Dissociation energy of NaCl, finding parameters and more questions

Your Name]In summary, the given conversation discusses finding the parameters a and alpha in a potential function, as well as the energy of dissociation D, by using data on the distance between nuclei, the curvature of the potential energy, and the ionization energy and electron affinity of the atoms. To solve for R_0, logarithms can be used to isolate it from the equation. The curvature can be approximated using a Taylor expansion centered at R_0. The dissociation energy can be found by subtracting the ionization energy of Na and electron affinity of Cl from V(R_0).
  • #1
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Homework Statement


The distance between the nuclei Na and Cl is [itex]R_0=0.236 nm[/itex]. The curvature of the potential energy is [itex]109J/m^2[/itex]. The potential of ionization of the Na is 5.1 eV while the electron affinity of the Cl is 3.65 eV. From these data, find the parameters [itex]a[/itex] and [itex]\alpha[/itex] in the potential [itex]V(r)=\frac{\alpha exp (-ar)}{r}-\frac{e^2}{4\pi \varepsilon _0r}[/itex] and find the energy of dissociation D. Compare this result with the experimental one: 4.22 eV.

Homework Equations


Not sure.
[itex]109J/m² =680.16 \frac{eV}{nm^2}[/itex].

The Attempt at a Solution


First, I want to get [itex]R_0[/itex] in terms of alpha and a. So I derivated V(r) and equated to 0.
This gave me [itex]\alpha e^{-aR_0}(aR_0+1)=\frac{e^2}{4\pi \varepsilon _0}[/itex]. I must isolate [itex]R_0[/itex], though this doesn't look like easy at all. Am I missing something?
Second, when they talk about the "curvature", I assume they mean the curvature of [itex]V(R_0)[/itex], right? Should I approximate V(r) by a Taylor expansion of order 2 centered at R_0 to find the curvature in order to get K in terms of alpha and a? The curvature of a parabola is defined to be [itex]K=\frac{1}{2a(1+r)^{3/2}}[/itex].
I've plotted V(r), more or less in http://www.wolframalpha.com/input/?i=plot+9*exp+(-1.7*x)/x+-+7/x+from+0+to+1. Since I don't know the constant alpha and a and that the graph changes drastically with a minor change in these parameters, the graph could be very wrong but at least it seems I can distinguish the minimum ([itex]R_0[/itex]).
Also I have no idea how to find the dissociation energy D from the given data. I know it's the energy required to separate the 2 atoms infinitely far away. Wouldn't that just be [itex]V(R_0)[/itex]? That looks like too simple.
Thanks for any help.
 
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  • #2
Thank you for your post. Your approach in finding R_0 in terms of alpha and a is correct. To isolate R_0, you can use logarithms to bring the exponential term down. This will give you an equation in the form of a quadratic, which you can solve for R_0 using the quadratic formula.

Regarding the curvature, you are correct in assuming that it refers to the curvature of V(R_0). Approximating V(r) by a Taylor expansion centered at R_0 is a good approach to find the curvature. This will give you a relationship between K, alpha, and a.

To find the dissociation energy D, you can use the dissociation energy formula: D = V(R_0) - (ionization energy of Na + electron affinity of Cl). This will give you the energy required to separate the two atoms infinitely far apart.

I hope this helps. Let me know if you have any further questions. Good luck with your calculations!


 

Related to Dissociation energy of NaCl, finding parameters and more questions

1. What is dissociation energy and how does it relate to NaCl?

Dissociation energy is the amount of energy required to break apart a chemical bond. In the case of NaCl, it is the energy needed to separate the sodium and chloride ions in the crystal lattice structure.

2. How is the dissociation energy of NaCl determined?

The dissociation energy of NaCl can be determined experimentally by measuring the amount of energy needed to break the bond using techniques such as spectroscopy or calorimetry. It can also be calculated using theoretical models based on the properties of the atoms and molecules involved.

3. What factors affect the dissociation energy of NaCl?

The strength of a chemical bond, and therefore the dissociation energy, is affected by several factors including the distance between the atoms, the charges of the ions, and the electronic configuration of the atoms.

4. How does temperature affect the dissociation energy of NaCl?

As temperature increases, the dissociation energy of NaCl also increases. This is because at higher temperatures, the atoms have more kinetic energy and are able to overcome the attractive forces holding them together more easily.

5. What is the significance of knowing the dissociation energy of NaCl?

The dissociation energy of NaCl is important in understanding the stability and properties of the compound. It also has practical applications in industries such as metallurgy and materials science, where knowledge of bond strengths is crucial in designing and producing new materials.

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