Finding the Equilibrium Separation in a Diatomic Molecule: A Scientific Approach

In summary, the potential energy of a diatomic molecule is given by U=(A/r^12)-(B/r^6) and is associated with the force that binds the two atoms together. To find the equilibrium separation, we need to set F(x)=0 and solve for r, which gives us the equation (2A/B)^(1/6)=r. Depending on whether the separation is larger or smaller than the equilibrium separation, the force between the atoms will be negative or positive, respectively. This means that a negative force will draw the atoms together and a positive force will push them apart. Sketching the potential energy function can help visualize this concept.
  • #1
Ginny Mac
17
0
Oh boy...this problem has stumped me. I am not even sure I know where to start:

The potential energy of a diatomic molecule (a two-atom system like H2 or O2) is given by
U= (A/r^12) - (B/r^6)

where r is the separation of the two atoms of the molecule and A and B are positive constants. This potential energy is associated with the force that binds the two atoms together.
(a) Find the equilibrium separation - that is, the distance between the atoms at which the force on each atom is zero. Is the force repulsive (atoms are pushed apart) or attractive (atoms are pulled together) if their separation is (b) smaller and (c) larger than the equilibrium separation?

Okay. I started out thinking that because we are looking for the force, we would need to use F(x) = - (d/dx)(U(x)), and after differentiation

F(x) = (12A/r^13) + (6B/r^7)

So here is the big "now what?" Please help. Thank you - any help is greatly appreciated.
~Gin
 
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  • #2
Ginny Mac said:
(a) Find the equilibrium separation - that is, the distance between the atoms at which the force on each atom is zero.
[snip]

Okay. I started out thinking that because we are looking for the force, we would need to use F(x) = - (d/dx)(U(x)), and after differentiation

F(x) = (12A/r^13) + (6B/r^7)

So here is the big "now what?" Please help. Thank you - any help is greatly appreciated.
~Gin

You have the next step: you said "Find ... the distance between the atoms at which the force on each atom is zero." (that is, find the value of r that makes F=0)
 
  • #3
okay...but there are three unknowns in my equation... Is there a step I am leaving out?? Even solving algebraically only reduces down to 2A-Br^6 = 0. Does something cancel out that I am not aware of? Or maybe we are supposed to be left with an equation?
 
  • #4
You are told "A and B are positive constants". So, you can expect that your answer is in terms of those positive [but otherwise unspecified] constants.
 
  • #5
Ahhh...so I got my equation down to (2A/B)^(1/6) = r. So that value of r is the equilibrium separation. Now for the next part, figuring out atomic behaviors based on larger/smaller values in relation to equilibrium sep. This should be interesting.

Equation: 2A-Br^6=0

If sep. is larger than equilibrium, we should have a negative force, and if it is smaller than eq., we should have a positive force. I think a negative force will draw the atoms together (?) Any thoughts are appreciated. Thank ya'll.

gin
 
  • #6
Can you sketch the potential energy function U= (A/r^12) - (B/r^6) vs separation r, and locate your special value of r?
 

1. What is equilibrium separation?

Equilibrium separation refers to the distance between two particles at which the attractive and repulsive forces between them are balanced, resulting in a stable position.

2. How is equilibrium separation determined?

Equilibrium separation is determined by various factors such as the mass, charge, and distance between the particles, as well as the strength of their attractive and repulsive forces.

3. What is the significance of equilibrium separation in chemical reactions?

In chemical reactions, equilibrium separation plays a crucial role in determining the stability and outcome of the reaction. It can affect the rate of reaction, the formation of products, and the overall energy of the system.

4. Can equilibrium separation be altered?

Yes, equilibrium separation can be altered by changing the factors that influence it, such as the mass, charge, or distance between particles. This can be done through external forces or by manipulating the properties of the particles themselves.

5. How is equilibrium separation related to phase transitions?

Equilibrium separation is closely related to phase transitions, as it determines the distance at which particles are in a stable state, whether it be a solid, liquid, or gas. Changes in equilibrium separation can lead to phase transitions, such as melting or vaporization.

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