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I try to solve the following equations for some system solvated in the water. The goal is to obtain value in kcal/mol. Unfortunately, I cannot reach the desired output. May someone help me?


Equation


The variables are:

n = 1 (single electron exchanged)
rA = 3.68Å
rD = 2.90Å
R = 6.58Å
ε∞ = 5.2 F/m
ε0 = 78.408 F/m

My try:

\begin{gather} \lambda \ =\ ( ne)^{2}\left(\frac{1}{2r_{A}} +\frac{1}{2r_{B}} -\frac{1}{R}\right)\left(\frac{1}{\epsilon _{\infty }} -\frac{1}{\epsilon _{0}}\right) \notag\\ \notag\\ \lambda \ -\ reorganization\ energy\ [ ?\ kcal/mol] \notag\\ \notag\\ n\ -\ number\ of\ exchanged\ electrons\ [ =1] \notag\\ e\ -\ electron\ charge\ \left[ =\ 1.602\times 10^{-19} C\right] \notag\\ r_{A} -\ radius\ of\ the\ acceptor\ cavity\ [ =3.68Å] \notag\\ r_{D} -\ radius\ of\ the\ donor\ cavity\ [ =2.50Å ]\\ R\ -\ intermolecular\ distance\ [ =6.18Å ] \notag\\ \epsilon _{\infty } \ -\ optical\ dielectric\ constant\ [ =5.2] \notag\\ \epsilon _{0} \ -\ static\ dielectric\ constant\ [ =78.408] \notag\\ \notag\\ \mathbf{Conversion:} \notag\\ \notag\\ r_{A} \ =\ 3.68Å \ =0.368nm \notag\\ r_{D} \ =\ 2.50Å =0.250nm \notag\\ R=6.18Å =0.618nm \notag\\ \epsilon _{\infty } \ =\ 5.2\times 8.854\times 10^{-12}\frac{F}{m} =4.604\times 10^{-11}\frac{F}{m} \notag\\ \epsilon _{0} \ =\ 78.408\times 8.854\times 10^{-12}\frac{F}{m} =6.942\times 10^{-10}\frac{F}{m} \notag\\ \notag\\ \notag\\ \mathbf{Calculation} : \notag\\ \notag\\ \lambda \ =\ 2.566\times 10^{-38} \ [ C] \ \left(\frac{1}{0.736\ [ nm]} +\frac{1}{0.500\ [ nm]} -\frac{1}{0.618\ [ nm]}\right)\left(\frac{1}{4.604\times 10^{-11}\frac{F}{m}} -\frac{1}{6.942\times 10^{-10}\frac{F}{m}}\right) \notag\\ \lambda \ =\ 2.566\times 10^{-38} \ [ C] \ \left( 1.359\left[\frac{1}{nm}\right] +2\left[\frac{1}{[ nm]}\right] -1.618\left[\frac{1}{nm}\right]\right)\left(\frac{1}{4.604\times 10^{-11}\frac{F}{m}} -\frac{1}{6.942\times 10^{-10}\frac{F}{m}}\right) \notag\\ \notag\\ \lambda \ =\ \left( 2.566\times 10^{-38}\right) \ [ C] \ \times 1.741\left[\frac{1}{nm}\right] \times \left( 2.028\times 10^{10}\right)\left[\frac{m}{F}\right] \notag\\ \lambda \ =\ 9.06\ \times 10^{-28}\left[\frac{C\times m}{nm\times F}\right] \ \notag\\ \lambda \ =\ 9.06\ \times 10^{-28}\left[ 10^{9}\frac{C\times nm}{nm\times F}\right] \ \notag\\ \lambda \ =\ 9.06\ \times 10^{-19}\left[\frac{C}{F}\right] \ \notag\\ \notag \end{gather}

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  • 2
    $\begingroup$ Change your units to SI, nm not angstrom then you need to multiply $e$ by $1/(4\pi \epsilon_0))$ where $\epsilon_0$ is the permittivity of free space $8.8\times 10^{-12}$ F/m. The answer will be in J then you can convert. $\lambda$ is typically $1 $eV $\sim 8000$ cm$^{-1}$. The dielectric constants are relative numbers, water about $78$. $\endgroup$
    – porphyrin
    Nov 27, 2023 at 8:53
  • $\begingroup$ @porphyrin Thanks for the answer. I will change units to SI, but why shall I multiply electron charge? Also, 𝜆 is the reorganization energy in kcal/mol, following Marcus theory. $\endgroup$
    – farmaceut
    Nov 27, 2023 at 9:18
  • 1
    $\begingroup$ Welcome to CH SE site! Note that using photos/screenshots of text instead of typing text itself is highly discouraged. The image text content cannot be indexed nor searched for, nor can be reused in answers. Specifically handwritten scripts can be difficult to decipher. Consider copy/pasting or rewriting of essential parts. // Optionally, here are formatting guides for texts and formulas/equations/expressions. $\endgroup$
    – Poutnik
    Nov 27, 2023 at 9:26
  • $\begingroup$ @Poutnik Done as required $\endgroup$
    – farmaceut
    Nov 27, 2023 at 9:42
  • $\begingroup$ @porphyrin May you elaborate? I have created online on mathcha, would appreciate help. Been struggling with that for few days already... mathcha.io/editor/L4KX2TX8UkvcJXtQ5e8x6IQ9jyZPTL4kdnJu98Q942 $\endgroup$
    – farmaceut
    Nov 27, 2023 at 14:19

1 Answer 1

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In SI units then conversion. see Harrison et al Chemical Physics 116 (1987) 429-448 for equations

nm  = 1e-9
q   = 1.60218e-19
eps0= 8.854e-12
rA  = 3.68/10*nm
rD  = 2.90/10*nm
R   = 6.58/10*nm
einf= 5.2 
e0  = 78.408 
lam= (q**2/(4*np.pi*eps0))*(1/(2*rA)+1/(2*rD)-1/R)*(1/einf-1/e0)
print('{:s}{:g}{:s}{:g}{:s}{:g}{:s}'.format('lambda ', lam, 'J or 
',lam*6.022e20,'kJ/mol or ', lam*6.022e20/4.18, ' kcal/mol') ) 

lambda 6.47513e-20J or 38.9932kJ/mol or 9.32852 kcal/mol

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  • $\begingroup$ Thank you ! I will try to decipher this now :) $\endgroup$
    – farmaceut
    Nov 27, 2023 at 17:06
  • $\begingroup$ And basically - if there are more electrons exchanged I just and only need to multiply q^2 by the number of electrons exchanged. Nowhere else? $\endgroup$
    – farmaceut
    Nov 27, 2023 at 17:56
  • $\begingroup$ I'm not sure about more that one electron. If two this could mean from the same orbital then the reorganisation energy one could argue would be the same as potentials are the same, from different orbitals then its probably a 2 -step process with different $\lambda$ anyway. BTW the code above is in python. $\endgroup$
    – porphyrin
    Nov 28, 2023 at 14:52

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