Supercooled water is liquid water that has been cooled below its normal freezing

Question

Supercooled water is liquid water that has been cooled below its normal freezing point. This state is thermodynamically unstable and tends to freeze into ice spontaneously. Suppose we have 2.0 moles of supercooled water turning into ice at -10 degree C and 1.0 atm. Calculate the values of Delta Ssys, DeltaSsurr, and Delta Suniv for this process. The values of water and ice for the temperature range between 0 and 10 degree C are 75.3 J K-1mol1 and 37.7 J K-1mol-1, respectively, and the molar heat of fusion of water is 6.01 kJ mol-1.

Explanation / Answer

deltaS(univ) = deltaS(sys) + deltaS(surr)

deltaS(sys) = deltaS(products) - deltaS(reactants)

We have,

H2O(water) ----> H2O(ice)

deltaS(system) = 2 x 37.7 - 2 x 75.3 = -75.2 J/K

deltaS(surr) = -deltaH(fus)/Teq

With,

deltaH(fus) of water = 6.01 kJ/mol = 6010 J/mol

Teq = -10 oC = 273 - 10 = 263 K

Feed values,

deltaS(surr) = -6.01/263

= -0.023 J/K.mol

= -0.023 x 2

= -0.046 J/K

Feed values in the above equation find deltaS(univ),

deltaS(univ) = -75.2 - 0.046 = -75.25 J/K

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