Consider a molecule of LiCI in the gas phase at room temperature (300K). The ion

Question

Consider a molecule of LiCI in the gas phase at room temperature (300K). The ionic radii of Li^+ and CI^- are listed in Wikipedia as being 0.090 and 0.167 nm, respectively. These radii should allow you to estimate the separation at which the short-ranged, sharp repulsive interactions are balanced by the attractive electrostatic interactions. Estimate the energy (kJ/mol) for dissociating this molecule into its constituent ions. What is the probability of observing the ions separated by a distance of 1 nm, relative to the probability of them being in contact What is the corresponding probability of observing the ions separated by a distance of 1 nm, relative to the probability of them being in contact, IN WATER

Explanation / Answer

Ionic radius of Li+ = 0.090 nm

ionic radius of Cl- = 0.167 nm

(a). Distance between the centres of the two ions = 0.090 + 0.167 = 0.257 nm

= 2.57x10-10 m

According to electrostatic interactions, electrostatic potential energy between two charges is given as -

E = k q1*q2 / r

where k = 9x109 Nm2C-2.

q1 = 3e- and q2 = 17e-

E = 9x109 * 3e- * 17e- / (2.57x10-10)

= 9x109 * 51(e-)2 / (2.57x10-10)

= 9x109 * 51(1.6x10-19)2 / (2.57x10-10)

= 4.57x10-17 J

This is the energy required to break the interaction between 1 LiCl atom.

The energy required to break 1 mole atoms = 4.57x10-17 * 6.023x1023

E = 2.75x107 J/mol

E = 2.75x104 kJ/mol

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