Entropy of activation

Entropy of activation determines the preexponential factor A of the Arrhenius equation for temperature dependence of reaction rates. The relationship depends on the molecularity of the reaction:

• for reactions in solution and unimolecular gas reactions

A = (ek_BT/h) exp(ΔS^‡/R),

• while for bimolecular gas reactions

A = (e²k_BT/h) (RT/p) exp(ΔS^‡/R).

In these equations e is the base of natural logarithms, h is the Planck constant, k_B is the Boltzmann constant and T the absolute temperature. R' is the ideal gas constant in units of (bar·L)/(mol·K). The factor is needed because of the pressure dependence of the reaction rate. R' = 8.3145 × 10⁻² (bar·L)/(mol·K).[1]

The value of ΔS^‡ provides clues about the molecularity of the rate determining step in a reaction, i.e. the number of molecules that enter this step.[2] Positive values suggest that entropy increases upon achieving the transition state, which often indicates a dissociative mechanism in which the activated complex is loosely bound and about to dissociate. Negative values for ΔS^‡ indicate that entropy decreases on forming the transition state, which often indicates an associative mechanism in which two reaction partners form a single activated complex.[3]

It is possible to obtain entropy of activation using Eyring equation. This equation is of the form

$${\displaystyle k={\frac {\kappa k_{\mathrm {B} }T}{h}}e^{\frac {\Delta S^{\ddagger }}{R}}e^{-{\frac {\Delta H^{\ddagger }}{RT}}}}$$

where:

• ${\displaystyle k}$ = reaction rate constant
• ${\displaystyle T}$ = absolute temperature
• ${\displaystyle \Delta H^{\ddagger }}$ = enthalpy of activation
• ${\displaystyle R}$ = gas constant
• ${\displaystyle \kappa }$ = transmission coefficient
• ${\displaystyle k_{\mathrm {B} }}$ = Boltzmann constant = R/N_A, N_A = Avogadro constant
• ${\displaystyle h}$ = Planck's constant
• ${\displaystyle \Delta S^{\ddagger }}$ = entropy of activation

This equation can be turned into the form

$${\displaystyle \ln {\frac {k}{T}}={\frac {-\Delta H^{\ddagger }}{R}}\cdot {\frac {1}{T}}+\ln {\frac {\kappa k_{\mathrm {B} }}{h}}+{\frac {\Delta S^{\ddagger }}{R}}}$$

The plot of ${\displaystyle \ln(k/T)}$ versus ${\displaystyle 1/T}$ gives a straight line with slope ${\displaystyle -\Delta H^{\ddagger }/R}$ from which the enthalpy of activation can be derived and with intercept ${\displaystyle \ln(\kappa k_{\mathrm {B} }/h)+\Delta S^{\ddagger }/R}$ from which the entropy of activation is derived.