homework - Problem#1 By using the combination of the...

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O O O (4) O O O O O O (2) (7) (6) O (8) O (5) (1) (3) O Problem #1 By using the combination of the Lorenz K Lorenz relation and the Clausius K Mosotti equation the dielelctric constant can calculated All the values needed are given in table 11.2 (handout) restart : with linalg with plots : u 2 3 \$ k \$ T = e K 1 e C 2 \$ 3 \$ M \$ 4 \$ pi \$ e 0 rho \$ 4 \$ pi K n 2 K 1 n C 2 \$ 3 \$ M \$ 4 \$ pi \$ e 0 rho \$ 4 \$ pi ; 1 3 u 2 k T = 3 e K 1 M e 0 e C 2 r K 3 n 2 K 1 M e 0 n C 2 r n d 1.446; r d 1480; M d 119.4 \$ 10 K 3 ; u d 1.06 \$ 3.336 \$ 10 K 30 ; k d 1.381 \$ 10 K 23 ; T d 293; e 0 d 8.854 \$ 10 K 12 ; n := 1.446 r := 1480 M := 0.1194000000 u := 3.536160000 10 -30 k := 1.381000000 10 -23 T := 293 e 0 := 8.854000000 10 -12 e chloroform = solve u 2 3 \$ k \$ T = e K 1 e C 2 \$ 3 \$ M \$ 4 \$ pi \$ e 0 rho \$ 4 \$ pi K n 2 K 1 n C 2 \$ 3 \$ M \$ 4 \$ pi \$ e 0 rho \$ 4 \$ pi , e ; e chloroform = 2.389652345 Problem #2 Two non-polar molecules 1 and 2 interact across polar medium 3. The interaction energy as a function of distance between molecule 1 and 2 can be calculated using the unified model for the vdW forces restart : with linalg with plots : w r = w r n = 0 C w r n O 0 ; w r = w r n = 0 C w r 0 ! n w d r / K 3 \$ k \$ T \$ e 1 K e 3 e 1 C 2 \$ e 3 \$ e 2 K e 3 e 2 C 2 \$ e 3 \$ a 1 3 \$ a 2 3 r 6 C K 3 \$ h \$ v e 2 \$ n 1 2 K n 3 2 \$ n 2 2 K n 3 2 n 1 2 C 2 \$ n 3 2 \$ n 2 2 C 2 \$ n 3 2 \$ n 1 2 C 2 \$ n 3 2 C n 2 2 C 2 \$ n 3 2 \$ a 1 3 \$ a 2 3 r 6 ; w := r / K 3 k T e 1 K e 3 e 2 K e 3 a 1 3 a 2 3 e 1 C 2 e 3 e 2 C 2 e 3 r 6 K 1 2 3 h v e n 1 2 K n 3 2 n 2 2 K n 3 2 a 1 3 a 2 3 n 1 2 C 2 n 3 2 n 2 2 C 2 n 3 2 n 1 2 C 2 n 3 2 C n 2 2 C 2 n 3 2 r 6 k d 1.381 \$ 10 K 23 ; T d 298; n 1 d 1.42; n 2 d 1.50; n 3 d 1.45; e 1 d 1.42 2 ; e 2 d 1.50 2 ; e 3 d 6.0; v e d 3 \$ 10 15 ; h d 6.634 \$ 10 K 34 ; k := 1.381000000 10 -23 T := 298 n 1 := 1.42 n 2 := 1.50 n 3 := 1.45 e 1 := 2.0164 e 2 := 2.2500 e 3 := 6.0 v e := 3000000000000000 h := 6.634000000 10 -34 w r ; K 9.233935602 10 -22 a 1 3 a 2 3 r 6 C 3.960729514 10 -22 3 a 1 3 a 2 3 r 6 it can be seen, that the term a 1 3 \$ a 2 3 will not change anything but the magnitude, and therefore not the nature of the interaction. simplify w r a 1 3 \$ a 2 3 ; K 2.373750848 10 -22 r 6

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O (10) O (9) O O O O O (11) O plot K 2.373750848 10 -22 r 6 , r = 2 \$ 10 K 10 ...4 \$ 10 K 9 , caption = "Interaction energy between molecule 1 and two across medium 3", legend = "w(r)/(a_1^3 \$ a_2^3)" ; w(r)/(a_1^3*a_2^3) r 2 # 10 - 10 2,5 # 10 - 10 3 # 10 - 10 3,5 # 10 - 10 4 # 10 - 10 K 3,5 # 10 36 K 3 # 10 36 K 2,5 # 10 36 K 2 # 10 36 K 1,5 # 10 36 K 1 # 10 36 K 5 # 10 35 Interaction energy between molecule 1 and two across medium 3 From the graph it can be seen that the nature of the interction is always attractive because w(r)<0, the dependence on seperation distance can be seen from the plot. Problem # 3
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This note was uploaded on 03/29/2008 for the course CHEM 747 taught by Professor Abbott during the Spring '08 term at University of Wisconsin.

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homework - Problem#1 By using the combination of the...

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