Force constants of DCl and HCl

[Steven Hanna]

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?

 

[Bystander]

Think in terms of percentages; a 2 % variation isn't that much.

 

[Steven Hanna]

Think in terms of percentages; a 2 % variation isn't that much.

true, but the difference is more pronounced than for other isotopomers

 

[Bystander]

true, but the difference is more pronounced than for other isotopomers

... for a 100 % increase in one of the masses.

 

[DrClaude]

... for a 100 % increase in one of the masses.

The increase in mass shows up in the vibrational frequency, not in the force constant.

I am also a bit surprised by the size of the difference. I'll try to look at the numbers.

 

[TeethWhitener]

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?

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.

 

[DrClaude]

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]

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.

These numbers are from an experimental protocol for my lab report. I can check on the actual references. In the meantime, I'll give another example that I can actually cite. The force constants of H₂ (k = 510 N/m) and D₂ (k = 527 N/m) differ by ca. 17 N/m. The reference for these numbers is McQuarrie, D. A. Quantum Chemistry, 2 ed.; University Science Books, Mill Valley, CA, 2008, p 221.
I've posted a page from the textbook here: http://imgur.com/TBmpHNT

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.

 

[DrClaude]

I'll check my calculations again and will post them here.

 

[Steven Hanna]

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
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)).

 

[TeethWhitener]

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)).

$$k=\mu \omega^2$$
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.

 

[Steven Hanna]

$$k=\mu \omega^2$$
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.

OK I think I get it. If I differentiate the Morse equation twice I get
dV/dx = 2D[1-exp(-ax)]*a = 2aD -2aD*exp(-2ax)
dV²/dx² = 4a²D*exp(-2ax)=4(μω²/2D)D*exp(-2ax) = 4μω²exp[-2x*sqrt(μω²/2D)]

This final result is not simply equal to μω², contains an additional mu term, and decreases exponentially with x, as does the restoring force of a covalent bond near the dissociation limit. Agree?

 

[TeethWhitener]

Agree?

Sounds reasonable.