PS2_08-Solution - Chem 444B Homework Set #2 - Solutions...

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Problem Set #2 -1- Chem 444B Homework Set #2 - Solutions DUE: Friday, February 1, 2008 1) The energy spacing between a ground rotational state and the 1 st excited state is 0.05 kJ/mol. Calculate the ratio of the number of particles in the 1 st excited state to the number of particles in the ground state at room temperature (298K). Remember that the degeneracy for a rigid-rotator is g J =2J+1. () 10 11 00 23 20 20 1 23 0 Modify Eq. 17.10 to include degeneracies or take a look at the end of Lecture 5: 1 0.05 / 8.303 10 6.022 10 / 2.432 10 1.38 10 / 298 2(0) 1 1; EE B ng e J k Jm o l J kJ mol J kT JK K g β −− = −= × = × × == = × × =+ = ( ) 20 1 23 1 2.432 10 8.303 10 1 0 2(1) 1 3 3 3(0.98) 2.94 1 JJ g n e n −× × = = 2) Given that the first excited electronic level of O 2 is 15.72x10 -20 J above the ground level, calculate n 1 /n 0 at a)298K and b)1500K. The degeneracies are g 0 =3 and g 1 =2. a) ( ) 20 23 / 1 0 15.72 10 / 1.38 10 298 17 1 0 Modify Eq. 17.10 to include degeneracies or take a look at the end of Lecture 5: 2 1.67 10 3 B EE kT K g ee g n e n × × b) ( ) ( ) ( ) 20 23 15.72 10 / 1.38 10 1500 4 1 0 2 3.36 10 3 K n e n × × 3) A hypothetical system is made of 3 doubly degenerate energy states. Assume the ground state energy is zero, and the separation between successive states is 60.0 cm -1 . At a temperature of 850K, calculate the following:
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Problem Set #2 -2- a) q, the molecular partition function () ( ) ( ) ( ) 0 12 11 / / // 012 0 / 590.75 60 / 590.75 120 / 590.75 Eq. 17.54 0.695 / 850 590.75 22 2 5.439 jB B BB kT j j B cm cm cm cm cm cm qg e e g e g e cm K K qe e e q ε εε −− = =+ + == + = b) <E>/N ( ) ( ) 0 / / / 00 2 2 0 / 590.75 60 / 590.75 Eq. 17.42 and Eq. 17.43 (with degeneracy)
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This homework help was uploaded on 04/13/2008 for the course CHEM 444 taught by Professor Jameslis during the Spring '08 term at University of Illinois at Urbana–Champaign.

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PS2_08-Solution - Chem 444B Homework Set #2 - Solutions...

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