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# Calculate the entropy change of 75.2 g of water that freezes into ice at 273.15 K. ΔHfus for water is 6.01 kJ/mol.?

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- Steve4PhysicsLv 72 years agoBest Answer
Molar mass of water M = 2x1 + 16 = 18g

Amount of water: n = m/M = 75.2/18 = 4.178mol

Heat transferred when water freezes = -nΔHfus = -4.178mol x 6.01kJ/mol = -25.11kJ

(Negative as heat is removed, not added.)

Entropy change (ΔS) at temperature T is defined as:

ΔS = ΔQ/T = -25.11kJ / 273.15K = -0.0919 kJ/K = -91.9J/K

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