Simple rules

Consider the reaction:

aA + bB <===> cC + dD

Here, a moles of the chemical species A react with b moles of B to give c moles C plus d moles D in a reversible reaction.

A and B are commonly called reactants, and C and D are called products, and the reaction is taken as proceeding from left to right in the spontaneous direction, so that the change in number of moles is negative for reactants, and positive for products.

The free-energy change for this reaction is given by the equation below, in which the curly brackets indicate activities:

DG' = DGo + RT ln ({C}c . {D}d / {A}a . {B}b )

At equilibrium, for any reaction with defined species, and under known reaction conditions (temperature, pressure), the ratio of activities of reactant and products reaches a fixed value, defined by the equilibrium constant. For the reaction above:

Keq = {C}ceq . {D}deq / {A}aeq . {B}beq

were the subscript (eq) indicates that these are values found at equilibrium.

The equilibrium constant provides a link to both energetic and kinetic aspects of a reaction.

Energetic aspects are spelt out more completely in the pages on thermodynamics. At equilibrium, a chemical process is unable to provide any work, or more formally, DG' for the reaction is zero. From the standard equation:

DG' = DGo + RT ln ({C}ceq . {D}deq / {A}aeq . {B}beq) = DGo + RT ln Keq = 0

where the activities are those under equilibrium conditions. From this it follows that:

Go = - RT ln Keq

Relation to rate constants

Kinetic aspects come directly from the fact that at equilibrium, the net flux through the reaction is zero, and the forward and reverse reactions are therefore equal in rate. For the reaction above, at equilibrium:

vforwards = kf. {A}aeq . {B}beq = vbackwards = kb. {C}ceq . {D}deq

and by substitution:

Keq = kf / kb

Here, kf and kb are rate constants for the forward and reverse reactions. Rates of reaction, and rate constants will be dealt withh separately when we discuss kinetics.

These simple rules become more complicated when we consider concentrations rather than activities, and when we refer to reference states other than standard conditions. Some of these complications are covered in the pages on Derivation of Thermodynamic Functions.

A technical point about Keq

An important point to remember is that Keq as defined in the context of thermodynamic equations is a dimensionless term, because each term in the ratio is referred to the standard state of 1 M activity, and the units cancel (see previous lecture notes). Because the standard state activity is defined as 1 M, its inclusion doesn't effect the numerical value, - just the units, which cancel to give for each species a dimensionless value equal to the M concentration (activity). For this reason, the standard state activities are ommitted from the ratio term, because otherwise the equations get a bit unwieldy:

Keq = ({C}ceq/{C}css . {D}deq/{D}dss) / ({A}aeq /{A}ass. {B}beq/{B}bss)

When used in the context of kinetics, Keq has dimensions and units as determined by use of the concentrations (or activities) of the terms in the ratio. For example, in the reaction:

A <==> B + C

Keq = kf / kb = [B][C] / [A] (units M)

at equilibrium:
vforwards = kf. {A}aeq (units M.s-1)
(kf is a first-order rate constant, with units s-1)
= vbackwards
            = kb . {B}beq . {C}ceq (units M-1.s-1.M2 = M.s-1)
(kb is a second-order rate constant, with units M-1.s-1)

Although this distinction is important from the formal point of view, in practice, the numerical value of the kinetic and thermodynamic equilibrium constants will always be the same when the concentrations (activities) are expressed in M units.

©Copyright 1996, Antony Crofts, University of Illinois at Urbana-Champaign,