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Steven Hanna
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The force (spring) constant of HCl 477.8 N/m, and the force constant of DCl is 487.95. Why should these values differ by 10 N/m? Is a D-Cl bond somehow stronger than an H-Cl bond?
true, but the difference is more pronounced than for other isotopomersBystander said:Think in terms of percentages; a 2 % variation isn't that much.
... for a 100 % increase in one of the masses.Steven Hanna said:true, but the difference is more pronounced than for other isotopomers
The increase in mass shows up in the vibrational frequency, not in the force constant.Bystander said:... for a 100 % increase in one of the masses.
Steven Hanna said:The force (spring) constant of HCl 477.8 N/m, and the force constant of DCl is 487.95. Why should these values differ by 10 N/m? Is a D-Cl bond somehow stronger than an H-Cl bond?
I've checked the numbers, and these don't make sense. The actual force constant is 2 orders of magnitude smaller, and the relative difference between HCl and DCl is of the order of 4×10-5.Steven Hanna said:The force (spring) constant of HCl 477.8 N/m, and the force constant of DCl is 487.95. Why should these values differ by 10 N/m? Is a D-Cl bond somehow stronger than an H-Cl bond?
DrClaude said:I've checked the numbers, and these don't make sense. The actual force constant is 2 orders of magnitude smaller, and the relative difference between HCl and DCl is of the order of 4×10-5.
TeethWhitener said:Out of curiosity, where are these numbers coming from?
In the harmonic approximation, there should be no difference in the force constants of HCl and DCl. However, adding anharmonic terms changes things. The most straightforward explanation that I can think of is that the zero-point energy for DCl is smaller, so the energy levels sit further down in the well. For a harmonic oscillator, this doesn't make a difference because the force constant is the second derivative of the potential with respect to the coordinate: [itex]k = \frac{\partial^2V}{\partial x^2}[/itex], which is a constant (obviously) for [itex]V=kx^2[/itex]. However, if the potential is anharmonic (as it is in a real molecule), ##k## is in general not constant, and so lowering the zero-point energy has the effect of changing ##k##. There are probably other higher order effects at play here as well, but the main reason is the anharmonicity of the covalent bond.
kTeethWhitener said:Out of curiosity, where are these numbers coming from?
In the harmonic approximation, there should be no difference in the force constants of HCl and DCl. However, adding anharmonic terms changes things. The most straightforward explanation that I can think of is that the zero-point energy for DCl is smaller, so the energy levels sit further down in the well. For a harmonic oscillator, this doesn't make a difference because the force constant is the second derivative of the potential with respect to the coordinate: [itex]k = \frac{\partial^2V}{\partial x^2}[/itex], which is a constant (obviously) for [itex]V=kx^2[/itex]. However, if the potential is anharmonic (as it is in a real molecule), ##k## is in general not constant, and so lowering the zero-point energy has the effect of changing ##k##. There are probably other higher order effects at play here as well, but the main reason is the anharmonicity of the covalent bond.
$$k=\mu \omega^2$$Steven Hanna said:k
Sorry to dredge up an old thread, but I'm taking a physorg class and we're talking about isotope effects. So anyway, I understand why the ZPE of DCl is lower than that of HCl, but I'm not quite sure how anharmonicity causes k to not be constant. As I understand it, the series expansion of V(x), where x is the displacement of the bond from equilibrium, is V(x) = ∑(1/n!)(d(n)V/dx)*xn. However, these terms don't depend on the reduced mass. Since you can differentiate taylor series term-by-term, d2V/dx2 should also be independent of reduced mass too, right? Also, the equation for a Morse potential, V(x) = D[1-exp(-ax)]2 doesn't contain any mass terms (a = sqrt(k/2D)).
OK I think I get it. If I differentiate the Morse equation twice I getTeethWhitener said:$$k=\mu \omega^2$$
So the Morse potential does contain mass terms; they're just hidden in the force constant. Ditto for a generic potential, since even in the harmonic case, ##V(x)=\frac{1}{2}\mu \omega^2 x^2##. So for an anharmonic example, if you have ##V=\frac{1}{2}\mu \omega^2 x^2 + \varepsilon x^3##, then your force constant
$$k=\frac{d^2V}{dx^2}$$
will be ##k(x)=\mu \omega^2 + 6\varepsilon x##, both mass-dependent and non-constant.
Sounds reasonable.Steven Hanna said:Agree?
Force constants are numerical values that represent the strength of the bonds between different atoms in a molecule. They are used to calculate the energy required to stretch or bend a bond, and can provide information about the stability and reactivity of a molecule.
Force constants for DCl and HCl can be determined experimentally using techniques such as infrared spectroscopy or theoretical calculations based on quantum mechanics. These methods allow us to measure the vibrations and movements of the atoms in the molecule and determine the force constants that govern these movements.
The force constants of DCl and HCl can be influenced by various factors, including the masses of the atoms, the bond lengths, and the strength of the intermolecular forces between the molecules. Temperature and pressure can also affect the force constants by altering the molecular vibrations and bond lengths.
The force constants of DCl and HCl may differ due to the difference in mass between deuterium (D) and hydrogen (H) atoms. The heavier mass of deuterium can result in stronger bonds and higher force constants compared to hydrogen. Additionally, the force constants for DCl may be slightly higher than HCl due to the larger size of the deuterium atom.
The force constants of DCl and HCl play a crucial role in understanding the properties and behavior of these molecules. They can provide information about the bond strength, molecular structure, and reactivity of these compounds. Force constants are also essential in predicting the vibrational spectra of these molecules, which can be useful in chemical analysis and identification.