JEE Chemical Kinetics 1

Since the second step is the limiting step (slow) of the reaction, we can derive the overall rate law from its chemical equation: $v = k_{2}[X][Y_{2}]$

We don't want X in the rate law since it does not appear in the overall reaction, thus we can try to obtain its concentration in terms of either $X_{2}$ or $Y_{2}$ . The first step is an equilibrium reaction, and so we can write the equilibrium constant $K_{1}$ as follows, and isolate $[X]$ :

$K_{1} = \frac{[X]^{2}}{[X_{2}]}$ $[X]=\sqrt{K_{1}[X_{2}]}$

Substituting the $[X]$ term in the original rate law with the expression above, we get:

$v = k\sqrt{K_{1}[X_{2}]}[Y_{2}]$

We see that the overall reaction order is thus 0.5 + 1 = 1.5, where 0.5 and 1 are the exponents of the $[X_{2}]$ and $[Y_{2}]$ terms respectively.