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

Explanation:

It is given that r = 0.283 nm. As 1 nm = $10^{-9} m$.

Hence, 0.283 nm = $0.283 \times 10^{-9} m$

• Formula for coulombic energy is as follows.

             $U_{coulomb} = -1.748 \frac{e^{2}}{4 \pi \epsilon_{o} r}$

where,   e = $1.6 \times 10^{-19}$ C

            $\epsilon_{o}$ = $8.85 \times 10^{-12}$

          $U_{coulomb} = -1.748 \frac{(1.6 \times 10^{-19}^{2}}{4 \times 3.14 \times 8.85 \times 10^{-12} \times 0.283 \times 10^{-9}}$

                         = $1.423 \times 10^{-18} J$

• As 1 eV = $1.6 \times 10^{-19} J$

So,       1 J = $\frac{1 eV}{1.6 \times 10^{-19}}$

Hence,    U = $\frac{1.423 \times 10^{-18} J}{1.6 \times 10^{-19} J}$

                   = 8.9 eV

• Also,   1 J = $\frac{10^{-3} kJ}{6.022 \times 10^{23}mol}$

                = $1.67 \times 10^{-27}$ kJ/mol

Therefore, U = $1.423 \times 10^{-18} J \times 1.67 \times 10^{-27}$ kJ/mol

                     = $2.37 \times 10^{-45} kJ/mol$