What is the equilibrium equation in chemistry? Is it different from the regular equation?

We call the short-hand notation for a chemical reaction an “equation” because it does involve “balancing” both sides and it is thus related to the mathematical equation. HOWEVER, it is not an equation in the mathematical sense, but a description of the equilibrium conditions of a chemical reaction.

In many cases the “reaction” proceeds primarily to the products, but even in “complete” reactions there will remain residual amounts of the reactants due to the “equilibrium” requirements of chemical interactions.

“Equilibrium constants”, including pH and solubilities are measures of the degree of “completion” or shift of a reaction from one side of the equation to the other.

ALL chemical "equations" (reactions) are reversible, although the energy requirements may be excessive and make the actual event practically impossible.

There is a #"rate forward"=K_f[A][B]#.....where, as usual, #[A], [B]# are the concentrations of the reactants.......

And a #"rate backward"=K_r[C][D]#. The condition of equilibrium specifies that #"rate forwards "=" rate backwards"#. It is usually specified for a given set of conditions.

So at equilibrium,

#"rate forward"=K_f[A][B]-="rate backward" =K_r[C][D]#

And on rearrangement, #K_f/K_r="rate forward"/"rate backward"#.

And we call the quotient, #K_f/K_r#, the thermodynamic equilibrium constant, i.e. #K_"eq"#.

And thus we write the familiar equilibrium expression:

#K_"eq"=([C][D])/([A][B])#

#K_"eq"# is a constant for a given set of conditions. It cannot be altered, but it certainly can be manipulated. For instance, if the reaction has reached equilibrium, if we can somehow remove the products, the reaction will have to re-establish the equilibrium, by shifting to the right.

And thus large values of #K_"eq"# mean that the products are favoured at equilibrium. Small values mean that the reactants are favoured. #K_"eq"# can be related to the #"Gibbs Free Energy"# of the reaction:

#lnK_"eq"=-(DeltaG^@)/(RT)=DeltaH^@-TDeltaS^@#