Molecular Simulation/Charge-Charge Interactions

All intramolecular and intermolecular interactions are the result of electrostatic interactions between charged particles. All molecules are comprised the three subatomic particles: protons, neutrons, and electrons. Neutrons do not carry a charge, but protons and electrons carry charges with equal magnitude but opposite sign. The magnitude of these charges are fixed. This value is elementary charge, e. By convention, protons are defined as having positive charges and electrons are defined as having negative charges. The magnitude of these charges have a constant value known as the elementary charge, e=1.602176565(35) × 10^-19 C. ε₀ is the vacuum permittivity constant, which is equal to 8.854187817... 10^-12 F/m (farads per metre).

The potential energy surface of the interaction of two charged particles, calculated using Coulomb's law.

The force between two charged particles due to this electrostatic interactions is,

Coulomb's law (force) $F(r)=\frac{-1}{4\pi {ϵ}_0}\frac{q_A{q}_B}{r_{AB}^2}$

In this equation, $r_{AB}$ is the distance between particles A and B. The charge of a particle is given by the variable q. A charge is a scalar quantity with a sign and a magnitude.

It is often more convenient to discuss intermolecular forces in terms of the potential energy of the interaction. The potential energy of the interaction of two charged particles, labelled A and B, can be determined by integrating the force experienced between the particles if they were moved from infinite separation where the intermolecular interaction is zero, to the distance ($r_{AB}$) they are actually separated by,

$𝒱(r)=\underset{\mathrm{\infty }}{\overset{r}{\int }}\frac{-1}{4\pi {ϵ}_0}\frac{q_A{q}_B}{r_{AB}^2}dr$

$=\frac{1}{4\pi {ϵ}_0}\frac{q_A{q}_B}{r_{AB}}{|}_{\mathrm{\infty }}^{r_{AB}}$

$=\left[\frac{1}{4\pi {ϵ}_0}\frac{q_A{q}_B}{r_{AB}}\right]-\left[\frac{1}{4\pi {ϵ}_0}\frac{q_A{q}_B}{\mathrm{\infty }}\right]=\frac{1}{4\pi {ϵ}_0}\frac{q_A{q}_B}{r_{AB}}$

Coulomb's law (potential energy) $𝒱(r)=\frac{1}{4\pi {ϵ}_0}\frac{q_A{q}_B}{r_{AB}}$

Ionic molecules have charge-charge Coulombic interactions. If the charges have the same sign (e.g., two positive ions), the interaction is repulsive. If the charges have the opposite sign, the interaction is attractive.

Example

What is the potential energy of the electrostatic interaction between an Na⁺ ion and a Cl^- ion separated by 5.00 Å?

Solution: These are both ions at intermediate distances, so their electrostatic interaction can be approximated using Coulomb's law. The charge-charge distance is given (5 Å). The charges, q_A and q_B, are the elementary charge of a proton/electron for the Na+ cation and the Cl- anion, respectively. These distances and charges must be converted to the SI units of m and C, respectively.

$𝒱(r)=\frac{1}{4\pi {ϵ}_0}\frac{q_A{q}_B}{r_{AB}}$

$=\frac{1}{4\pi {ϵ}_0}\frac{(1e)(-1e)}{5.00\text{Å}}$

$=\frac{1}{4\pi (8.8541\times 10^{-12}{\text{ F m}}^{-1})}\frac{(1.602\times 10^{-19}C)(-1.602\times 10^{-19}C)}{5.00\times 10^{-10}m}$

$=-4.61\times 10^{-19}\text{ J}$

$=-278\text{ kJ/mol}$

Further Reading

Wikipedia:Coulomb's law Template:BookCat