​a reaction takes place in two steps with equilibrium constant 10^-2 for slow step and 10^2 for fast step the equilibrium constant of the overall reaction will be.
The answer is 1.

Dear Student,

If two or more reactions are added to give another, the equilibrium constant for the reaction is the product of the equilibrium constants of the equation added. 
K = K₁ x K₂
Here, K₁=10^-2
K₂= 10²
Therefore, K = 10^-2 x 10² = 10⁰=1

Hence, the equilibrium constant for the overall reaction is 1.