Use standard enthalpies of formation to calculate the standard change in enthalpy for the melting of ice. kj/mol use this value to calculate the mass of ice required to cool 480. ml of water from room temperature (25.0°c) to 0.0 °c. g

To use standard enthalpies of formation to calculate the standard change in enthalpy for the melting of ice. kJ/mol

∆Hf° = standard heat of formation  can be use to claculate the std enthalpy change as follows;

∆Hf°reaction = ∆Hf° p - ∆Hf° r

if the reaction is H₂O (s) H₂O(l)

But

∆Hf°reaction = ∆Hf° H₂O(l)  - ∆Hf° H₂O (s), which were -285kJ/mol and -291.8kJ/mol (obtained from the back of chemical process analysis text)

∆Hf°reaction = -285.8 kJ/mole - (-291.8 kJ/mole) = 6.0 kJ/mole = 6000 J/mole

heat gained by ice melting = heat lost by cooling water

(mass x ∆Hf°) ice = (mass x Cp x dT) liquid water

mass ice = (mass x Cp x dT) liquid water / (∆Hf°) ice

mass ice = 480 mL x 1.00 g/mL x 4.18 J/gC x (25-0 C) / 6000 J/mole

= 5.66 mole ice = 5.66 mole x 18.0 g /mole = 150 g  of ice required to cool the 480mL of water.