Question

Calculate the zero point energy of^1H^127 I and^2 H^127 I given that the force constant is 291 N/m. If the potential energy minimum E_min is -294.7 kJ/mol for both cases what is the dissociation energies E_diss of each of these molecules? Using the results of Q2 assume that a chemical reaction depends on the dissociation of these molecules such that the rate constant k follows the Arrhenius formula k = A middot exp(-E_diss/RT) What is the ratio of the rate constants of these two molecules? (You can evaluate the ratio at T 300 K) This difference in the rate constants is known as the kinetic isotope effect.

Answer 1

2.

Assuming Harmonic oscillator models for HI, the zero point energy is related to the force constant k as follows

Eo

Where

2 6.626 x 103 kg m. S 1.054 x 1034 ka m2 S 2T 2T

k = force constant =

291 N m 291 kg s2

Lreduced mass

Now, before we calculate the zero point energy for the molecules, we must calculate their reduced mass

m(H)1.00 u = 1.66 x 10 2 kg

m(2 H) 2.00 u

m(127 I127 u

Hence,

т(Н) (12 Г) 1.00 и х 127 и х т 0.99 и дн127 I) 127 I т('Н) + m 1.00 и + 127 и и(н) — 0.99 х 1.66 х 10-2 kg3 1.647 х 10 2 kg

Similarly,

т("H) I) т(?Н) + m 2.00 и х 127 и х т 1.968 и н?н127 127 I 2.00 и + 127 и 7 и("н? ) — 1.968 х 1.66 х 10-2 kg — 3.268 х 10 2 kg

Now, the zero point energies can be calculated as

h k Eo(H127I) 2 H(H127 I 1.054 x 10-34 kg m2 s-1 291 kq s-2 S 127 1.647 x 10-27 kg 2 127 -1 35 kg m2 x 4.203 x 1014 2.215 x 1020 J -1 S Eo(H I) 5.27 x 10 20 2 Eo(H12 I)2.215 x 102 kg m2

Hence, the zero point energy of 1H127I in kJ per mole units can be calculated as

1 kJ Eg(H12I)2.215x10-20 JX103X6.022 x 1023 mol1 213.34 kJmol-1

Similarly for the other molecules

$E_0 (^2H^{127}I) = \frac{\hbar}{2} \sqrt\frac{k}{\mu(^2H^{127}I)} \\\Rightarrow E_0 (^2H^{127}I) = \frac{1.054 \times 10^{-34} \ kg \ m^2 \ s^{-1}}{2} \times \sqrt \frac{291 \ kg \ \ s^{-2}}{3.268 \times 10^{-27}\ kg} \\ \\ \Rightarrow E_0 (^2H^{127}I) = 5.27 \times 10^{-35} \ kg \ m^2 \ s^{-1}\times 2.984 \times 10^{14} \ s^{-1} \\ \Rightarrow E_0 (^2H^{127}I) = 1.572 \times 10^{-20} \ \ kg \ m^2 \ s^{-2} = 1.572 \times 10^{-20} \ J$

Now, the zero point energy in kJ/mol is

$E_0 (^2H^{127}I) = 1.572 \times 10^{-20 } \ J \times \frac{1 \ kJ }{10^3 \ J} \times 6.022 \times 10^{23} \ mol^{-1} \approx {\color{Red} 9.47 \ kJ \ mol^{-1}}$

Now, the dissociation energy  , minimum potential energy  and zero point energy, $E_0$ are related as

Hence, given that for both the molecules

The dissociation energies are

$E_{diss} (^1H^{127}I) = -294.7 \ kJ /mol + 13.34 \ kJ /mol = {\color{Red} -281.36 \ kJ/mol}$

$E_{diss}(^2H^{127}I) = -294.7 \ kJ /mol + 9.47 \ kJ /mol = {\color{Red} -285.23 \ kJ/mol}$

Note that dissociation energies are always positive quantities as it describes how much energy needs to be provided to break the bond into respective atoms.

Hence, the dissociation energies are

EdissH12 I) -281.36 kJ /mol

Ediss(2H12 I) -285.23 kJ/mol

3.

Now using the Arrhenius equation at T = 300 K

-281.36x103 J mol-/8.314 J K mol-x300 K k(H127 I k(2H127 I k(H127 I) k(2H127I) A exp(-Ediss RT) A exp(-Ediss/RT) 1.02 x 10-49 -285,23x103 J mol-1/8.314 J K-1 mol-1x300 K 2.16 x 10-50 4.71

Hence, the rate of dissociation of HI is about 4.71 times faster than DI.