Question

Calculate the energy needed to dissociate 1.00 mol of crystalline RbCl into its gaseous ions if the Madelung constant for its structure is 1.7476 and the radii of ${\mathbf{\text{Rb}}}^{\mathbf{+}}$and${\text{Cl}}^{-}$ are $1.48\stackrel{°}{\text{A}}$and$1.81\stackrel{°}{\text{A}}$ , respectively. Assume that the repulsive energy reduces the lattice energy by 10% from the pure Coulomb energy.

Short answer

${\text{664kJmol}}^{\text{-1}}$

Step-by-step solution

The internuclear separation between rubidium and chlorine ions.

As the radii of ${\text{Rb}}^{\text{+}}$and ${\text{Cl}}^{-}$are$1.48\stackrel{\text{o}}{\text{A}}$ and $1.81\stackrel{\text{o}}{\text{A}}$, respectively.

Therefore;

$\begin{array}{c}{\text{R}}_{\text{o}}=1.48\stackrel{\text{o}}{\text{A}}+1.81\stackrel{\text{o}}{\text{A}}\\ =3.29\stackrel{\text{o}}{\text{A}}\\ =3.29\times {10}^{-10}\text{m}\end{array}$

The energy needed to dissociate 1.00 mol of crystalline RbCl into its gaseous ions.

Using the Madelung constant of 1.7476 for its structure;

$\text{latticeenergy=}\frac{{\text{N}}_{\text{A}}{e}^2\text{M}}{4\pi {\epsilon }_0{R}_0}$

$\begin{array}{l}\text{latticeenergy=}\frac{(6.02\times {10}^{23}mo{l}^{-1}){(1.602\times {10}^{-19})}^2\left(1.7476\right)}{4\left(3.14\right)(8.854\times {10}^{-12}{\text{C}}^2{\text{J}}^{-1}{\text{m}}^{-1})(3.29\times {10}^{-10}\text{m})}\\ =7.38\times {10}^5{\text{Jmol}}^{-1}\\ =738{\text{kJmol}}^{-1}\end{array}$

The 10% of the lattice energy is;

$\frac{\text{738}}{100}\times \text{10\%=73.8}$

As the repulsive energy reduces the lattice energy by 10% from the pure Coulomb energy.

$738{\text{-73.8=664kJmol}}^{-1}$