# Equilibrium constant for a multi-step reaction [closed]

If a reaction proceeds in steps, is the equilibrium constant of every step taken into account to arrive at the final constant? Or can we just calculate equilibrium constant for the overall reaction with reactants on one side and the products on the other?

## closed as unclear what you're asking by Buck Thorn, andselisk♦, Jon Custer, Todd Minehardt, MithoronMay 13 at 18:52

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• It might help if you provide a concrete example. In general, you can compute an equilibrium constant between any reagents or products, irrespective of the number of intervening steps in the mechanism, as long as they are in equilibrium. – Buck Thorn May 13 at 7:09

To give a specific example, let's take this set of reactions:

$$\tag{1}\ce{A + B <=> I_1}$$

$$\tag{2}\ce{I_1 + C <=> I_2}$$

$$\tag{3}\ce{I_2 <=> I_3 + D}$$

$$\tag{4}\ce{I_3 <=> E + F}$$

To get the net reaction, we add up the equations and cancel the intermediates:

$$\ce{A + B + C <=> D + E + F}$$

When the system is a equilibrium, all the reactions are at equilibrium. You can write the equilibrium constant expression based on the net reaction:

$$K_\text{net} = \frac{[\ce{D}][\ce{E}][\ce{F}]}{[\ce{A}][\ce{B}][\ce{C}]}$$

You would arrive at the same expression if you take the product of the individual equilibrium constants (and cancel):

$$K_\text{net} = K_1 K_2 K_3 K_4 =\frac{[\ce{I_1}]}{[\ce{A}][\ce{B}]} \times \frac{[\ce{I_2}]}{[\ce{I_1}][\ce{C}]} \times \frac{[\ce{I_3}][\ce{D}]}{[\ce{I_2}]} \times \frac{[\ce{E}][\ce{F}]}{[\ce{I_3}]}$$

The concentrations of the intermediates will not be zero because they are in equilibrium with the other species. They are intermediates in the sense that we care a bit less about them since they don't appear in the net equation of interest. However, they are part of the mass balance and may well be among the major species.