T7-Answers - TUTORIAL#9 November 19th 2007 Solutions Chapter 7 Correction to Q 4 CHEM 1A03 1 From the course notes we had an expression for work w

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CHEM 1A03 TUTORIAL #9 November 19 th , 2007 Solutions – Chapter 7 Correction to Q 4 _____________________________________________________________________ 1. From the course notes, we had an expression for work: gases w P V n RT = ! " = !" From this, and our expressions for enthalpy and internal energy, we can re-write Δ H as: V U q w ! = + H U P V ! = ! + ! gases H U n RT !" = " + " Based on this expression, calculate the difference between Δ H and Δ U for the combustion of octane at 298K. To find Δ n gases RT, we need the balanced equation: C 8 H 18 (l) + 25/2 O 2 (g) 8 CO 2 (g) + 9 H 2 O(l) 2 2 25 9 8 ( ) ( ) 2 2 gases n mol CO mol O moles !" = # = # 2. Nitric Acid is produced by the Ostwald Process in the gas phase at elevated temperature and pressure, which is an industrially important chemical widely used in the fertilizer industry, as discussed in Chapter 15. Based on the balanced net reaction below, and the given heats of formation, all in kJ/mol, determine whether this overall process is exothermic or endothermic, and estimate the energy change per mole of HNO 3 formed. 12 NH 3 (g) + 21 O 2 (g) 8 HNO 3 + 4 NO(g) + 14 H 2 O(g) Δ H f o NH 3 (g): -45.9; HNO 3 (g): -133.9; H 2 O(g): -241.8; NO(g): +91.3 ( )( ) ( )( ) ( )( ) ( )( ) ( )( ) 8 133.9 / 4 91.3 / 12 45.9 / 21 0 / 14 241.8 / rxn mol kJ mol mol kJ mol H mol kJ mol mol kJ mol mol kJ mol ! " # + ! " $ = # # + % ( + # % ( [ ] [ ] 3 4091.2 550.8 3540.4 3.54 10 rxn kJ kJ H kJ kJ = ! ! ! " = ! = ! # 8moles HNO 3 produced per mole of HNO 3 Δ H rxn = -443 kJ/mol rxn rxn gases H U n RT ! "! = + ! ( ) 1 1 9 8.314 298 2 11.1 / mol Jmol K K kJ mol ! ! = ! " = !
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CHEM 1A03 TUTORIAL #9 November 19 th , 2007 Solutions – Chapter 7 Correction to Q 4 _____________________________________________________________________ 3. When treating a hypothermia victim, it is important to both raise the body temperature, and dry the body. Calculate the enthalpy change required to for a person to warm their body (mass 70kg) from 34 o C to 37 o C, and dry it, using body heat, by evaporating 500g of water from the skin surface (phase change). Heat Capacity of body ~ heat capacity of H 2 O = 4.18J/g, Molar Enthalpy of Vapourization for Water = 44.0 kJ/mol at 298K Part 1: q = m x specific heat x Δ T 1000 4.18 70.0 3 878 o o g J kg C kg g C kJ = ! ! ! " = Part 2: 2 2 1 44.0 500 1,220 18.02 / 1 molH O kJ g kJ g mol molH O ! ! = Total Enthalpy changed: (878 1,220) 2,100 H kJ kJ ! = + = 4. Calculate the total calorie content for each burger patty (no condiments, oil, bread, etc.) Assume an average energy content of 16.74 kJ/g for both protein & carbohydrates, and 37.66 kJ/g for fats. Which burger has the lowest total calorie content? Explain why. What is the percentage of the average
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This note was uploaded on 06/18/2008 for the course CHEM 1a03 taught by Professor Landry during the Spring '08 term at McMaster University.

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T7-Answers - TUTORIAL#9 November 19th 2007 Solutions Chapter 7 Correction to Q 4 CHEM 1A03 1 From the course notes we had an expression for work w

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