Question

The molar enthalpy of fusion (ΔH°) of bromine is 10.570 kJ/mol, and the molar entropy of fusion (ΔS°) of bromine is 39.750 J/mol K. Using this information, estimate the freezing point of bromine. (You must answer in Kelvin.)

Answer 1

Answer:

265.91194 K

Explanation:

$\Delta H^{\circ}$ = Molar enthalpy of fusion = 10.57 kJ/mol

$\Delta S^{\circ}$ = Molar entropy of fusion = 39.75 J/mol

$T_f$ = Freezing point

Molar entropy of fusion is given by

$\Delta S^{\circ}=\frac{\Delta H^{\circ}}{T_f}\\\Rightarrow T_f=\frac{\Delta H^{\circ}}{\Delta S^{\circ}}\\\Rightarrow T_f=\frac{10.57\times 10^3}{39.75}\\\Rightarrow T_f=265.91194\ K$

The freezing point of bromine is 265.91194 K