Reactions study :
kinetiks and thermodynamics

  1. 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
K = e- G°r/RT = e- ( H°r -TS°r )/RT (2')


    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)

  1. Kinetics
    1. Introduction
Reagents k1

Products (4)
    k1 and k2 are the rate constants for this reversible reaction. We have the following relation between equilibrium and rate constants :
K = k1

    1. Arrhenius Equation
    Most of the chemical reactions, either in gaseous phase or in solution, follow the Arrhenius empirical equation for the rate constant k (k1):
k=A.e-Ea/RT (6)
    where : Two experimental measurements of k at two different temperatures let us calculate the activation energy and the frequence factor.

    1. 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 - TS (9)

    1. Calculations vs Experiment
    Ea 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 :

Ea = H + RT (10)
    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 Ea and A.

  1. Reaction competition : Kinetic and Thermodynamic selectivity
  2. If one reactant A can yield two different products B and C, we have two equilibria :
B 1
A 2
C (11)

    1. Thermodynamic selectivity
    In order to be able to say which equilibrium is most favoured, we have to compare the two constants:
 = e-G°/RT = e-(H°-TS°)/RT (12)
    where G° = G1 - G°2. This relation shows that the kinetic selectivity depends on the temperature.

    1. Kinetic selectivity
    Reagent A can yield two different products B and C. We can define two different activation energies for the two pathways. If k1 and k2 are the two rate constants, we can write
= A1
e-(Ea1-Ea2)/RT =  A1
e-(H1-H2)/RT (13)
    With two mesurearements of the ratio k1/k2 at two different temperatures, we can estimate the activation energy difference between the two concurrent reactions and the ratio A1/A2. Here is a way to compare experiment with calculation : Ea1-Ea2 = H1-H2
 = e-G/RT = e-(H-TS)/RT (14)
    where G = G1 - G2. This relation shows that the kinetic selectivity depends on the temperature. With (13) and (14), we can estimate the activation entropy difference :
S = S1-S2 = -Rln(A1/A2) (15)

    *Proof[1] of (10)

    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
k = c.K.T (16)
    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 :
k = c.æ.e-G/RT (17)
    We will now differentiate with respect to T the two expressions of k. (17) and (8) gives
    and (6) gives
    Comparing (18) and (19) you can immediatly prove (10).

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