The Application of Coulombís Law to Chemistry:

Coulombís law is an equation describing the electric force between two charged objects.

Its original form is given as follows:

where F is the force one object exerts on the other, k = 1/4πεo is a constant equal to 9 x 10^9, each q is the electric charge of one of the two objects, and r is the distance between the two objects.

 

For chemistry, we only use a modified form of Coulombís law, derived through basic physics and calculus.  This formula is:

 

where E is the lattice energy between the two objects, and k, q, and r have the same meaning as in the original formula.

The most important application of Coulombís law is the energy between an atom with a positive charge and an atom with a negative charge.  A negative energy means that the particles attract each other.  A positive energy indicates that the particles repel each other.  As the atoms get closer, the energy between them grows greater as the force attracting the particles to each other also increases.

The equation suggests that two ions of the same charge will move as far away from each other as physically possible.  The equation also would suggest that two oppositely charged ions should be as close together as physically possible to achieve the smallest amount of energy (as nature tends to follow the path of least resistance).  However, at extremely short separations, two ions actually experience repulsive forces due to the strong positive charges of the ionsí nuclei Ė and nuclear forces are much stronger than electric forces, but only on very small scales.

 

The depth of the minimum in the potential energy curve above represents the bond strength, and the distance at the energy minimum is the bond length.  Using Coulomb's law and the bond length, one can predict the strength of an ionic bond.  Based on Coulombís law, we note that ionic compounds formed by ions with larger charges create stronger bonds and ionic compounds with shorter bond lengths form stronger bonds.

 

Hereís a worked example problem:

Find the energy of interaction between the pair of ions in NaCl using Coulombís Law given that the distance between the two ions is 0.25 nm.

 

Step one: Write out Coulombís Law

 

E = k(Q1Q2)/r

 

* E is the energy of interaction between a pair of ions and has units of joules

* k is a constant that equals 2.31 x 10-19J ∙ nm

* Q1 and Q2 are the numerical ion charges

* r is the distance between the ion centers in nanometers.

 

Step two: Plug in the given radius and corresponding charges to solve for E

 

E = 2.31 x 10-19J ∙ nm(+1)(-1)/0.25nm

E = -9.24 x 10-19 joules

And here are two more problems for you to try yourself:

 

  1. Given that the atomic radius of Na (plus) is .095 nm and Cl- is .181 nm, find the ionic energy of a single NaCl molecule.
  2. Two ions of with charges +1 and -1 are separated by a distance of 1.0 angstroms. By what factor is the potential energy increased when those ions are moved to 8.0 angstroms away?

The answers to these problems are at the bottom of the page.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Answers:


1.†††††††† Use Coulomb's law equation, then plug in values given, i.e. radius and known charges:
2.31x10^-19 J x nm ((-1)(+1)/(.095 + .181)nm= -8.37 x 10^-19 J

 

2. ††††††† Coulomb's law dictates that the energy should increase by 8 times.Since opposite charges attract, increasing the separation between the charges will increase the energy.Because the distance is increased by a factor of 8, the energy must therefore increase by a factor of 8.

 

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