phys 369 09 hw 6 solutions

phys 369 09 hw 6 solutions - Phys. 369, Fall 2009 Homework...

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1 Phys. 369, Fall 2009 Homework #6 Solutions 1. (10 points) 2. (14 points)
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2 3. (20 points)
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3
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4 4. (12 points) 5. (14 points)
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5 6. (30 points) a) The combustion reaction 81 8 2 2 2 2C H +25O 16CO +18H O ( 1 ) converts 2 moles of octane and 25 moles of oxygen into 16 moles of carbon dioxide and 18 moles of water. We obtain the standard heat of combustion (for two moles of octane) by subtracting the enthalpy of formation of the reactants from the enthalpy of formation of the products. So, () ( ) ( ) 00 0 0 22 8 1 82 16 CO 18 H O 2 C H 25 O 16 393.5 kJ 18 241.8 kJ 2 250.1 kJ 25(0 kJ) 10,148.2 kJ ff f f HH H H H Δ= + −− = (2) I used values for the heats of formation listed in the NIST Chemistry web book at http://webbook.nist.gov/chemistry/ and used the values for liquid octane, and gaseous water. The enthalpy change calculated in eq. (2) can be written as ( ) 2o c t a n e chem v thermal H UL U P V Δ =Δ ( 3 ) where ( ) 8 1 16 CO 18 H O 2C H 25 O chem chem chem UU U + + ( 4 ) is the change in the potential energy of the chemical bonds, ( ) octane v L is the latent heat of vaporization of liquid octane (per mole), and thermal U Δ is the change in the translational, vibrational, and rotational energy of the molecules. We are using standard heats, so every quantity is to be evaluated at standard conditions of 298 K and 1 bar (0.987 atm) pressure. The change in PV is approximately given by 9 9 8.315 298 2.48 kJ gas PV n RT RT Δ≈ Δ = = ×× = ( 5 )
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6 where the change in the number of moles of gas is 16 18 25 9 gas n Δ =+−= . The latent heat of vaporization of octane is listed in the NIST chemistry webbook as L v =41.6 kJ/mole. Finally, we need to evaluate thermal U Δ . We notice by comparing eq. (5) and eq. (2) that quantities of the order of nRT are going to be small compared to the heat of combustion. The thermal energies are of the order of nRT , so they are going to be small and so we won’t need them accurately. A rough approximation is ~3 758 kJ thermal atom Un R T = for both the initial and final state, where atom n = 102 is the number of moles of atoms. The difference in thermal U for the initial and final states should be a modest fraction of this number, roughly 100 kJ or so. So we can finally determine that the change in molecular bond energy is () ( ) 2 octane 10,148.2 kJ 2(41.6 kJ) 100 kJ 2.5 kJ 10,068 100 kJ chem v thermal UH L U P V Δ =Δ − =− ± + ± (6) where 100 kJ ±
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This note was uploaded on 11/06/2010 for the course PHY 369 taught by Professor Staff during the Fall '08 term at University of Texas.

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phys 369 09 hw 6 solutions - Phys. 369, Fall 2009 Homework...

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