Mechanism of a hypothetical reaction $${X_2} + {Y_2} \to 2XY$$    is given below
$$\eqalign{ & \left( {\text{i}} \right){X_2} \rightleftharpoons X + X\left( {{\text{fast}}} \right) \cr & \left( {{\text{ii}}} \right)X + {Y_2} \to XY + Y\left( {{\text{slow}}} \right) \cr & \left( {{\text{iii}}} \right)X + Y \to XY\,\left( {{\text{fast}}} \right) \cr} $$
The overall order of the reaction will be

Solution :
We know that, slowest step is the rate determining step.
$$\eqalign{ & \therefore \,\,{\text{Rater}}\left( r \right) = {K_1}\left[ X \right]\left[ {{Y_2}} \right]\,\,\,...\left( {\text{i}} \right) \cr & {\text{Now, from equation}}{\text{. (i), i}}{\text{.e}}{\text{.}} \cr & {X_2} \to 2X\left[ {{\text{fast}}} \right] \cr & {K_{eq}} = \frac{{{{\left[ X \right]}^2}}}{{\left[ {{X_2}} \right]}} \cr & \left[ X \right] = {\left\{ {{K_{eq}}\left[ {{X_2}} \right]} \right\}^{\frac{1}{2}}}\,\,\,...\left( {{\text{ii}}} \right) \cr} $$
Now, substitute the value of $$\left[ X \right]$$  from equation. (ii) in equation. (i), we get
$$\eqalign{ & {\text{Rate}}\left( r \right) = {K_1}{\left( {{K_{eq}}} \right)^{\frac{1}{2}}}{\left[ {{X_2}} \right]^{\frac{1}{2}}}\left[ {{Y_2}} \right] \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = K{\left[ {{X_2}} \right]^{\frac{1}{2}}}\left[ {{Y_2}} \right] \cr & \therefore \,\,{\text{Order of reaction}} \cr & = \frac{1}{2} + 1 \cr & = \frac{3}{2} \cr & = 1.5 \cr} $$