Sample Exercise 10.3 Calculating AH for Temperature and Phase Changes Calculate the total enthalpy change upon converting 1.00 mol of ice at -25 °C to water vapor (steam) at 125 °C under a constant pressure of 1 atm. The specific heats of ice, water, and steam are 2.03 J/g-K, 4.18 J/g-K, and 1.84 J/g-K, respectively. For H₂O, AHfus = 6.01 kJ/mol and AHvap = 40.67 kJ/mol. Solution Temperature (°C) 125 100 75 50 0 -25 BC Ice D Water vapor. Liquid water and vapor (vaporization) Liquid water Ice and liquid water (melting) F E Heat added (each division corresponds to 4 kJ) Format for AB, CD, or EF: AH = SH x Y x AT ** Format for BC or DE: AH = AH X Y ** **(where Y = moles or grams) AB: AH = (1.00 mol) (18.0 g/mol) (2.03 J/g-K) (25 K) = 914J = 0.91 kJ BC: AH = (1.00 mol)(6.01 kJ/mol) = 6.01 kJ CD: AH = (1.00 mol) (18.0 g/mol)(4.18 J/g-K)(100 K) = 7520 J = 7.52 DE: AH = (1.00 mol)(40.67 kJ/mol) = 40.7 kJ EF: AH = (1.00 mol)(18.0 g/mol) (1.84 J/g-K) (25 K) = 830 J= 0.83 kJ Thus, the total enthalpy change for the process is: AH = 0.91 kJ +6.01 kJ + 7.52 kJ + 40.7 kJ + 0.83 kJ = 56.0 kJ NOTE: Practice how to do this Exercise!

Chemistry
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ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
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Determine the enthalpy of a heating process for a pure substance at P = 1 atm. Equations and constants supplied.

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Sample Exercise 10.3 Calculating AH for Temperature and Phase Changes
Calculate the total enthalpy change upon converting 1.00 mol of ice at -25 °C to water
vapor (steam) at 125 °C under a constant pressure of 1 atm. The specific heats of ice,
water, and steam are 2.03 J/g-K, 4.18 J/g-K, and 1.84 J/g-K, respectively. For H₂O, AHfus
= 6.01 kJ/mol and Hap = 40.67 kJ/mol.
Solution
Temperature (°C)
125
100
75
50
25
0
-25
BC
Ice
D
Water vapor.
Liquid water and vapor
(vaporization)
Liquid water
Ice and liquid water (melting)
F
E
A
Heat added (each division corresponds to 4 kJ)
Format for AB, CD, or EF:
AH = SH X Y x AT **
Format for BC or DE:
ΔΗ = ΔΗ x Υ **
**(where Y = moles or grams)
|AB: AH = (1.00 mol) (18.0 g/mol) (2.03 J/g-K)(25 K) = 914 J = 0.91 kJ
BC: AH = (1.00 mol) (6.01 kJ/mol) = 6.01 kJ
CD: AH = (1.00 mol) (18.0 g/mol) (4.18 J/g-K) (100 K) = 7520 J = 7.52 kJ
DE: AH = (1.00 mol)(40.67 kJ/mol) = 40.7 kJ
EF: AH = (1.00 mol)(18.0 g/mol)(1.84 J/g-K) (25 K) = 830 J = 0.83 kJ
Thus, the total enthalpy change for the process is:
=
AH = 0.91 kJ + 6.01 kJ + 7.52 kJ + 40.7 kJ + 0.83 kJ :
NOTE: Practice how to do this Exercise!
56.0 kJ
Transcribed Image Text:Sample Exercise 10.3 Calculating AH for Temperature and Phase Changes Calculate the total enthalpy change upon converting 1.00 mol of ice at -25 °C to water vapor (steam) at 125 °C under a constant pressure of 1 atm. The specific heats of ice, water, and steam are 2.03 J/g-K, 4.18 J/g-K, and 1.84 J/g-K, respectively. For H₂O, AHfus = 6.01 kJ/mol and Hap = 40.67 kJ/mol. Solution Temperature (°C) 125 100 75 50 25 0 -25 BC Ice D Water vapor. Liquid water and vapor (vaporization) Liquid water Ice and liquid water (melting) F E A Heat added (each division corresponds to 4 kJ) Format for AB, CD, or EF: AH = SH X Y x AT ** Format for BC or DE: ΔΗ = ΔΗ x Υ ** **(where Y = moles or grams) |AB: AH = (1.00 mol) (18.0 g/mol) (2.03 J/g-K)(25 K) = 914 J = 0.91 kJ BC: AH = (1.00 mol) (6.01 kJ/mol) = 6.01 kJ CD: AH = (1.00 mol) (18.0 g/mol) (4.18 J/g-K) (100 K) = 7520 J = 7.52 kJ DE: AH = (1.00 mol)(40.67 kJ/mol) = 40.7 kJ EF: AH = (1.00 mol)(18.0 g/mol)(1.84 J/g-K) (25 K) = 830 J = 0.83 kJ Thus, the total enthalpy change for the process is: = AH = 0.91 kJ + 6.01 kJ + 7.52 kJ + 40.7 kJ + 0.83 kJ : NOTE: Practice how to do this Exercise! 56.0 kJ
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
AH fus = 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 AH fus = 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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