## Inversion rates and energies

### Half-Life Calculator

Half-life

Ea
kJ mol−1

ΔG
kJ mol−1

Use this calculator to observe variations in the half-life of a first-order process, starting from any combination of temperature T, activation enthalpy ΔH, and activation entropy ΔS for the process. Values for Ea and ΔG are also provided in the output. For the theory and equations underlying the calculator, see the following sections.

N.B. When the process of interest is racemisation, it should be borne in mind that a substance is racemic once 50% of the molecules have undergone inversion, i.e. the rate of racemisation is 2 x the rate of inversion.

### Half-life of a first-order reaction

For a first-order reaction of the form AB, the rate is given by:

$$-\frac{d[A]}{dt}=k{[A]} \tag{1} \\ \\$$

or

$$\frac{d[A]}{[A]}=-k.dt \tag{2} \\ \\$$

Integrating gives:

$$[A]_{t}=[A]_{0}e^{-kt} \tag{3} \\ \\$$

or

$$\ln \left (\frac{[A]_{t}}{[A]_{0}}\right)=-kt \tag{4} \\ \\$$

The half-life of a reaction is defined as the time required for the concentration of a reactant to reach 50% of its initial value. For a first-order process, the half-life is constant over time and independent of concentration. In calculating a half-life, equation (4) becomes:

$$\ln \left (\frac{0.5}{1}\right)=-kt \tag{5} \\ \\$$

The half-life is therefore given by

$$t_{1/2}=\frac{\ln 2}{k} \tag{6} \\ \\$$

### Evaluating inversion kinetics using the Arrhenius and Eyring equations

The kinetics of stereochemical inversion can be analysed in terms of the Arrhenius equation, which may be expressed as follows:

$$k = Ae^{-E_{a}/RT} \tag{7} \\ \\$$

or

$$\ln k=\ln A-\frac{E_{a}}{RT} \tag{8} \\ \\$$

where
k is the rate constant for the process (units of s−1 for a first-order rate constant)
T is the absolute temperature
Ea is the Arrhenius activation energy, the threshold energy that the reactant(s) require before reaching the transition state
A is the pre-exponential factor (or frequency factor)
R is the universal gas constant

Experimental values for Ea and A can be obtained from a plot of ln k against T−1. For a single-step unimolecular process, the enthalpy of activation, ΔH, is given by:

$$\Delta H^{\ddagger }=E_{a}-RT \tag{9} \\ \\$$

The free energy of activation, ΔG, is given by the Gibbs equation:

$$\Delta G^{\ddagger }=\Delta H^{\ddagger }-T\Delta S^{\ddagger } \tag{10} \\ \\$$

Rate constants are also accessible using transition state theory. This was developed in the 1930s by Eyring, Evans and Polanyi, and led to the emergence of the Eyring equation, one form of which is:

$$\ln k=\ln \left ( \frac{\kappa k_{B}T}{h} \right )-\frac{\Delta G^{\ddagger }}{RT} \tag{11} \\ \\$$

where
k is the rate constant for the process (units of s−1 for a first-order reaction)
T is the absolute temperature
ΔG is the Gibbs energy of activation
R is the universal gas constant
kB is Boltzmann's constant
h is Planck's constant
κ is the transmission coefficient (a probability factor, max. value 1.0)

The Eyring equation is similar in form to the Arrhenius equation — cf. equations (8) and (11) — and a comparison provides an opportunity to delve into the latter's pre-exponential factor, A, which is otherwise an empirical term. For this purpose, equation (11) can be modified by applying the Gibbs equation (10) and setting the transmission constant κ to unity, to provide a maximum value for the rate constant k in the form:

$$\ln k=\ln \left ( \frac{k_{B}}{h} \right )+\ln T+ \frac{-\Delta H^{\ddagger }}{RT}+\frac{\Delta S^{\ddagger }}{R} \tag{12} \\ \\$$

Similarly, for a first-order reaction the Arrhenius equation (8) can be expanded using equation (9) giving:

$$\ln k=\ln A-\frac{\left ( \Delta H^{\ddagger }+RT \right )}{RT} \tag{13} \\ \\$$

Combining equations (12) and (13) brings in the A term, thus:

$$\ln A-\frac{\left ( \Delta H^{\ddagger }+RT \right )}{RT}=\ln \left ( \frac{k_{B}}{h} \right )+\ln T+ \frac{-\Delta H^{\ddagger }}{RT}+\frac{\Delta S^{\ddagger }}{R} \tag{14} \\ \\$$

which simplifies to

$$\ln A=\ln \left ( \frac{k_{B}}{h} \right )+\ln T+1+\frac{\Delta S^{\ddagger }}{R} \tag{15} \\ \\$$

Using the Eyring equation thus allows the upper limit of an Arrhenius pre-exponential factor A to be represented by:

$$A=5.66397834\times 10^{10}Te^{\frac{\Delta S^{\ddagger }}{R}} \tag{16} \\ \\$$

The relationship shown in equation (16) was used in generating the table of half-lives shown in Section 4.1.2.

### Racemisation kinetics and half-life

For a unimolecular dynamic equilibrium of the form AB, the approach to equilibrium can be described in terms of the rate of disappearance of the starting material A:

$$-\frac{d[A]}{dt}=k_{f}{[A]}-k_{r}[B] \tag{17} \\ \\$$

where kf is the rate constant for the forward process and kr is the rate constant for the reverse process. Initially only A is present and [B] corresponds to the depletion in [A]. Equation (17) can therefore be rewritten as

$$-\frac{d[A]}{dt}=k_{f}{[A]}-k_{r}\left ( [A]_{0}-[A] \right ) \tag{18} \\ \\$$

which can be rearranged to

$$\frac{d[A]}{dt}=-\left ( k_{f}+k_{r} \right )[A]+k_{r}[A]_{0} \tag{19} \\ \\$$

The conditions at equilibrium provide a relationship between the rate constants:

$$K=\frac{k_{f}}{k_{r}}=\left ( \frac{[A]_{0}-[A]_{eq}}{[A]_{eq}} \right ) \tag{20} \\ \\$$

where K is the equilibrium constant and [A]eq is the equilibrium concentration of A. An expression for [A]0 is obtained by rearranging equation (20) to

$$k_{r}[A]_{0}=\left ( k_{f} +k_{r}\right )[A]_{eq} \tag{21} \\ \\$$

The equality in equation (21) can then be used to rewrite equation (19), thus:

$$\frac{d[A]}{dt}=-\left ( k_{f}+k_{r} \right )\left ( [A]-[A]_{eq} \right ) \tag{22} \\ \\$$

Integration then provides

$$\ln\left ( \frac{[A]-[A]_{eq}}{[A]_{0}-[A]_{eq}} \right )=-\left ( k_{f}+k_{r} \right )t \tag{23} \\ \\$$

In the special case of A and B being a pair of enantiomers, equation (23) simpifies – the forward and reverse rate constants must be equal and the equilibrium position is 1:1. When the concentrations of A and B are both halfway to their equilibrium values, and with the same forward and reverse rate constant k, equation (23) becomes

$$\ln\left ( \frac{0.75-0.5}{1-0.5} \right )=-2kt \tag{24} \\ \\$$

The half-life of the racemisation process is therefore given by

$$t_{1/2}=\frac{\ln 2}{2k} \tag{25} \\ \\$$