Bell–Evans–Polanyi principle

Consider the reaction

${\displaystyle AB+C\to A+BC.}$

The system is assumed to have two degrees of freedom: r_AB, the distance between atoms A and B, and r_BC, the distance between atoms B and C. The distance between A and C is assumed to be fixed such that

${\displaystyle r=r_{\text{AB}}=const-r_{\text{BC}}}$

As the A—B bond stretches, the energy of the system increases up to the activation energy associated with the transition state, whereupon the bond breaks. The energy then decreases as the B—C bond is formed. Evans and Polanyi approximated the two energy functions between reactants, the transition state, and the products by two straight lines (with slopes m₁ and m₂ respectively) that intersect at the transition state.

For the AB molecule, the energy is given as a function of bond distance r:

${\displaystyle E_{\text{AB}}(r)=m_{1}(r-r_{1}).}$

(1)

At the transition state, r = r^‡ and E = E_a. Therefore, we can write that

${\displaystyle E_{\text{a}}=m_{1}(r^{\ddagger }-r_{1}),}$

(2)

which rearranges to give

${\displaystyle r^{\ddagger }={\frac {E_{\text{a}}}{m_{1}}}+r_{1}.}$

(3)

For the BC molecule, a similar expression of energy as a function of r is given by

${\displaystyle E_{\text{BC}}(r)=m_{2}(r-r^{\ddagger })+E_{\text{a}}.}$

(4)

The overall enthalpy change ΔH of the system can thus be expressed as

${\displaystyle \Delta H=m_{2}(r_{2}-r^{\ddagger })+E_{\text{a}}.}$

(5)