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6.75 When 1 mol of CS,(I) forms from its elements at 1 atm and 25°C, 89.7 kJ of heat is absorbed, and it takes 27.7 kJ to vaporize 1 mol of the liquid. How much heat is absorbed when 1 mol of CS2(g) forms from its elements at these conditions?

6.75 When 1 mol of CS,(I) forms from its elements at 1 atm and 25°C, 89.7 kJ of heat is absorbed, and it takes 27.7 kJ to vaporize 1 mol of the liquid. How much heat is absorbed when 1 mol of CS2(g) forms from its elements at these conditions?

Chemistry: Principles and Practice
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 5.86QE: One of the components of jet engine fuel is n-dodecane, C12H26(), which has a standard enthalpy of...
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6.84 Cop Cu2O B Given A. 6.85 Ace CH2(g) Given A -241.8 Figure P6.72 Figure P6.73 6.74 At a given set of conditions, 241.8 kJ of heat is released when 1 mol of H,O(g) forms from its elements. Under the same conditions, 285.8 kJ is released when 1 mol of H,O(l) forms from its elements. Find AH for the vaporization of water at these Conditions. Probler 6.86 TI current. tion tha 6.75 When 1 mol of CS,(1) forms from its elements at 1 atm and 25°C, 89.7 kJ of heat is absorbed, and it takes 27.7 kJ to vaporize 1 mol of the liquid. How much heat is absorbed when1 mol of CS2(g) forms from its elements at these conditions? 2F (a) Use (b) Us (1) Pb Problems in Context 6.76 Diamond and graphite are two crystalline forms of carbon. At 1 atm and 25°C, diamond changes to graphite so slowly that the enthalpy change of the process must be obtained indirectly. Using equations from the list below, determine AH for (2) SC Comp C(diamond)- C(graphite) CO2(8) 6.87 AH = -395.4 kJ (1) C(diamond) + O2(8) (2) 2CO2(g) (3) C(graphite) + O2(g) → CO2(8) (4) 2CO(8) : hydrc It is u 2CO(g) + O2(8) AH = 566.0 kJ AH = -393.5 kJ foun C(graphite) + CO2(g) AH=-172.5 kJ you e (a) V Standard Enthalpies of Reaction (AHn) (Sample Problems 6.10 and 6.11) gase (b) -94 (c) Concept Review Questions stea 6.77 What is the difference between the standard enthalpy of for- mation and the standard enthalpy of reaction? (d) fat; 6.78 How are AH? values used to calculate AHn? Enthalpy, H Enthalpy, H
Transcribed Image Text:6.84 Cop Cu2O B Given A. 6.85 Ace CH2(g) Given A -241.8 Figure P6.72 Figure P6.73 6.74 At a given set of conditions, 241.8 kJ of heat is released when 1 mol of H,O(g) forms from its elements. Under the same conditions, 285.8 kJ is released when 1 mol of H,O(l) forms from its elements. Find AH for the vaporization of water at these Conditions. Probler 6.86 TI current. tion tha 6.75 When 1 mol of CS,(1) forms from its elements at 1 atm and 25°C, 89.7 kJ of heat is absorbed, and it takes 27.7 kJ to vaporize 1 mol of the liquid. How much heat is absorbed when1 mol of CS2(g) forms from its elements at these conditions? 2F (a) Use (b) Us (1) Pb Problems in Context 6.76 Diamond and graphite are two crystalline forms of carbon. At 1 atm and 25°C, diamond changes to graphite so slowly that the enthalpy change of the process must be obtained indirectly. Using equations from the list below, determine AH for (2) SC Comp C(diamond)- C(graphite) CO2(8) 6.87 AH = -395.4 kJ (1) C(diamond) + O2(8) (2) 2CO2(g) (3) C(graphite) + O2(g) → CO2(8) (4) 2CO(8) : hydrc It is u 2CO(g) + O2(8) AH = 566.0 kJ AH = -393.5 kJ foun C(graphite) + CO2(g) AH=-172.5 kJ you e (a) V Standard Enthalpies of Reaction (AHn) (Sample Problems 6.10 and 6.11) gase (b) -94 (c) Concept Review Questions stea 6.77 What is the difference between the standard enthalpy of for- mation and the standard enthalpy of reaction? (d) fat; 6.78 How are AH? values used to calculate AHn? Enthalpy, H Enthalpy, H
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