Midterm2-Solutions2003

Midterm2-Solutions2003 - Midterm #2 / Chemistry 4511 /...

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Unformatted text preview: Midterm #2 / Chemistry 4511 / Physical Chemistry for Engineers Helpful Information: PV=nRT 760 Torr = 1 atm R = 0.08206 atm dm3Klmorl R 8.3145 J KI morl kB= 1.381 x 10-23J K-I h = 6.626 X 10-34Js 1 J = 1 kg m2 S-2 1 amu = 1.66 x 10-27kg 1 amux Avogadro'sNumber = Massof molein grams Avogadro's Number = 6.022 x 1023morl dm3 = 1000 cm3 = 1 L dA/dt =-kAn t1l2 = ln2/k A(t) = A(O)exp(-kt) t1/2 = lI[A(O)k] IfA(t)- lIA(O) = kt Steady State Approximation, d[I]/dt = k = (kBT/h) exp[-iltOo/RT] iltoo = iltHo- Tiltso First-order unimolecular gas reaction: Ea =iltHo + RT First-orderunimoleculargasreaction: k = e(kBT/h)exp[il tSo/R]exp[-Ea/RT] e = 2.718 a2 + b2 = c2 nt..= 2d sine Cubic Lattice: dhkl = a/(h2 + k2 + 12)1/2-------------------------------------------------------------------------------------------------------------- 1. A third-order reaction (n=3) obeys the following rate equation: dA/dt =-kA3 What is the integrated rate equation for A(t)? 2. The second-order rate constant for the reaction: 2N02-7 2NO + O2 is k=6.3 X102cm3/(mol s) at 600 K. If the initial pressure of N02 is 760 Torr, how much time is required for 20% of the NO to decompose at 600 K? 3. The rate constant for a first-order unimolecular gas reaction is k=7.40 x 10-9s-1at 25 °c and the activation energy is Ea = 112.0 kllmo!. At 25 °C, calculate the Gibbs energy of activation, ~:j:Go,the enthalpy of activation, ~:j:Ho,the entropy of activation, ~:j:So,and the preexponential factor, A. 4. Aluminum crystallizes in a face-centered cubic (fee) lattice with the edge length of the unit cell equal to a= 4.0491 A. (a) Calculate the atomic radius of aluminum. (b) Determine the density (g/cm3)of aluminum. The molar mass of aluminum is 26.98 g/mo!....
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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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Midterm2-Solutions2003 - Midterm #2 / Chemistry 4511 /...

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