Which of the following can we predict from an equilibrium constant for a reaction?

Which of the following can we predict from an equilibrium constant for a reaction?

1. the extent of a reaction

2. which direction a reaction will proceed by measuring the concentrations of reactants and products

3. whether a system is at equilibrium by measuring the concentrations of the reactants and products.

A) 1 only
B) 2 only
C) 3 only
D) 1 and 2 only
E) 1, 2, and 3

1 Answer
Apr 24, 2018

Well, arguably, none of the above are actually useful aspects of a reaction that require an equilibrium constant to obtain...


But the intention of the question is probably that only #("1")# is correct.

1. the extent of a reaction

This can be determined by using #Q# OR #K#. The extent of reaction is simply given by #x#, i.e. the drop in concentration of the aqueous reactant(s) or the drop in partial pressure of the gaseous reactants.

Since #x# can be determined whether at equilibrium or not, #Q# can be used to find the extent of the reaction.

However, #K# is a measure of the maximum extent of the reaction.

[It's not all that useful for that purpose because by definition, at equilibrium #Q = K#, so you know the extent of reaction is #100%# at this point without ever using #K# directly.]

2. which direction a reaction will proceed by measuring the concentrations of reactants and products

This is the job of #Q#, the not-yet-equilibrium constant.

  • When #bb(Q < K)#, the reaction is reactant-heavy and wishes to proceed forward, and vice versa.
  • When #bb(Q = K)#, the reaction is at equilibrium, and thus the reaction has equal forward and reverse rates; it will not shift any more because the extent of reaction is #100%# (#Deltax = 0#).

3. whether a system is at equilibrium by measuring the concentrations of the reactants and products.

Again, this is the job of #Q#. This is mentioned in #(1)# and #(2)# in that at equilibrium, #Q = K#.

If you choose to measure concentrations at some arbitrary time, then if #Q# happens to be equal to #K#, the reaction is already at equilibrium.