Question

In solid NaCl, the equilibrium separation between neighboring Na+ and Cl- ions is 0.283 nm. Calculate the coulombic energy between Na+ and Cl???? at this distance. Give your answer in each of J, eV, and kJ/mol units.

Answer 1

Step-by-step explanation:

It is given that r = 0.283 nm. As 1 nm =
`10^(-9) m`.

Hence, 0.283 nm =
`0.283 * 10^(-9) m`

• Formula for coulombic energy is as follows.

`U_(coulomb) = -1.748 (e^(2))/(4 \pi \epsilon_(o) r)`

where, e =
`1.6 * 10^(-19)` C

`\epsilon_(o)` =
`8.85 * 10^(-12)`

`U_(coulomb) = -1.748 ((1.6 * 10^(-19)^(2))/(4 * 3.14 * 8.85 * 10^(-12) * 0.283 * 10^(-9))`

=
`1.423 * 10^(-18) J`

• As 1 eV =
  `1.6 * 10^(-19) J`

So, 1 J =
`(1 eV)/(1.6 * 10^(-19))`

Hence, U =
`(1.423 * 10^(-18) J)/(1.6 * 10^(-19) J)`

= 8.9 eV

• Also, 1 J =
  `(10^(-3) kJ)/(6.022 * 10^(23)mol)`

=
`1.67 * 10^(-27)` kJ/mol

Therefore, U =
`1.423 * 10^(-18) J * 1.67 * 10^(-27)` kJ/mol

=
`2.37 * 10^(-45) kJ/mol`