Calculate the entropy change of 84.5 g of water that freezes into ice at 273.15 K. ΔHfus for water is 6.01 kJ?
If anyone could please write out the steps needed in order to answer this question, it would help a lot!
- hfshawLv 78 years agoBest Answer
Assuming that the water is freezing at 1 atm pressure (i.e., the pressure at which water freezes at 273.15K), then this process is occurring at the equilibrium freezing temperature. If the reaction is occurring at equilibrium then the Gibbs free energy change for the reaction:
H2O(liquid) -> H2O(solid)
is equal to zero.
You should know that:
ΔG = ΔH - T*ΔS
and at equilibrium,
ΔG = ΔH - T*ΔS = 0, so:
ΔS = ΔH/T
In this case (note that ΔH_fus must have units of 6.01 kJ/mol, not just kJ!):
ΔS = (6,010 J/mol)/(273.15 K) = 22.0 J/(mol*K)
The molecular mass of water is 18.02 gm/mol, so 84.5 gm of water is 4.69 moles. The entropy change of the 84.5 gm of water that freezes is therefore 22.0 J/(mol*K) * 4.69 moles = 103.2 J/K
- duroseauLv 43 years ago
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