Chapter 8 - 8.8 (a)The quantities H vap and S vap can be...

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8.8 (a) The quantities vap vap and H S ∆ ° ∆ ° can be calculated using the relationship vap vap 1 ln H S P R T R ∆ ° ∆ ° = - + Because we have two temperatures with corresponding vapor pressures (we know that the vapor pressure = 1 atm at the boiling point), we can set up two equations with two unknowns and solve for vap vap and H S ∆ ° ∆ ° . If the equation is used as is, P must be expressed in atm, which is the standard reference state. Remember that the value used for P is really activity that, for pressure, is P divided by the reference state of 1 atm so that the quantity inside the ln term is dimensionless. vap 1 1 vap vap 1 1 vap 8.314 J K mol ln 1 311.6 K 13 Torr 8.314 J K mol ln 760 Torr 227.94 K H S H S - - - - ∆ ° × = - + ∆ ° ∆ ° × = - + ∆ ° which give, upon combining terms, 1 1 1 vap vap 1 1 1 vap vap 0J K mol 0.003209K 33.9 J K mol 0.004 3871K H S H S - - - - - - = - × ∆ ° + ∆ ° - = - × ∆ ° + ∆ ° Subtracting one equation from the other will eliminate the vap S ∆ ° term and allow us to solve for vap : H ∆ ° 1 1 1 vap 1 vap 33.9 J K mol 0.001178 K 28.8 kJ mol H H - - - - + = + × ∆ ° ∆ ° = + (b) We can then use vap H ∆ ° to calculate vap S ∆ ° using either of the two equations: 1 1 vap 1 1 vap 1 1 1 1 vap 1 1 vap 0 0.003 209 K ( 28800 J mol ) 92.4 J K mol 33.9 J K mol 0.004 3871K ( 28 800 J mol ) 92.4 J K
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This note was uploaded on 03/09/2009 for the course CHEM CHEM 6C taught by Professor Hoeger during the Winter '08 term at UCSD.

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Chapter 8 - 8.8 (a)The quantities H vap and S vap can be...

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