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For an M+2-N-2 ion pair, attractive and repulsive energies EA and ER

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For an \(\mathrm{M}^{+2}-\mathrm{N}^{-2}\) ion pair, attractive and repulsive energies \(\mathrm{E}_{\mathrm{A}}\) and \(\mathrm{E}_{\mathrm{R}}\), respectively, depend on the distance between the ions \(r\), according to

$$ \begin{array}{c} E_{A}=\frac{-4.3}{r} \\ E_{R}=\frac{8.5 \times 10^{-6}}{r^{9}} \end{array} $$

(a) Plot EA, ER and EN on the same graph.

(b) Determine r0 and E0 from the graph.

(c) Mathematically determine r0 and E0.

(d) Plot FA, FR and FN on the same graph.


You need to show all graphs and tables. If you are reading a value from graph, you need to show it on the graph.


Chemical-Engineering   chemistry   Material-Science   
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Answer #1

:: Solution ::

r(in nm) EA ER EN
0 ? ? ?
0.1 -14.36 732 717.64
0.2 -7.18 2.859375 -4.320625
0.3 -4.7866667 0.111568358 -4.675098308
0.4 -3.59 0.011169434 -3.578830566
0.5 -2.872 0.00187392 -2.87012608
0.6 -2.3933333 0.000435814 -2.392897519
0.7 -2.0514286 0.000126977 -2.051301594
0.8 -1.795 4.36306E-05 -1.794956369

  

800 4 3 700 ER graph 2 by removing r=0.1 nm values. 2 600 1 500 0 0 0.1 0.2 ois 0.4 0.5 0.6 0.7 0.8 0.9 -1 400 Ед са Interato

so, from graph, the equilibrium spacing ro between Na+ and Cl- ions = 0.3nm

and the magnitude of the bond energy Eo between the two ions = -5eV per Na+-Cl- pair

From -|-436 A = 1.436 & ER 7-32 x 10-6 B = 7-32 x 10-6 88 8 n = 8 n So Xo (A) - ( 1.436 8 X7:32 X106 ro 0.236 nm Calculated v

i hope it helps..

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Answer #2

how do i find the values of Fa Fr and Fn?

answered by: Abdullah Raihan
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