly.

If we increase the 1st step, though even more I is formed but the 2nd step is still the bottleneck, no significant increase on the overall reaction rate.