Why isn't bond dissociation energy/bond enthaply measured in Newtons?

AI Thread Summary
The discussion centers on the relationship between bond dissociation energy and the forces involved in breaking chemical bonds. It clarifies that bond dissociation energy is not simply the product of the force needed to break a bond and the bond length. Instead, it represents the energy required to separate atoms to an infinite distance, where the force is zero at equilibrium. The concept of equilibrium is highlighted, indicating that at the bond's equilibrium length, the energy is minimized, and no net force acts on the atoms. The conversation also touches on the modeling of atomic bonds as harmonic oscillators, which is an approximation that can simplify calculations but may not capture all quantum mechanical behaviors. The practical aspect of bond dissociation is noted, as atoms are effectively dissociated when separated by a few angstroms, rather than needing to be infinitely apart.
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Why isn't average bond dissociation energy/bond enthalpy measured in units of force/Newtons (kg*m/s^2)?
I understand every bond chemically has a length and energy to break, and energy is Newton*meters.
Is the Bond enthaply/Bond disassociation energy equivalent to the force needed to break the bond * the bond length?

Why don't we say, to break the bond from O to H we need to put magnets on left of the O and right of the H and apply some pulling force of XYZ?
 
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adf89812 said:
Is the Bond enthaply/Bond disassociation energy equivalent to the force needed to break the bond * the bond length?
No, it’s the energy required to move the atoms infinitely far apart. Think of it this way: force is the gradient of energy (derivative of energy with respect to distance)
$$F = \nabla E \left(=\frac{dE}{dx}\right)$$
At the equilibrium bond length, energy is at a minimum, meaning that the gradient (and therefore the force) is zero—this makes sense because a system at equilibrium has no net force acting on it.
 
TeethWhitener said:
No, it’s the energy required to move the atoms infinitely far apart. Think of it this way: force is the gradient of energy (derivative of energy with respect to distance)
$$F = \nabla E \left(=\frac{dE}{dx}\right)$$
At the equilibrium bond length, energy is at a minimum, meaning that the gradient (and therefore the force) is zero—this makes sense because a system at equilibrium has no net force acting on it.
It's an accepted model to represent atoms diatomic as ball attached to spring attached to ball so equilibrium is false. They can't be infinitely far apart because I can disassociate hydroxide in a small vial of very small size.
 
adf89812 said:
It's an accepted model to represent atoms diatomic as ball attached to spring attached to ball so equilibrium is false. They can't be infinitely far apart because I can disassociate hydroxide in a small vial of very small size.
I have no idea where you’re getting this from. I think you’re trying to say that bonds can be modeled as harmonic oscillators. And of course harmonic oscillators have an equilibrium point. It’s at the bottom of the potential well.

Also, the bond dissociation energy is a limit as the distance between atoms goes to infinity. Practically, with most bonds, once atoms are separated by more than a few angstroms they’re essentially dissociated.
 
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adf89812 said:
It's an accepted model to represent atoms diatomic as ball attached to spring attached to ball so equilibrium is false. They can't be infinitely far apart because I can disassociate hydroxide in a small vial of very small size.
The harmonic oscillator is an approximation. And you can in principle model a molecule as a balls on sticks that oscillate like a spring, but it's usually not sufficient for quantum mechanical calculations. Molecular dynamics force fields often model bonds with Hooke's law.
 
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