Question

The following information is given for antimony at 1 atm:

boiling point = 1440 °C	Hvap(1440 °C) = 195 kJ/mol
melting point = 631 °C	Hfus(631 °C) = 19.6 kJ/mol
specific heat solid = 0.209 J/g°C
specific heat liquid = 0.259 J/g°C

What is H in kJ for the process of freezing a 45.3 g sample of liquid antimony at its normal melting point of 631 °C.

Answer 1

mass of sample = 45.3 g

normal melting point of 631 °C

Hfus(631 °C) = 19.6 kJ/mol

mole of Sb = 45.3 g / 121.75 g / mole = 0.3721 mole.

delta H = 0.3721 mole * 19.6 kJ/mole = 7.29 KJ

this is exothermic and so answer should be - 7.29 KJ.