The above equation can be use for uni-molecular directly. The simplified interpretation is that the reaction rate constant is proportional to the frequency of the reaction mode (kBT/h) and the relative concentrations of the transition state and the reactant(s):
[P(TS)/1 bar] / [P(reactant)/1 bar] = e^(-delta G/RT), where delta G is the standard Gibbs energy difference between the transition state and the reactant at temperature T and the standard state.
For a bi-molecular reaction, [P(TS)/1 bar] / {[P(A)/1 bar][P(B)/1 bar]} = e^(-delta G/RT),and again, the delta G is the standard Gibbs energy difference between the transition state and the two reactants at temperature T and the standard state..
For the bi- and tri-molecular reactions, we have learned that the reaction rate constants should be s^(-1)M^(-1) and s^(-1)M^(-2). Why? And how can we still use the Eyring equation to obtain the reaction rate constant given the Gibbs energy of activation?
Note that delta G refers to the difference between the STANDARD Gibbs energy of the transitions state (TS) and that of the reactant(s). The STANDARD state for gases is 1 bar!
Therefore, the Eyring equation can still be used to calculate the rate constant k for the bi- and tri-molecular reactions as long as we use PRESSURE/(1 bar) consistently in the expression of the reaction rate.
For example, A + B -> C
Reaction rate = d[P(C)/1 bar)]/dt = -d[P(A)/1 bar)]/dt = -d[P(B)/1 bar)]/dt = k[P(A)/1 bar)][P(B)/1 bar)]
Note that [P(A)/1 bar)], [P(B)/1 bar)], [P(C)/1 bar)] are unit-less. Pressures are divided by 1 bar because the delta G are usually determined at 1 bar. GAMESS calculations also give the values of delta G at 1 bar by default.
Therefore, we can see the left-hand-side (lhs) of the equation is in s^(-1) and so is the rhs.
d[P(C)/1 bar)]/dt = k[P(A)/1 bar)][P(B)/1 bar)]
What if the users of the above equation prefer to use M (mol/L)?
Again, we can simply derive the equation in terms of concentrations.
d[P(C)/1 bar)]/dt = k[P(A)/1 bar)][P(B)/1 bar)]
d([C]RT/1 bar)/dt = k ([A]RT/1 bar) ([B]RT/1 bar)
d([C])/dt = k(RT/1 bar) [A] [B]
Therefore the rate constant of a bi-molecular reaction becomes the product of 3 terms:
the rate constant in s^(-1)M^(-1) = (kBT/h) e^(-delta G/RT) (RT/1 bar)
For a tri-molecular reaction, the rate constant = (kBT/h) e^(-delta G/RT) (RT/1 bar)^2
Note that many rate constants of bi-molecular reactions are in the units of (s, cm, molecule),
We simply need to find the value of (RT/1 bar) in (s, cm, molecule)
For example, at 298.15K,
(RT/1 bar) = (8.3145 * 298.15) / 100000 = 0.024790 m^3/mol = 4.1164 * 10^(-20) cm^3/molecule
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