Midterm3-Solutions2003

Midterm3-Solutions20 - Midterm#3 Chemistry 4511 Physical Chemistry for Engineers Helpful Information e = K[A(l K[A PV=nRT 1 mole = 6.022 x

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Unformatted text preview: Midterm #3 / Chemistry 4511 / Physical Chemistry for Engineers Helpful Information: e = K[A]/(l + K[A]) PV=nRT 1 mole = 6.022 x 1023molecules 1000 mL = 1 dm3 h = 6.626 X 10-34 Js 1 atm = 760 Torr 1 bar = 750 Torr R = 0.08206 atm dm3 Klmorl R = 8.3145 J Kl mol-1 kB= 1.381 x 10-23J K-1 1 J = 1 kg m2 S-2 1 eV = 1.602 x 10-19J 1 amu = 1.66 x 10-27kg 1 amu x Avogadro's Number = Mass of mole in grams c =VA ~E = hv hv = (l/2)mv2 + <I> A= hip p=mv Energyoperator= i(hl2n)d/dt Momentumoperator =-i(h/2n)d/dx E = n2h2/8ma2 mass of electron, m=9.1 09 x 10-31kg 1.0 A= 1.0 X 10-10 m c = 3.0 X 108 m sol- - - - - --- - - - - - - - - - - - - - - - --- - - - - - - - - - - - - - - - - --- - - - - - - - --- --- - - - - - - - - - - - - - - --- - - --- - - ------ - - - - - - - - - - - - - - - - --- - --- - - -- 1. The following data for the adsorption of N2 on alumina (AhO3) obey the Langmuir adsorption isotherm equation. peN 2) in bar N2 adsorbed in mmol/g of AhO3 0.0422 83.1 0.0860 90.3 0.1306 104.5 0.22E 111.8 (a) Calculate the amount of nitrogen that would form one complete monolayer on the alumina. Adsorbed N2is given in units of millimoles (mmol) per gram of AbO3. (b) If a single N2 molecule occupies an area of 16 x 10-20m2, what is the surface area pi gram of alumina? 2. Benjamin Franklin showed that 1.0 cm3of oil poured on the surface of water would cover one-half acre. One'-halfacre is equivalent to 2.024 x 107cm2. If the density of oilOne'-halfacre is equivalent to 2....
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This note was uploaded on 02/26/2008 for the course CHEM 4521 taught by Professor Bierbaum during the Spring '03 term at Colorado.

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Midterm3-Solutions20 - Midterm#3 Chemistry 4511 Physical Chemistry for Engineers Helpful Information e = K[A(l K[A PV=nRT 1 mole = 6.022 x

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