Question

Calculate the potential energy of interaction between Na and Cl separated by 1 nm in water at 298K. Dielectric constant of water is 78.54.

a) 2.83 x 10⁻¹⁸ J
b) 2.31 x 10⁻¹⁸ J
c) 4.63 x 10⁻¹⁸ J
d) 9.26 x 10⁻¹⁸ J

Answer 1

The potential energy of interaction between Na and Cl separated by

`1 nm` in water at
`298K` can be calculated using Coulomb's law and the given dielectric constant. The potential energy is

approximately
`-1.15 x 10⁻¹⁸ J`.

Step-by-step explanation:

The potential energy of interaction between Na and Cl separated by

`1 nm` in water can be calculated using Coulomb's law. Coulomb's law states that the potential energy between two charged particles is given by ,

the equation
`U = (k*q1*q2)/(r*d)` , where k is the Coulomb's constant,
`q1 and q2` are the charges of the particles,
`r` is the distance between them, and d is the dielectric constant of the medium. In this case, the charges of
`Na and Cl` ions are
`+1 and -1` , respectively.

The distance between them is
`1 nm`, which is equal to
`1e-9 m`. The dielectric constant of water is
`78.54`. Plugging in these values, we get:

`U = (9e9 * 1 * -1) / (1e-9 * 78.54)`

`U = -1.15e-18 J`

The potential energy of interaction between
`Na and Cl` separated by

`1 nm` in water at
`298K` is,

approximately
`-1.15 x 10⁻¹⁸ J.`