Entropy of activation

In chemical kinetics, the entropy of activation of a reaction is one of the two parameters (along with the enthalpy of activation) that are typically obtained from the temperature dependence of a reaction rate constant, when these data are analyzed using the Eyring equation of the transition state theory. The standard entropy of activation is symbolized ΔS^‡ and equals the change in entropy when the reactants change from their initial state to the activated complex or transition state (Δ = change, S = entropy, ‡ = activation).

Importance

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. The factor is needed because of the pressure dependence of the reaction rate. R′ = 8.3145×10⁻² (bar·L)/(mol·K).

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. 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.

Derivation

It is possible to obtain entropy of activation using Eyring equation. This equation is of the form $$k={\frac {\kappa k_{\mathrm {B} }T}{h}}e^{\frac {\Delta S^{\ddagger }}{R}}e^{-{\frac {\Delta H^{\ddagger }}{RT}}}$$ where:

• $k$ = reaction rate constant
• $T$ = absolute temperature
• $\Delta H^{\ddagger }$ = enthalpy of activation
• $R$ = gas constant
• $\kappa$ = transmission coefficient
• $k_{\mathrm {B} }$ = Boltzmann constant = R/N_A, N_A = Avogadro constant
• $h$ = Planck constant
• $\Delta S^{\ddagger }$ = entropy of activation

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