Question

Question 4 The molar enthalpy of fusion of ice at 0 °C and 1 atm pressure is 6024 J mol. The molar heat capacities at constant pressure of ice and water are 37.65 J Kmol and 75.30 J Kmol respectively, and may be taken as constant over the temperature range 0 to -20°C. Consider 2 mole of liquid water supercooled to -20 °C, which is allowed to freeze isothermally from liquid to solid at-20 °C. (a) Construct an alternative reversible pathway for this process and use this to calculate the entropy change of the system for 2 mole of water.

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Answer 1

For a reversible process,

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G_universe =
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H_system - T
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S_system = 0.

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S_universe =
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S_system +
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S_surroundings    =
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S_system -
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H_system/T = 0

Thus,
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S_system = -
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S_surroundings

S_surroundings = (heat given off to the surroundings)_reversible/T

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At constant pressure, H = (heat given off to the surroundings)_reversible

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Hence, we need to determine the heat that is involved in every step of the process as well as the temperature at which this heat is transferred between the system and its surroundings.

For a constant temperature process, such as melting, it is straightforward since melting occurs at a constant temperature. Thus,

S_system = heat_melting/T_{freezing point}   = (6024 J/mole ice)( 2 mole ice) / 273.15 K = 44.10 J/K

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Heat here is positive since the system, ice, is taking heat from the surroundings. For the heating of the ice and the liquid water, the process is no longer at constant temperature. Thus, one needs to take an integral over the range of temperature:

S_system = (heat capacity)(change in temperature) / T   = C_p ln(T₂/T₁)

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Heating of ice: S_system = (37.68 J/K mole)(2 mole)(ln (273/253)) = 5.727 J/K

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Heating of liquid water: S_system = (75.30 J/K mole)(2 mole)(ln (293/273)) = 10.54 J/K

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Thus, total S_system = 44.10 + 5.727 + 10.54 = 49.8 J/K

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