Question

Calculate the lattice energy of TlBr(s) using the following thermodynamic data (all data is in kJ/mol). Note that the data given has been perturbed, so looking up the answer is probably not a good idea.

Tl(s)	ΔHsublimation = 161 kJ/mol
Tl(g)	IE = 569 kJ/mol
Br-Br(g)	DBr-Br = 173 kJ/mol
Br(g)	EA = -345 kJ/mol
TlBr(s)	ΔH°f = -193 kJ/mol

_____________Kj/mol

Do you expect this value to be larger or smaller than the lattice energy of TlCl(s)? _________

Answer 1

Given Thermodynamic Data (all in kJ/mol):

• ΔH°sublimation (Tl): 161

• Ionization Energy (IE) of Tl(g): 569

• Bond Dissociation Energy (Br-Br): 173

• Electron Affinity (EA) of Br(g): -345

• ΔH°f (TlBr(s)): -193

• 
  

Steps to Calculate Lattice Energy ($U$):

Use the Born-Haber Cycle for TlBr(s):

$$\mathrm{\Delta }{H}_f^{\circ }=\mathrm{\Delta }{H}_{\text{sub}}+\text{IE}+\frac{1}{2}{D}_{\text{Br-Br}}+\text{EA}+U$$

Rearrange to solve for $U$:

$$U=\mathrm{\Delta }{H}_f^{\circ }-(\mathrm{\Delta }{H}_{\text{sub}}+\text{IE}+\frac{1}{2}{D}_{\text{Br-Br}}+\text{EA})$$

Plug in Values:

$$U=-193-(161+569+\frac{1}{2}\times 173+(-345))$$$$U=-193-(161+569+86.5-345)$$$$U=-193-(471.5)=-664.5\text{kJ/mol}$$

Answer:

• Lattice Energy of TlBr(s): −665 kJ/mol (rounded to 3 sig figs).

• 
  

• Comparison to TlCl(s): The lattice energy of TlBr(s) is smaller (less negative) than TlCl(s) because Br⁻ is larger than Cl⁻, reducing electrostatic attraction.