The mass–action ratio,[1] [2] often denoted by
\Gamma
\Gamma=
p1p2\ldots | |
s1s2\ldots |
This assumes that the stoichiometric amounts are all unity. If not, then each concentration must be raised to the power of its corresponding stoichiometric amount. If the product and reactant concentrations are at equilibrium then the mass–action ratio will equal the equilibrium constant. At equilibrium:
\Gamma=Keq
The ratio of the mass–action ratio to the equilibrium constant is often called the disequilibrium ratio, denoted by the symbol
\rho
\rho=
\Gamma | |
Keq |
and is a useful measure for indicating how from equilibrium a given reaction is. At equilibrium
\rho=1
\rho ≠ 1
\rho<1
For a uni-molecular reaction such as
A\rightleftharpoonsB
v=k1A-k2B=vf-vr
At thermodynamic equilibrium the rate equals zero, that is . Rearranging gives:
k1 | |
k2 |
=
Beq | |
Aeq |
=Keq
but , therefore
\rho=\Gamma
k2 | |
k1 |
\rho=
B | |
A |
k2 | |
k1 |
=
vr | |
vf |
In other words the disequilibrium ratio is the ratio of the reverse to the forward rate. When the reverse rate, is less than the forward rate, the ratio is less than one, , indicating that the net reaction is from left to right.
If the natural log is taken on both sides of the disequilibrium ratio and both sides is multiplied by RT, one obtains:
ln(\rho)=ln(\Gamma)-ln\left(Keq\right)
RTln(\rho)=-RTlnKe+RTln\Gamma
However,
-RTlnKe
RTln(\rho)
This shows that the free energy of a reaction is just an alternative way of expressing the disequilibrium ratio and as such gives a more intuitive interpretation of free energy. That is if the free energy for a reaction is less than zero then it indicates that and hence , i.e the net reaction rate is from left to right.