Reactions study :
kinetiks and thermodynamics
 Thermodynamics
 Reagents

 Products
 (1)

A chemical reaction is an equilibrium between reactants and products. It is characterized by an equilibrium constant K :
 K =
 [ Products ]_{ eq} [ Reactants ]_{ eq}

 (2)

 K = e^{ G°r/RT} = e^{ ( H°r TS°r )/RT
}  (2')

where
 K^{‡}=k^{‡}/k^{‡} is the equilibrium constant
 G°_{r} is the reaction Gibbs standard free energy
 H°_{r} is the reaction enthalpy
 S°_{r} is the reaction entropy
The Heat of Formation (H°_{f}) of a compound is defined as the enthalpy of the reaction of elements in their standard states to form the compound. The standard states of each element is the most stable state of that element at 25° (298K). Hess' law (3) is the link between reaction enthalpy and heats of formation. For the reaction (1), the reaction enthalpy is:
 H°_{r} = H°_{f}(Products)  H°_{f}(Reactants)
 (3)

 Kinetics
 Introduction
 Reagents
 k_{1}
k_{2}
 Products
 (4)

k1 and k2 are the rate constants for this reversible reaction. We have the following relation between equilibrium and rate constants :
 Arrhenius Equation
Most of the chemical reactions, either in gaseous phase or in solution, follow the Arrhenius empirical equation for the rate constant k (k_{1}):
where :
 A is the frequence factor. It is correlated to the number of efficient collisions between molecules
 E_{a} is the Arrhenius Activation Energy
 R is the gas constant and T is the temperature
Two experimental measurements of k at two different temperatures let us calculate the
activation energy and the frequence factor.
 Eyring activated complex
We suppose that there is an equilibrium between reagents and an activated complex called the Transition State [ ]^{‡}
 Reagents
 k^{‡}
k^{ ‡}
 [^{ Transition State }]^{‡}

 Products
 (7)

For this equilibrium the thermodynamical laws give :
 G^{‡} = RTln(K^{‡})
 (8)

 G^{‡} = H^{‡}  TS^{‡}
 (9)

where
 K^{‡}=k^{‡}/k^{‡} is the equilibrium constant
 G^{‡} is the free activation enthalpy
 H^{‡} is the activation enthalpy
 S^{‡} is the activation entropy
 Calculations vs Experiment
E_{a} can be measured, because it is directly linked to the rate constant k. H^{‡} can be calculated by finding the transition state which is a first order saddle point on the energy map.
For Perfect Gases and under some approximations we can prove* the following result :
This will allow us to compare experiments and calculations, and check the pertinence of the models used for the computations. Two measures of the reaction rate at two different temperatures let us estimate E_{a} and A.
 Reaction competition : Kinetic and Thermodynamic selectivity
If one reactant A can yield two different products B and C, we have two equilibria :
 B
 1
 A
 2
 C
 (11) 
 Thermodynamic selectivity
In order to be able to say which equilibrium is most favoured, we have to compare the two constants:
 K_{1} K_{2}  ^{ = eG°/RT = e(H°TS°)/RT}
 (12)

where G° = G^{‡}_{1}  G°_{2}. This relation shows that the kinetic selectivity depends on the temperature.
 Kinetic selectivity
Reagent A can yield two different products B and C. We can define two different activation energies for the two pathways. If k_{1} and k_{2} are the two rate constants, we can write
 k_{1} k_{2}
 =
 A_{1} A_{2}
 ^{e(Ea1Ea2)/RT = }
 A_{1} A_{2}
 ^{e(H‡1H‡2)/RT}
 (13)

With two mesurearements of the ratio k_{1}/k_{2} at two different temperatures, we can estimate the activation energy difference between the two concurrent reactions and the ratio A_{1}/A_{2}. Here is a way to compare experiment with calculation : E_{a}^{1}E_{a}^{2} = H^{‡}_{1}H^{‡}_{2}
 k_{1} k_{2}  ^{ = eG‡/RT = e(H‡TS‡)/RT}
 (14)

where G^{‡} = G^{‡}_{1}  G^{‡}_{2}. This relation shows that the kinetic selectivity depends on the temperature. With (13) and (14), we can estimate the activation entropy difference :
 S^{‡} = S^{‡}_{1}S^{‡}_{2} = Rln(A_{1}/A_{2})
 (15)

k^{ ‡} is only dependent on the temperature : the transition state does not "see" the reagents and the products whereas k^{‡} is dependent on the free activation enthalpy. For perfect gases there is a simple proportional relation between k^{ ‡} and the temperature: k^{ ‡} = c.T. Then
The next hypothesis is the relation between k (the rate constant) and k^{‡} : k = æ.k^{‡}
with æ < 1. A fraction of the molecules that reach the transition state went down to the products. Then (6) becomes :
We will now differentiate with respect to T the two expressions of k. (17) and (8) gives
 dk dT  =  k T  .  (RT+H^{‡}) RT
 (18)

and (6) gives
Comparing (18) and (19) you can immediatly prove (10).