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

Chemistry: Principles and Practice
3rd Edition
ISBN:9780534420123
Author:Daniel L. Reger, Scott R. Goode, David W. Ball, Edward Mercer
Publisher:Daniel L. Reger, Scott R. Goode, David W. Ball, Edward Mercer
Chapter5: Thermochemistry
Section: Chapter Questions
Problem 5.24QE
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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
Transcribed Image Text: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
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