What is the total enthalpy change during the process in which 100.0 g of water at 50.0 °C is cooled to ice at -30.0 °C ? (Assume P= 1.0 atm). Note: COOLING is an EXOTHERMIC PROCESS!!! AH WILL BE NEGATIVE! Solution S.H. water = 4.18 J/g °C Temperature AHfus = 6.01 kJ/mol 50.0 °C 0.0 °C 1 -30.0 °C 2 3 S.H. ice = 2.03 J/g °C 100.0 g H₂O = 5.552 mol H₂O 18.01 g/mol ΔΗ, = SH x Υ x ΔΤ AH₁ = (4.18 J/g °C) (100.0 g)(0.0 - 50.0 °C) = - 20900 J = - 20.9 kJ Energy (J) Thus the total enthalpy change is: AH₂ = -AHfus XY AH₂ = -(6.01 kJ/mol)(5.552 mol) = -33.4 kJ ΔΗ, = SH x Υ x ΔΤ AH3 = (2.03 J/g °C)(100.0 g)(-30.0 -0.00 °C) = 6090 J = -6.09 kJ -20.9 kJ - 33.4 kJ - 6.09 kJ = - 60.4 kJ
Thermochemistry
Thermochemistry can be considered as a branch of thermodynamics that deals with the connections between warmth, work, and various types of energy, formed because of different synthetic and actual cycles. Thermochemistry describes the energy changes that occur as a result of reactions or chemical changes in a substance.
Exergonic Reaction
The term exergonic is derived from the Greek word in which ‘ergon’ means work and exergonic means ‘work outside’. Exergonic reactions releases work energy. Exergonic reactions are different from exothermic reactions, the one that releases only heat energy during the course of the reaction. So, exothermic reaction is one type of exergonic reaction. Exergonic reaction releases work energy in different forms like heat, light or sound. For example, a glow stick releases light making that an exergonic reaction and not an exothermic reaction since no heat is released. Even endothermic reactions at very high temperature are exergonic.

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