For a C-H bond with a stretching frequency 3000 cm^-1, what is the expected isotope (deuterium) effect k_H/k_D at 298 K for a full bond homolysis?

Given h = 6.63 × 10^-34Js, c = 3 × 10¹⁰cm/s, kg = 1.38 × 10^-23J/K

Concept:

The Kinetic Isotope Effect (KIE) is a phenomenon observed in chemistry and chemical kinetics where the rate of a chemical reaction is altered when one or more of the atoms in the reacting molecules are replaced with isotopes of the same element. This effect is particularly pronounced when hydrogen atoms (protons) are replaced with deuterium atoms or tritium atoms.

Explanation:

The isotope effect (k_H/k_D) for a C-H bond during full bond homolysis can be calculated using the following formula:

\({k_H \over k_D }= {e^{1/2 hc(1-1/√2) ṽ_H \over k_gT}}\)

Where:

k_H/k_D = Isotope effect for hydrogen/deuterium (H/D) substitution

h = Planck's constant = 6.63 × 10^-34 J·s

ṽ = wave number = 3000 cm^-1

k_g = Boltzmann's constant = 1.38 × 10^-23 J/K

T = Temperature in Kelvin

Putting the values in above equation

\({k_H \over k_D }= {e^{1/2 (6.63 × 10^{-34})(3 × 10^{10})(1-1/√2) (3000 ) \over (1.38 ×10^{-23})(298)}}\)

∴ \({k_H \over k_D}={e^{2.1}}\)

Conclusion:

The expected isotope effect for full bond homolysis of deuterium is e².