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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 125 Format for AB, CD, or EF: AH = SH x Y x AT ** 100 Temperature (°C) 75 25 50 50 D Format for BC or DE: F Water vapor AH = AH x Y** E Liquid water and vapor (vaporization) Liquid water **(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 25 25 B C 0 A -25 Ice and liquid water (melting) Ice 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 Heat added (each division corresponds to 4 kJ) Thus, the total enthalpy change for the process is: EF: AH (1.00 mol)(18.0 g/mol)(1.84 J/g-K)(25 K) = 830 J = 0.83 kJ AH = 0.91 kJ + 6.01 kJ + 7.52 kJ + 40.7 kJ + 0.83 kJ = 56.0 kJ

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 125 Format for AB, CD, or EF: AH = SH x Y x AT ** 100 Temperature (°C) 75 25 50 50 D Format for BC or DE: F Water vapor AH = AH x Y** E Liquid water and vapor (vaporization) Liquid water **(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 25 25 B C 0 A -25 Ice and liquid water (melting) Ice 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 Heat added (each division corresponds to 4 kJ) Thus, the total enthalpy change for the process is: EF: AH (1.00 mol)(18.0 g/mol)(1.84 J/g-K)(25 K) = 830 J = 0.83 kJ AH = 0.91 kJ + 6.01 kJ + 7.52 kJ + 40.7 kJ + 0.83 kJ = 56.0 kJ

Chemistry & Chemical Reactivity
Chemistry & Chemical Reactivity
9th Edition
ISBN:9781133949640
Author:John C. Kotz, Paul M. Treichel, John Townsend, David Treichel
Publisher:John C. Kotz, Paul M. Treichel, John Townsend, David Treichel
Chapter5: Principles Of Chemical Reactivity: Energy And Chemical Reactions
Section: Chapter Questions
Problem 109SCQ
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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 125 Format for AB, CD, or EF: AH = SH x Y x AT ** 100 Temperature (°C) 75 25 50 50 D Format for BC or DE: F Water vapor AH = AH x Y** E Liquid water and vapor (vaporization) Liquid water **(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 25 25 B C 0 A -25 Ice and liquid water (melting) Ice 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 Heat added (each division corresponds to 4 kJ) Thus, the total enthalpy change for the process is: EF: AH (1.00 mol)(18.0 g/mol)(1.84 J/g-K)(25 K) = 830 J = 0.83 kJ AH = 0.91 kJ + 6.01 kJ + 7.52 kJ + 40.7 kJ + 0.83 kJ = 56.0 kJ
Transcribed Image Text: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 125 Format for AB, CD, or EF: AH = SH x Y x AT ** 100 Temperature (°C) 75 25 50 50 D Format for BC or DE: F Water vapor AH = AH x Y** E Liquid water and vapor (vaporization) Liquid water **(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 25 25 B C 0 A -25 Ice and liquid water (melting) Ice 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 Heat added (each division corresponds to 4 kJ) Thus, the total enthalpy change for the process is: EF: AH (1.00 mol)(18.0 g/mol)(1.84 J/g-K)(25 K) = 830 J = 0.83 kJ AH = 0.91 kJ + 6.01 kJ + 7.52 kJ + 40.7 kJ + 0.83 kJ = 56.0 kJ
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