calculate the entropy change when 36.0 g of ice melts at 273 K and 1 atm.?
calculate the entropy change when 36.0 g of ice melts at 273 K and 1 atm.
- TimothyLv 68 years agoFavorite Answer
(36.0 g)/(18.0 g/mol H₂O) = 2 mol H₂O
∆H(fus) = 6.04 kJ/mol
∆S(fus) = ∆H(fus)/Tm = (6040 J/mol)/(273 K) = 21.99 J/mol•K
⇒ ∆S = 2 × 21.99 J/mol•K = 43.98 J/K
- Anonymous4 years ago
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- Simonizer1218Lv 78 years ago
If you supply additional information, this will be easy to answer We need values for heat of fusion, etc.
- JohnLv 78 years ago
The molar heat of fusion for ice is 6.02-kJ/mol - that is, 6.02-kJ of thermal energy is absorbed for each mole of water that melts.
36.0-g x 1 mol H₂O / 18.0-g H₂O x 6.02kJ/mol = 12.0-kJ
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- Johnny DeppLv 58 years ago
I believe it is 0.5 grams after the ice melts.